4.4 Sums and direct sums · Matrix Theory

In Section 1.4.5 we met the sum $𝑈 + 𝑊$ of two subspaces, the smallest subspace that contains both. This section answers two questions about it: how large it is, and when a vector of $𝑈 + 𝑊$ can be written as $𝐮 + 𝐰$ in only one way. Two pictures in $ℝ^{3}$ already suggest the answers. Every vector of $ℝ^{3}$ is, in exactly one way, a vector of the $𝑥 𝑦$ -plane plus a vector of the $𝑧$ -axis, and $2 + 1 = 3$ . The $𝑥 𝑦$ -plane and the $𝑥 𝑧$ -plane also add up to $ℝ^{3}$ , but now $(1, 0, 0) =$ $(1, 0, 0) +$ $𝟎 =$ $𝟎 +$ $(1, 0, 0)$ has two decompositions, and $2 + 2$ overshoots $3$ by exactly $1$ , the dimension of the $𝑥$ -axis, where the two planes overlap. Both questions are answered by looking at the intersection.

Sums in which decompositions are unique, the direct sums, are to subspaces what bases are to vectors. A basis cuts $𝑉$ into lines, and the coordinates of a vector name its pieces; a direct sum cuts $𝑉$ into subspaces of any dimension. Later chapters understand an operator by cutting $𝑉$ into pieces that it maps into themselves. A diagonalizable operator is one for which $𝑉$ is the direct sum of its eigenspaces (Section 7.4), and the Jordan form starts from a direct sum of generalized eigenspaces (Section 10.2). Both need direct sums of more than two subspaces, and that is where the subtleties are.

Throughout, $𝑉$ is a vector space over a field $𝐹$ , and $𝑈$ , $𝑊$ , $𝑋$ , $𝑈_{1}$ , $𝑈_{2}$ , $\dots$ denote subspaces of $𝑉$ .

4.4.1Sums of several subspaces

The sum of two subspaces extends to any finite number of them.

Definition 4.4.1 (Sum of subspaces). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ , where $𝑘 \geq 1$ . Their sum is

𝑈_{1} + \dots + 𝑈_{𝑘} = {𝐮_{1} + \dots + 𝐮_{𝑘} | 𝐮_{𝑖} \in 𝑈_{𝑖} for 𝑖 = 1, \dots, 𝑘},

also written $\sum_{𝑖 = 1}^{𝑘} 𝑈_{𝑖}$ . An expression $𝐯 = 𝐮_{1} + \dots + 𝐮_{𝑘}$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ for each $𝑖$ is a decomposition of $𝐯$ along $𝑈_{1}, \dots, 𝑈_{𝑘}$ , with components $𝐮_{1}, \dots, 𝐮_{𝑘}$ .

For $𝑘 = 2$ this is the sum $𝑈_{1} + 𝑈_{2}$ of Proposition 1.4.24, and for $𝑘 = 1$ it is $𝑈_{1}$ . By convention, a sum of no subspaces is ${𝟎}$ , just as an empty sum of vectors is $𝟎$ . The sum of $𝑘$ subspaces is the sum of two, applied repeatedly, so the two-subspace facts carry over.

Proposition 4.4.2 (Sums are the smallest containing subspaces). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ .

(a)If $𝑘 \geq 2$ , then $𝑈_{1} +$ $\dots +$ $𝑈_{𝑘} =$ $(𝑈_{1} + \dots + 𝑈_{𝑘 - 1}) +$ $𝑈_{𝑘}$ .
(b) $𝑈_{1} + \dots + 𝑈_{𝑘}$ is a subspace of $𝑉$ that contains every $𝑈_{𝑖}$ , and it is contained in every subspace of $𝑉$ that contains every $𝑈_{𝑖}$ .
(c)If $𝑈_{𝑖} = span ({ℒ︀}_{𝑖})$ for lists ${ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘}$ , then $𝑈_{1} +$ $\dots +$ $𝑈_{𝑘} =$ $span ({ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘})$ .

Proof.routine

Uses. Definition 4.4.1 (Sum of subspaces), Proposition 1.4.24 (The sum of two subspaces), Exercise 1.3.36

Techniques. Double containment (Technique 1.3.14), Induction on the number of summands (Technique 4.4.10)

Hint 1. For (a), an element of the left side is

(𝐮_{1} + \dots + 𝐮_{𝑘 - 1}) +

𝐮_{𝑘}

. Check both containments.

Hint 2. For (b) and (c), induct on

𝑘

. By (a), the case of

𝑘

subspaces is a sum of two,

𝑆 + 𝑈_{𝑘}

with

𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}

, where Proposition 1.4.24 and Exercise 1.3.36 apply.

Sketch. (a) is regrouping. By (a), the sum is

𝑆 + 𝑈_{𝑘}

with

𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}

, so (b) and (c) follow by induction on

𝑘

from the two-subspace facts:

𝑆 + 𝑈_{𝑘}

is the smallest subspace containing

𝑆

and

𝑈_{𝑘}

, and

span (ℒ︀) +

span (ℳ︀) =

span (ℒ︀, ℳ︀)

(a) Let $𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}$ . An element of $𝑈_{1} + \dots + 𝑈_{𝑘}$ has the form $𝐮_{1} +$ $\dots +$ $𝐮_{𝑘} =$ $(𝐮_{1} + \dots + 𝐮_{𝑘 - 1}) +$ $𝐮_{𝑘}$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ , and this lies in $𝑆 + 𝑈_{𝑘}$ . Conversely, an element of $𝑆 + 𝑈_{𝑘}$ is $𝐬 + 𝐮_{𝑘}$ with $𝐬 =$ $𝐮_{1} +$ $\dots +$ $𝐮_{𝑘 - 1} \in$ $𝑆$ and $𝐮_{𝑘} \in 𝑈_{𝑘}$ , which is an element of $𝑈_{1} + \dots + 𝑈_{𝑘}$ .

(b) We induct on $𝑘$ . For $𝑘 = 1$ the sum is $𝑈_{1}$ , and there is nothing to prove. Let $𝑘 \geq 2$ , and assume that $𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}$ is a subspace containing $𝑈_{1}, \dots, 𝑈_{𝑘 - 1}$ and contained in every subspace that contains them. By (a) and Proposition 1.4.24(a), $𝑈_{1} +$ $\dots +$ $𝑈_{𝑘} =$ $𝑆 +$ $𝑈_{𝑘}$ is a subspace containing $𝑆$ and $𝑈_{𝑘}$ , hence every $𝑈_{𝑖}$ . If $𝑋$ is a subspace containing every $𝑈_{𝑖}$ , then $𝑆 \subseteq 𝑋$ by the inductive hypothesis and $𝑈_{𝑘} \subseteq 𝑋$ , so $𝑆 + 𝑈_{𝑘} \subseteq 𝑋$ by Proposition 1.4.24(b).

(c) Again we induct on $𝑘$ , and $𝑘 = 1$ is trivial. For $𝑘 \geq 2$ , the inductive hypothesis gives $𝑈_{1} +$ $\dots +$ $𝑈_{𝑘 - 1} =$ $span ({ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘 - 1})$ . So by (a) and Exercise 1.3.36, applied to the lists ${ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘 - 1}$ and ${ℒ︀}_{𝑘}$ ,

𝑈_{1} + \dots + 𝑈_{𝑘} = span ({ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘 - 1}) + span ({ℒ︀}_{𝑘}) = span ({ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘}) .

□

By (b), the sum depends only on the subspaces involved, not on their order or grouping. For instance $𝑈 + 𝑊 = 𝑊 + 𝑈$ and $(𝑈 + 𝑊) +$ $𝑋 =$ $𝑈 +$ $(𝑊 + 𝑋)$ , since each side is the smallest subspace containing $𝑈$ , $𝑊$ (and $𝑋$ ).

Example 4.4.3 (Sums).

(a)In $𝐹^{3}$ , $span (𝐞_{1}) +$ $span (𝐞_{2}) +$ $span (𝐞_{3}) =$ $span (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) =$ $𝐹^{3}$ by Proposition 4.4.2(c). In $ℝ^{2}$ , the three lines $span (𝐞_{1})$ , $span (𝐞_{2})$ , and $span (𝐞_{1} + 𝐞_{2})$ also add up to $ℝ^{2}$ , although the third is not needed.
(b)In $𝐹^{𝑛 \times 𝑛}$ , let $𝑌$ be the subspace of upper triangular matrices and $𝑍$ the subspace of lower triangular matrices. Every matrix is the sum of its part on and above the diagonal, which lies in $𝑌$ , and its part below the diagonal, which lies in $𝑍$ . So $𝑌 + 𝑍 = 𝐹^{𝑛 \times 𝑛}$ . The two subspaces overlap in the diagonal matrices, so many matrices have more than one decomposition: $𝐼_{𝑛} =$ $𝐼_{𝑛} +$ $0 =$ $0 +$ $𝐼_{𝑛}$ .
(c)For $𝑆, 𝑇 \in L (𝑉, 𝑊)$ we have $(𝑆 + 𝑇) (𝐯) =$ $𝑆 (𝐯) +$ $𝑇 (𝐯)$ , so $im (𝑆 + 𝑇) \subseteq$ $im 𝑆 +$ $im 𝑇$ . Together with the bound $dim (im 𝑆 + im 𝑇) \leq$ $dim im 𝑆 +$ $dim im 𝑇$ of Theorem 4.4.13 below, this is the reason that rank is subadditive (Proposition 4.3.20). The containment can be strict: for $𝑇 \neq 0$ , $im (𝑇 + (- 𝑇)) = {𝟎}$ is smaller than $im 𝑇 + im (- 𝑇) = im 𝑇$ .

4.4.2Direct sums

A basis $𝐯_{1}, \dots, 𝐯_{𝑛}$ of $𝑉$ writes every vector in exactly one way as $𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}$ . Each term $𝑎_{𝑖} 𝐯_{𝑖}$ lies in the line $span (𝐯_{𝑖})$ , so the basis exhibits $𝑉$ as a sum of lines in which every vector has exactly one decomposition. Direct sums keep the uniqueness and allow pieces of any dimension.

Definition 4.4.4 (Direct sum). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ . The sum $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct if every vector of $𝑈_{1} + \dots + 𝑈_{𝑘}$ has exactly one decomposition along $𝑈_{1}, \dots, 𝑈_{𝑘}$ . In that case we also say that $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent, and we write the sum as

𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘} or ⨁_{𝑖 = 1}^{𝑘} 𝑈_{𝑖} .

If moreover $𝑈_{1} + \dots + 𝑈_{𝑘} = 𝑉$ , we write $𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘}$ and say that $𝑉$ is the direct sum of $𝑈_{1}, \dots, 𝑈_{𝑘}$ .

So writing $𝑈 \oplus 𝑊$ is an assertion, not a new operation: it denotes the subspace $𝑈 + 𝑊$ , together with the claim that the sum is direct. (The external direct sum, which builds a new space out of two unrelated ones, is the product of Section 4.6.) Some texts define independent subspaces by condition (II) of Theorem 4.4.7 below instead; the theorem shows that the two definitions agree.

Example 4.4.5 (Direct sums).

(a)Bases. If $𝐯_{1}, \dots, 𝐯_{𝑛}$ is a basis of $𝑉$ , then $𝑉 =$ $span (𝐯_{1}) \oplus$ $\dots \oplus$ $span (𝐯_{𝑛})$ . A decomposition of $𝐯$ has components $𝑎_{𝑖} 𝐯_{𝑖}$ with $𝑎_{𝑖} \in 𝐹$ , and the component determines the scalar: if $𝑎 𝐯_{𝑖} = 𝑏 𝐯_{𝑖}$ , then $(𝑎 - 𝑏) 𝐯_{𝑖} = 𝟎$ , so $𝑎 = 𝑏$ by Proposition 1.2.35, since $𝐯_{𝑖} \neq 𝟎$ . So decompositions of $𝐯$ correspond to representations $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ , and there is exactly one. The components of $𝐯$ are its coordinates times the basis vectors.
(b)Triangular parts. Let $𝑌$ be the upper triangular matrices in $𝐹^{𝑛 \times 𝑛}$ and $𝑍^{○}$ the strictly lower triangular ones, whose diagonal entries are $0$ . Then $𝐹^{𝑛 \times 𝑛} = 𝑌 \oplus 𝑍^{○}$ : in a decomposition $𝐴 = 𝐵 + 𝐶$ with $𝐵 \in 𝑌$ and $𝐶 \in 𝑍^{○}$ , the entries of $𝐴$ on and above the diagonal must be those of $𝐵$ and the entries below must be those of $𝐶$ .
(c)Even and odd functions. In $ℝ^{ℝ}$ , the even functions $𝐸$ and the odd functions $𝑂$ satisfy $𝐸 \cap 𝑂 = {𝟎}$ and $𝐸 + 𝑂 = ℝ^{ℝ}$ (Exercise 1.4.35). By Theorem 4.4.7 below, this says $ℝ^{ℝ} = 𝐸 \oplus 𝑂$ .
(d)Even and odd polynomials. In $𝐹 [𝑥]$ , let $𝐸$ be the polynomials whose coefficients of $𝑥, 𝑥^{3}, 𝑥^{5}, \dots$ vanish and $𝑂$ those whose coefficients of $1, 𝑥^{2}, 𝑥^{4}, \dots$ vanish. Every polynomial is the sum of its even-degree terms and its odd-degree terms, and in a decomposition $𝑝 = 𝑞 + 𝑟$ with $𝑞 \in 𝐸$ and $𝑟 \in 𝑂$ , each coefficient of $𝑝$ belongs to exactly one of $𝑞$ and $𝑟$ . So $𝐹 [𝑥] = 𝐸 \oplus 𝑂$ over every field, including $𝔽_{2}$ . Compare Exercise 1.4.35(b), where even and odd functions on $𝔽_{2}$ coincide.

Non-example 4.4.6 (Sums that are not direct).

(a)The $𝑥 𝑦$ -plane and the $𝑥 𝑧$ -plane add up to $ℝ^{3}$ , but $𝟎 =$ $(1, 0, 0) +$ $(- 1, 0, 0)$ is a second decomposition of $𝟎$ , besides $𝟎 + 𝟎$ .
(b)In $ℝ^{2}$ , the three lines $ℓ_{1} = span (𝐞_{1})$ , $ℓ_{2} = span (𝐞_{2})$ , $ℓ_{3} = span (𝐞_{1} + 𝐞_{2})$ meet pairwise only in $𝟎$ . Their sum is still not direct:
$𝟎 = 𝐞_{1} + 𝐞_{2} + (- (𝐞_{1} + 𝐞_{2}))$
is a decomposition of $𝟎$ with nonzero components.

Part (b) is the trap of this section. For three or more subspaces, “no two of them overlap” is not the right notion of independence. The analogy with lists explains why: three vectors of $ℝ^{2}$ , no two of them parallel, are still dependent. Independence of a list compares each vector with the span of the others (Corollary 1.5.13), and the right condition for subspaces compares each subspace with the sum of the others.

Intuition (What the criteria say). The theorem below gives three tests for directness. Condition (II) is uniqueness at

𝟎

; as for lists, uniqueness everywhere follows by subtracting two decompositions. Conditions (III) and (IV) are the subspace versions of “no vector is a combination of the others” and “no vector is a combination of the earlier ones”. Condition (IV) is the economical one to check: it asks for

𝑘 - 1

intersections to be

{𝟎}

, each with a sum that is built up one subspace at a time.

Theorem 4.4.7 (Criteria for direct sums). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ . The following are equivalent.

(I)The sum $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct.
(II) $𝟎$ has only the trivial decomposition: if $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ for each $𝑖$ , then $𝐮_{1} = \dots = 𝐮_{𝑘} = 𝟎$ .
(III)Each $𝑈_{𝑗}$ meets the sum of the others only in $𝟎$ : $𝑈_{𝑗} \cap$ $\sum_{𝑖 \neq 𝑗} 𝑈_{𝑖} =$ ${𝟎}$ for $𝑗 = 1, \dots, 𝑘$ .
(IV)Each $𝑈_{𝑗}$ meets the sum of the earlier ones only in $𝟎$ : $𝑈_{𝑗} \cap$ $(𝑈_{1} + \dots + 𝑈_{𝑗 - 1}) =$ ${𝟎}$ for $𝑗 = 2, \dots, 𝑘$ .

In particular, for two subspaces, $𝑈 + 𝑊$ is direct if and only if $𝑈 \cap 𝑊 = {𝟎}$ .

Here $\sum_{𝑖 \neq 𝑗} 𝑈_{𝑖}$ is the sum of all $𝑈_{𝑖}$ except $𝑈_{𝑗}$ . (For $𝑘 = 1$ it is the empty sum ${𝟎}$ , and all four conditions hold.) The first proof is a cycle of implications.

Technique 4.4.8 (Compare components). In a direct sum, an equation between decompositions splits into one equation per summand. If $𝐮_{1} +$ $\dots +$ $𝐮_{𝑘} =$ $𝐮_{1}^{'} +$ $\dots +$ $𝐮_{𝑘}^{'}$ with $𝐮_{𝑖}, 𝐮_{𝑖}^{'} \in 𝑈_{𝑖}$ and the sum is direct, then $𝐮_{𝑖} = 𝐮_{𝑖}^{'}$ for every $𝑖$ . This is the analogue of comparing coefficients with respect to a basis. To use it, sort every term of an equation into the summand it belongs to, then read off the components one at a time.

Proving that a sum is direct runs the same move backwards. Take two decompositions of one vector, subtract them, and move terms across so that the two sides of the resulting equation lie in different subspaces. Then both sides lie in the intersection, and if the intersection is ${𝟎}$ , both sides vanish. For two subspaces this is the whole argument: $𝐮 + 𝐰 = 𝐮^{'} + 𝐰^{'}$ gives $𝐮 -$ $𝐮^{'} =$ $𝐰^{'} -$ $𝐰 \in$ $𝑈 \cap$ $𝑊$ .

Reach for it whenever a vector is given as a sum of pieces from known subspaces. The canonical example is Theorem 4.4.7. Later uses include orthogonal decompositions (Section 6.4), the independence of eigenvectors with distinct eigenvalues (Section 7.2), and the generalized eigenspace decomposition (Section 10.2).

Used in 12 places and 1 practice exercise

Theorem 4.4.7, Proposition 4.4.9, Exercise 4.4.11, Proposition 4.4.12, Proposition 4.4.26, Theorem 4.4.27, Proposition 4.4.29, Proposition 4.4.31, Exercise 4.4.36, Exercise 4.4.44, Exercise 4.4.51, Problem 4.7; 1 practice exercise.

Proof (by a cycle of implications).standard

Uses. Definition 4.4.4 (Direct sum), Definition 4.4.1 (Sum of subspaces)

Techniques. Proving TFAE (Technique 2.7.1), Compare components (Technique 4.4.8), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. Prove (I)

\Leftrightarrow

(II) first, by subtracting two decompositions. Then prove (II)

\Rightarrow

(III)

\Rightarrow

(IV)

\Rightarrow

(II).

Hint 2. For (II)

\Rightarrow

(III), a vector

𝐯 \in 𝑈_{𝑗}

that also equals

\sum_{𝑖 \neq 𝑗} 𝐮_{𝑖}

gives a decomposition of

𝟎

whose

𝑗

th component is

- 𝐯

Hint 3. For (IV)

\Rightarrow

(II), take a decomposition

𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎

. If some

𝐮_{𝑖} \neq 𝟎

, look at the largest such index

𝑗

and solve for

𝐮_{𝑗}

Sketch. (I)

\Leftrightarrow

(II):

𝟎 + \dots + 𝟎

is one decomposition of

𝟎

, and the difference of two decompositions of

𝐯

is a decomposition of

𝟎

. (II)

\Rightarrow

(III): if

𝐯 \in 𝑈_{𝑗}

equals

\sum_{𝑖 \neq 𝑗} 𝐮_{𝑖}

, then

\sum_{𝑖 \neq 𝑗} 𝐮_{𝑖} +

(- 𝐯) =

𝟎

, so

𝐯 = 𝟎

. (III)

\Rightarrow

(IV): the sum of the earlier subspaces is contained in the sum of the others. (IV)

\Rightarrow

(II): if

𝑗

is the largest index with

𝐮_{𝑗} \neq 𝟎

, then

𝐮_{𝑗} =

- (𝐮_{1} + \dots + 𝐮_{𝑗 - 1}) \in

𝑈_{𝑗} \cap

(𝑈_{1} + \dots + 𝑈_{𝑗 - 1})

, a contradiction.

We prove (I) $\Leftrightarrow$ (II), and then (II) $\Rightarrow$ (III) $\Rightarrow$ (IV) $\Rightarrow$ (II).

(I) $\Rightarrow$ (II). Since $𝟎 \in 𝑈_{𝑖}$ for each $𝑖$ , $𝟎 = 𝟎 + \dots + 𝟎$ is a decomposition of $𝟎$ . By (I) it is the only one.

(II) $\Rightarrow$ (I). Suppose $𝐯 =$ $𝐮_{1} +$ $\dots +$ $𝐮_{𝑘} =$ $𝐮_{1}^{'} +$ $\dots +$ $𝐮_{𝑘}^{'}$ with $𝐮_{𝑖}, 𝐮_{𝑖}^{'} \in 𝑈_{𝑖}$ . Then

(𝐮_{1} - 𝐮_{1}^{'}) + \dots + (𝐮_{𝑘} - 𝐮_{𝑘}^{'}) = 𝐯 - 𝐯 = 𝟎,

and $𝐮_{𝑖} - 𝐮_{𝑖}^{'} \in 𝑈_{𝑖}$ , since $𝑈_{𝑖}$ is a subspace. By (II), $𝐮_{𝑖} - 𝐮_{𝑖}^{'} = 𝟎$ for each $𝑖$ , so the two decompositions agree.

(II) $\Rightarrow$ (III). Fix $𝑗$ , and let $𝐯 \in$ $𝑈_{𝑗} \cap$ $\sum_{𝑖 \neq 𝑗} 𝑈_{𝑖}$ , say $𝐯 = \sum_{𝑖 \neq 𝑗} 𝐮_{𝑖}$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ . Then

𝐮_{1} + \dots + 𝐮_{𝑗 - 1} + (- 𝐯) + 𝐮_{𝑗 + 1} + \dots + 𝐮_{𝑘} = 𝟎

is a decomposition of $𝟎$ , as $- 𝐯 \in 𝑈_{𝑗}$ . By (II) all its components vanish. In particular $- 𝐯 = 𝟎$ , so $𝐯 = 𝟎$ .

(III) $\Rightarrow$ (IV). Let $2 \leq 𝑗 \leq 𝑘$ . Every element $𝐮_{1} + \dots + 𝐮_{𝑗 - 1}$ of $𝑈_{1} + \dots + 𝑈_{𝑗 - 1}$ is also an element of $\sum_{𝑖 \neq 𝑗} 𝑈_{𝑖}$ : take the components $𝟎$ in $𝑈_{𝑖}$ for $𝑖 > 𝑗$ . So

𝑈_{𝑗} \cap (𝑈_{1} + \dots + 𝑈_{𝑗 - 1}) \subseteq 𝑈_{𝑗} \cap \sum_{𝑖 \neq 𝑗} 𝑈_{𝑖} = {𝟎} .

(IV) $\Rightarrow$ (II). Let $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ . Suppose, for a contradiction, that some $𝐮_{𝑖} \neq 𝟎$ , and let $𝑗$ be the largest index with $𝐮_{𝑗} \neq 𝟎$ . Then $𝐮_{𝑖} = 𝟎$ for $𝑖 > 𝑗$ , so

𝐮_{𝑗} = - (𝐮_{1} + \dots + 𝐮_{𝑗 - 1}) = (- 𝐮_{1}) + \dots + (- 𝐮_{𝑗 - 1}) .

If $𝑗 = 1$ , the right side is the empty sum $𝟎$ , which contradicts $𝐮_{1} \neq 𝟎$ . If $𝑗 \geq 2$ , the left side lies in $𝑈_{𝑗}$ and the right side in $𝑈_{1} + \dots + 𝑈_{𝑗 - 1}$ , so $𝐮_{𝑗} \in$ $𝑈_{𝑗} \cap$ $(𝑈_{1} + \dots + 𝑈_{𝑗 - 1}) =$ ${𝟎}$ , again a contradiction. Hence every $𝐮_{𝑖} = 𝟎$ .

For $𝑘 = 2$ , condition (III) says $𝑈_{1} \cap 𝑈_{2} = {𝟎}$ , which proves the last sentence. □

The step (IV) $\Rightarrow$ (II) chose an extremal index, the largest $𝑗$ with $𝐮_{𝑗} \neq 𝟎$ , exactly as the linear dependence lemma does for lists. The other natural route is induction on the number of summands. It rests on a fact that is useful in its own right: independent subspaces can be grouped.

Proposition 4.4.9 (Grouping summands). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ , let $1 \leq 𝑚 < 𝑘$ , and write $𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑚}$ and $𝑆^{'} = 𝑈_{𝑚 + 1} + \dots + 𝑈_{𝑘}$ . Then $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct if and only if both of the following hold.

(i)The sums $𝑈_{1} + \dots + 𝑈_{𝑚}$ and $𝑈_{𝑚 + 1} + \dots + 𝑈_{𝑘}$ are direct.
(ii) $𝑆 \cap 𝑆^{'} = {𝟎}$ .

Moreover, independence of subspaces does not depend on their order, and if $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent, then so is any selection of them.

Proof.routine

Uses. Theorem 4.4.7 (Criteria for direct sums), Definition 4.4.1 (Sum of subspaces)

Techniques. Compare components (Technique 4.4.8)

Hint 1. Use condition (II) of Theorem 4.4.7 throughout. Only its equivalence with (I) is needed.

Hint 2. For the backward direction, given

𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎

, set

𝐬 = 𝐮_{1} + \dots + 𝐮_{𝑚}

and

𝐬^{'} = 𝐮_{𝑚 + 1} + \dots + 𝐮_{𝑘}

. Then

𝐬 = - 𝐬^{'}

Sketch. (

\Rightarrow

) A decomposition of

𝟎

along

𝑈_{1}, \dots, 𝑈_{𝑚}

, padded with zero components, is one along all

𝑘

subspaces, so (i) follows from (II). If

𝐯 \in 𝑆 \cap 𝑆^{'}

, write

𝐯 =

𝐮_{1} +

\dots +

𝐮_{𝑚} =

𝐮_{𝑚 + 1} +

\dots +

𝐮_{𝑘}

; moving everything to one side gives a decomposition of

𝟎

, so all

𝐮_{𝑖} = 𝟎

and

𝐯 = 𝟎

. (

\leq

) If

\sum_{𝑖} 𝐮_{𝑖} = 𝟎

, then

𝐬 =

- 𝐬^{'} \in

𝑆 \cap

𝑆^{'} =

{𝟎}

, and (i) kills each block. The last sentence holds because (II) does not refer to the order, and padding with zeros works for any selection.

We use the equivalence of (I) and (II) in Theorem 4.4.7.

( $\Rightarrow$ ) Suppose the sum of all $𝑘$ subspaces is direct. If $𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ , then $𝐮_{1} +$ $\dots +$ $𝐮_{𝑚} +$ $𝟎 +$ $\dots +$ $𝟎 =$ $𝟎$ is a decomposition of $𝟎$ along $𝑈_{1}, \dots, 𝑈_{𝑘}$ , so every $𝐮_{𝑖} = 𝟎$ . Hence $𝑈_{1} + \dots + 𝑈_{𝑚}$ is direct, and likewise $𝑈_{𝑚 + 1} + \dots + 𝑈_{𝑘}$ . For (ii), let $𝐯 \in 𝑆 \cap 𝑆^{'}$ , say $𝐯 =$ $𝐮_{1} +$ $\dots +$ $𝐮_{𝑚} =$ $𝐮_{𝑚 + 1} +$ $\dots +$ $𝐮_{𝑘}$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ . Then

𝐮_{1} + \dots + 𝐮_{𝑚} + (- 𝐮_{𝑚 + 1}) + \dots + (- 𝐮_{𝑘}) = 𝟎

is a decomposition of $𝟎$ , so all components vanish and $𝐯 = 𝟎$ .

( $\leq$ ) Suppose (i) and (ii), and let $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ . Put $𝐬 = 𝐮_{1} + \dots + 𝐮_{𝑚} \in 𝑆$ and $𝐬^{'} =$ $𝐮_{𝑚 + 1} +$ $\dots +$ $𝐮_{𝑘} \in$ $𝑆^{'}$ . Then $𝐬 = - 𝐬^{'}$ lies in $𝑆$ and in $𝑆^{'}$ , so $𝐬 = 𝟎 = 𝐬^{'}$ by (ii). Now $𝐬 = 𝟎$ is a decomposition of $𝟎$ along $𝑈_{1}, \dots, 𝑈_{𝑚}$ , so $𝐮_{1} = \dots = 𝐮_{𝑚} = 𝟎$ by (i), and likewise $𝐮_{𝑚 + 1} = \dots = 𝐮_{𝑘} = 𝟎$ . So the sum is direct.

Finally, condition (II) does not refer to the order of $𝑈_{1}, \dots, 𝑈_{𝑘}$ , since a sum of vectors does not depend on the order of its terms. If $𝑈_{𝑖_{1}}, \dots, 𝑈_{𝑖_{𝑟}}$ are some of the $𝑈_{𝑖}$ (with distinct indices), a decomposition of $𝟎$ along them becomes a decomposition of $𝟎$ along $𝑈_{1}, \dots, 𝑈_{𝑘}$ once we insert the component $𝟎$ for every other $𝑈_{𝑖}$ . If $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent, all its components vanish. □

With $𝑚 = 𝑘 - 1$ , the proposition says: $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct exactly when $𝑈_{1} + \dots + 𝑈_{𝑘 - 1}$ is direct and $(𝑈_{1} + \dots + 𝑈_{𝑘 - 1}) \cap$ $𝑈_{𝑘} =$ ${𝟎}$ . That is an inductive description of directness, one new subspace at a time.

Technique 4.4.10 (Induction on the number of summands). To prove a statement about $𝑘$ subspaces $𝑈_{1}, \dots, 𝑈_{𝑘}$ , group the first $𝑘 - 1$ of them into $𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}$ and treat $𝑆 + 𝑈_{𝑘}$ as a sum of two subspaces. Proposition 4.4.2(a) says that $𝑆 + 𝑈_{𝑘}$ is the whole sum, and Proposition 4.4.9 says that it is direct exactly when $𝑆$ is direct and $𝑆 \cap 𝑈_{𝑘} = {𝟎}$ . So the two-subspace case and the inductive hypothesis for $𝑆$ combine into the case of $𝑘$ subspaces.

Reach for it when the statement for two subspaces is easy, typically because it reduces to $𝑈 \cap 𝑊 = {𝟎}$ , and the general case looks messy. Watch the hypothesis in the inductive step: it needs $𝑆 \cap 𝑈_{𝑘} = {𝟎}$ , an intersection with a sum, and not just $𝑈_{𝑖} \cap 𝑈_{𝑘} = {𝟎}$ for each $𝑖 < 𝑘$ (Non-example 4.4.6(b)). The canonical example is the second proof of Theorem 4.4.7. The same induction shows that eigenspaces for distinct eigenvalues are independent (Section 7.2).

Used in 4 places

Proposition 4.4.2, Theorem 4.4.7, Exercise 4.4.37, Exercise 4.4.40.

Proof of Theorem 4.4.7 (of (IV) $\Rightarrow$ (II), by induction on $𝑘$ ).standard

Uses. Proposition 4.4.9 (Grouping summands)

Techniques. Induction on the number of summands (Technique 4.4.10)

Hint 1. Induct on

𝑘

. Condition (IV) for

𝑈_{1}, \dots, 𝑈_{𝑘}

contains condition (IV) for

𝑈_{1}, \dots, 𝑈_{𝑘 - 1}

Hint 2. With

𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}

, condition (IV) for

𝑗 = 𝑘

says

𝑆 \cap 𝑈_{𝑘} = {𝟎}

. Apply Proposition 4.4.9 with

𝑚 = 𝑘 - 1

Sketch. For

𝑘 = 1

, (II) is trivial. If

𝑈_{1}, \dots, 𝑈_{𝑘}

satisfy (IV), so do

𝑈_{1}, \dots, 𝑈_{𝑘 - 1}

, which are therefore independent by induction. Condition (IV) for

𝑗 = 𝑘

(𝑈_{1} + \dots + 𝑈_{𝑘 - 1}) \cap

𝑈_{𝑘} =

{𝟎}

, and Proposition 4.4.9 with

𝑚 = 𝑘 - 1

turns these two facts into directness of the whole sum.

This proof uses only (I) $\Leftrightarrow$ (II), which was proved directly above, through Proposition 4.4.9, whose proof uses nothing else. We induct on $𝑘$ . For $𝑘 = 1$ , condition (II) says that $𝐮_{1} = 𝟎$ implies $𝐮_{1} = 𝟎$ .

Let $𝑘 \geq 2$ , assume the implication for $𝑘 - 1$ subspaces, and suppose $𝑈_{1}, \dots, 𝑈_{𝑘}$ satisfy (IV). The conditions of (IV) for $𝑗 = 2, \dots, 𝑘 - 1$ are condition (IV) for $𝑈_{1}, \dots, 𝑈_{𝑘 - 1}$ , so by the inductive hypothesis these subspaces satisfy (II), and their sum $𝑆 = 𝑈_{1} + \dots + 𝑈_{𝑘 - 1}$ is direct. The sum $𝑈_{𝑘}$ of one subspace is trivially direct. Condition (IV) for $𝑗 = 𝑘$ says $𝑆 \cap 𝑈_{𝑘} = {𝟎}$ . By Proposition 4.4.9 with $𝑚 = 𝑘 - 1$ , the sum $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct, which is (II). □

The second proof explains where (IV) comes from: it is exactly what the induction consumes, and each step asks only that the new subspace avoid everything built so far.

Exercise 4.4.11.routine Let $𝑈_{1} = span ((1, 1, 0))$ , $𝑈_{2} = span ((1, 0, 1))$ , and $𝑈_{3} = span ((0, 1, 1))$ in $𝐹^{3}$ .

(a)Show that $𝑈_{𝑖} \cap 𝑈_{𝑗} = {𝟎}$ whenever $𝑖 \neq 𝑗$ , over every field $𝐹$ .
(b)Show that $𝑈_{1} + 𝑈_{2} + 𝑈_{3}$ is direct if $1 + 1 \neq 0$ in $𝐹$ , and not direct if $𝐹 = 𝔽_{2}$ .

Solution.

Uses. Theorem 4.4.7 (Criteria for direct sums), Proposition 1.1.17 (No zero divisors)

Techniques. Compare components (Technique 4.4.8)

Hint 1. For (a), compare coordinates: each of the three vectors has a coordinate equal to

1

where another has

0

Hint 2. For (b), use (IV) of Theorem 4.4.7: it remains to decide when

𝑈_{3} \cap

(𝑈_{1} + 𝑈_{2}) =

{𝟎}

. Solve

𝑐 (0, 1, 1) =

𝑎 (1, 1, 0) +

𝑏 (1, 0, 1)

Sketch. (a) If

𝑎 (1, 1, 0) =

𝑏 (1, 0, 1)

, the second coordinates give

𝑎 = 0

and the third give

𝑏 = 0

; the other pairs are similar. (b)

𝑐 (0, 1, 1) =

(𝑎 + 𝑏, 𝑎, 𝑏)

forces

𝑎 = 𝑏 = 𝑐

and

2 𝑐 = 0

. If

1 + 1 \neq 0

, then

𝑐 = 0

, so (IV) holds. Over

𝔽_{2}

(1, 1, 0) +

(1, 0, 1) +

(0, 1, 1) =

𝟎

violates (II).

(a) If $𝑎 (1, 1, 0) =$ $𝑏 (1, 0, 1)$ , then comparing second coordinates gives $𝑎 = 0$ , and comparing third coordinates gives $𝑏 = 0$ , so the common vector is $𝟎$ . In the same way, comparing coordinates shows $𝑈_{1} \cap 𝑈_{3} = {𝟎}$ (from $𝑎 (1, 1, 0) =$ $𝑐 (0, 1, 1)$ , the first coordinates give $𝑎 = 0$ and the third give $𝑐 = 0$ ) and $𝑈_{2} \cap 𝑈_{3} = {𝟎}$ (from $𝑏 (1, 0, 1) =$ $𝑐 (0, 1, 1)$ , the first coordinates give $𝑏 = 0$ and the second give $𝑐 = 0$ ).

(b) We check (IV) of Theorem 4.4.7. By (a), $𝑈_{2} \cap 𝑈_{1} = {𝟎}$ . Suppose $𝑐 (0, 1, 1) \in 𝑈_{1} + 𝑈_{2}$ , say

(0, 𝑐, 𝑐) = 𝑎 (1, 1, 0) + 𝑏 (1, 0, 1) = (𝑎 + 𝑏, 𝑎, 𝑏) .

Then $𝑎 = 𝑐$ , $𝑏 = 𝑐$ , and $𝑎 + 𝑏 = 0$ , so $(1 + 1) 𝑐 = 0$ . If $1 + 1 \neq 0$ , then $𝑐 = 0$ (Proposition 1.1.17), so $𝑈_{3} \cap$ $(𝑈_{1} + 𝑈_{2}) =$ ${𝟎}$ and the sum is direct. Over $𝔽_{2}$ , taking $𝑐 = 1$ gives

(1, 1, 0) + (1, 0, 1) + (0, 1, 1) = (0, 0, 0),

a decomposition of $𝟎$ with nonzero components, so (II) fails and the sum is not direct.

So the pairwise intersections are trivial over every field, but directness depends on the field. Over $𝔽_{2}$ , the three lines lie in the plane $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ , like the three lines of Non-example 4.4.6(b).

4.4.3Bases and dimension

For lists, independence is what makes a spanning list a basis. For subspaces it plays the same role one level up: bases of the summands concatenate to a basis of the sum exactly when the summands are independent. The lists version is a recitation exercise.

Proposition 4.4.12 (Concatenating independent lists). Let

{ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘}

be linearly independent lists in

𝑉

. The concatenation

{ℒ︀}_{1}, \dots, {ℒ︀}_{𝑘}

is linearly independent if and only if

span ({ℒ︀}_{1}),

\dots,

span ({ℒ︀}_{𝑘})

are independent subspaces. In particular, for linearly independent lists

ℒ︀

and

ℳ︀

, the concatenation

ℒ︀, ℳ︀

is linearly independent if and only if

span (ℒ︀) \cap

span (ℳ︀) =

{𝟎}

Proof.routine

Uses. Theorem 4.4.7 (Criteria for direct sums), Theorem 1.5.5 (Independence is decided at zero), Definition 1.3.10 (Span)

Techniques. Compare components (Technique 4.4.8)

Hint 1. A linear combination of the concatenation is a sum of

𝑘

pieces, the

𝑖

th of which is a combination of

{ℒ︀}_{𝑖}

and so lies in

span ({ℒ︀}_{𝑖})

Hint 2. So a linear relation among the concatenation is the same thing as a decomposition of

𝟎

along the spans, together with a representation of each component. Compare (II) of Theorem 1.5.5 with (II) of Theorem 4.4.7.

Sketch. Write

{ℒ︀}_{𝑖} =

𝐯_{𝑖 1},

\dots,

𝐯_{𝑖 𝑛_{𝑖}}

. A relation

\sum_{𝑖} \sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗} =

𝟎

is a decomposition

𝟎 = 𝐮_{1} + \dots + 𝐮_{𝑘}

with

𝐮_{𝑖} =

\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗} \in

span ({ℒ︀}_{𝑖})

, and since

{ℒ︀}_{𝑖}

is independent,

𝐮_{𝑖} = 𝟎

exactly when all

𝑎_{𝑖 𝑗} = 0

. So “only the trivial relation” and “only the trivial decomposition of

𝟎

” say the same thing.

Write ${ℒ︀}_{𝑖} =$ $𝐯_{𝑖 1},$ $\dots,$ $𝐯_{𝑖 𝑛_{𝑖}}$ and $𝑊_{𝑖} = span ({ℒ︀}_{𝑖})$ .

( $\leq$ ) Suppose $𝑊_{1}, \dots, 𝑊_{𝑘}$ are independent, and let $\sum_{𝑖 = 1}^{𝑘} \sum_{𝑗 = 1}^{𝑛_{𝑖}} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗} =$ $𝟎$ . Put $𝐮_{𝑖} =$ $\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗} \in$ $𝑊_{𝑖}$ . Then $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ , so every $𝐮_{𝑖} = 𝟎$ by (II) of Theorem 4.4.7. Since ${ℒ︀}_{𝑖}$ is independent, $𝑎_{𝑖 1} =$ $\dots =$ $𝑎_{𝑖 𝑛_{𝑖}} =$ $0$ (Theorem 1.5.5). So the concatenation has only the trivial relation and is independent.

( $\Rightarrow$ ) Suppose the concatenation is independent, and let $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ with $𝐮_{𝑖} \in 𝑊_{𝑖}$ , say $𝐮_{𝑖} =$ $\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗}$ . Then $\sum_{𝑖} \sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐯_{𝑖 𝑗} =$ $𝟎$ , so all $𝑎_{𝑖 𝑗} = 0$ and every $𝐮_{𝑖} = 𝟎$ . By (II) of Theorem 4.4.7, $𝑊_{1}, \dots, 𝑊_{𝑘}$ are independent.

The case $𝑘 = 2$ follows from the last sentence of Theorem 4.4.7. □

Theorem 4.4.13 (Bases and dimension of a sum). Let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be finite-dimensional subspaces of $𝑉$ , and for each $𝑖$ let ${ℬ︀}_{𝑖}$ be a basis of $𝑈_{𝑖}$ .

(a)The concatenation ${ℬ︀}_{1}, \dots, {ℬ︀}_{𝑘}$ spans $𝑈_{1} + \dots + 𝑈_{𝑘}$ . Hence $𝑈_{1} + \dots + 𝑈_{𝑘}$ is finite-dimensional, and
$dim (𝑈_{1} + \dots + 𝑈_{𝑘}) \leq dim 𝑈_{1} + \dots + dim 𝑈_{𝑘} .$
(b)The sum $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct if and only if ${ℬ︀}_{1}, \dots, {ℬ︀}_{𝑘}$ is a basis of $𝑈_{1} + \dots + 𝑈_{𝑘}$ .
(c)The sum $𝑈_{1} + \dots + 𝑈_{𝑘}$ is direct if and only if $dim (𝑈_{1} + \dots + 𝑈_{𝑘}) =$ $dim 𝑈_{1} +$ $\dots +$ $dim 𝑈_{𝑘}$ .

Proof.routine

Uses. Proposition 4.4.2 (Sums are the smallest containing subspaces), Proposition 4.4.12 (Concatenating independent lists), Definition 1.6.23 (Finite-dimensional), Corollary 4.1.11 (Spanning lists are long), Theorem 4.1.4 (All bases have the same length)

Techniques. Glue bases of the summands (Technique 4.4.14), Dimension counting (Technique 4.2.24)

Hint 1. (a) is Proposition 4.4.2(c), together with the fact that a spanning list is at least as long as the dimension.

Hint 2. For (b), the concatenation already spans, so it is a basis exactly when it is independent. For (c), when is a spanning list whose length equals the dimension a basis?

Sketch. (a) By Proposition 4.4.2(c), the concatenation, of length

𝑛_{1} + \dots + 𝑛_{𝑘}

with

𝑛_{𝑖} = dim 𝑈_{𝑖}

, spans the sum, so the sum is finite-dimensional with dimension at most

𝑛_{1} + \dots + 𝑛_{𝑘}

. (b) A spanning list is a basis iff it is independent, iff the sum is direct by Proposition 4.4.12. (c) A spanning list is a basis iff its length is the dimension, so (c) reduces to (b).

Let $𝑛_{𝑖} = dim 𝑈_{𝑖}$ , so that ${ℬ︀}_{𝑖}$ has length $𝑛_{𝑖}$ , and let $𝑁 = 𝑛_{1} + \dots + 𝑛_{𝑘}$ be the length of the concatenation $ℬ︀ = {ℬ︀}_{1}, \dots, {ℬ︀}_{𝑘}$ .

(a) By Proposition 4.4.2(c), $span (ℬ︀) =$ $span ({ℬ︀}_{1}) +$ $\dots +$ $span ({ℬ︀}_{𝑘}) =$ $𝑈_{1} +$ $\dots +$ $𝑈_{𝑘}$ . So the sum is spanned by a finite list and is finite-dimensional (Definition 1.6.23), and its dimension is at most $𝑁$ by Corollary 4.1.11(a).

(b) By (a), $ℬ︀$ spans the sum, so $ℬ︀$ is a basis of the sum if and only if it is linearly independent. Each ${ℬ︀}_{𝑖}$ is independent with span $𝑈_{𝑖}$ , so by Proposition 4.4.12, $ℬ︀$ is independent if and only if $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent.

(c) Let $𝑑 =$ $dim (𝑈_{1} + \dots + 𝑈_{𝑘})$ . If $ℬ︀$ is a basis of the sum, then $𝑁 = 𝑑$ by Theorem 4.1.4. Conversely, if $𝑁 = 𝑑$ , then $ℬ︀$ is a spanning list of length $𝑑$ , hence a basis by Corollary 4.1.11(b). So $𝑑 = 𝑁$ if and only if $ℬ︀$ is a basis, and (b) finishes the proof. □

In particular, if $𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘}$ is finite-dimensional, then $dim 𝑉 =$ $dim 𝑈_{1} +$ $\dots +$ $dim 𝑈_{𝑘}$ . Taking $𝑈_{𝑖} = span (𝐯_{𝑖})$ with $𝐯_{𝑖} \neq 𝟎$ , whose basis is the list $𝐯_{𝑖}$ , part (b) also gives the converse of Example 4.4.5(a): $𝑉 =$ $span (𝐯_{1}) \oplus$ $\dots \oplus$ $span (𝐯_{𝑛})$ if and only if $𝐯_{1}, \dots, 𝐯_{𝑛}$ is a basis. And part (c) gives a second look at Non-example 4.4.6(b): three lines in $ℝ^{2}$ have dimensions adding up to $3 > 2$ , so their sum cannot be direct.

Technique 4.4.14 (Glue bases of the summands). When $𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘}$ , choose a basis of each $𝑈_{𝑖}$ and concatenate. The result is a basis of $𝑉$ (Theorem 4.4.13) adapted to the decomposition: each basis vector lies in one summand, so the coordinates of a vector split into blocks, one block for each component. In the other direction, to show that subspaces are independent, it suffices to check that bases of them concatenate to an independent list, or just to compare dimensions.

Reach for it to compute the dimension of a direct sum, or to turn a statement about subspaces into one about lists. The canonical example is Theorem 4.4.13. Later, a basis glued from bases of eigenspaces is an eigenbasis (Section 7.4), and Jordan bases are glued from chains (Section 10.3).

Used in 7 places

Theorem 4.4.13, Proposition 4.4.16, Theorem 7.4.16, Theorem 7.6.34, Exercise 7.6.49, Exercise 8.1.36, Exercise 8.2.36.

The span explorer shows the case of a plane and a line in $ℝ^{3}$ .

With $𝑈 = span (𝐞_{1}, 𝐞_{2})$ and $𝑊 = span ((1, 1, 1))$ :

(2, 1, 3) = (- 1, - 2, 0) + (3, 3, 3) with (- 1, - 2, 0) \in 𝑈, (3, 3, 3) \in 𝑊 .

The first two vectors span a plane

𝑈

and the third spans a line

𝑊

. By Proposition 4.4.12, the list is independent exactly when

𝑈 \cap 𝑊 = {𝟎}

, and then the coefficients of the target

𝐰

split it as

(𝑎_{1} 𝐯_{1} + 𝑎_{2} 𝐯_{2}) +

𝑎_{3} 𝐯_{3} \in

𝑈 \oplus

𝑊

. Drag

𝐯_{3}

into the plane: the list becomes dependent, and the dependence relation, rearranged, exhibits a nonzero vector of

𝑈 \cap 𝑊

4.4.4Complements

Given a subspace $𝑈$ of $𝑉$ , can $𝑉$ be split as $𝑈 \oplus 𝑊$ ? In finite dimensions it always can, by extending a basis.

Definition 4.4.15 (Complement). Let

𝑈

be a subspace of

𝑉

. A complement of

𝑈

𝑉

is a subspace

𝑊

𝑉

with

𝑉 = 𝑈 \oplus 𝑊

By Theorem 4.4.7, $𝑊$ is a complement of $𝑈$ if and only if $𝑈 + 𝑊 = 𝑉$ and $𝑈 \cap 𝑊 = {𝟎}$ . The relation is symmetric: $𝑊$ is a complement of $𝑈$ if and only if $𝑈$ is a complement of $𝑊$ .

Proposition 4.4.16 (Complements exist). Let $𝑉$ be finite-dimensional, and let $𝑈$ be a subspace of $𝑉$ .

(a)If $𝐮_{1}, \dots, 𝐮_{𝑘}$ is a basis of $𝑈$ and $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐰_{1},$ $\dots,$ $𝐰_{𝑟}$ is a basis of $𝑉$ , then $span (𝐰_{1}, \dots, 𝐰_{𝑟})$ is a complement of $𝑈$ . In particular, $𝑈$ has a complement.
(b)Every complement of $𝑈$ arises as in (a), and every complement $𝑊$ of $𝑈$ has $dim 𝑊 = dim 𝑉 - dim 𝑈$ .

Proof.routine

Uses. Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.1.16 (Extending to a basis), Proposition 4.4.2 (Sums are the smallest containing subspaces), Proposition 4.4.12 (Concatenating independent lists), Theorem 4.4.13 (Bases and dimension of a sum)

Techniques. Extend a basis of a subspace (Technique 4.1.18), Glue bases of the summands (Technique 4.4.14)

Hint 1. For (a), the extended list spans

𝑉

and is independent. Translate these two facts into statements about

𝑈 + 𝑊

and

𝑈 \cap 𝑊

Hint 2. For (b), concatenate a basis of

𝑈

with a basis of a complement

𝑊

and apply Theorem 4.4.13.

Sketch. (a) Let

𝑊 =

span (𝐰_{1}, \dots, 𝐰_{𝑟})

. The concatenation spans

𝑉

, so

𝑈 + 𝑊 = 𝑉

(Proposition 4.4.2(c)), and it is independent, so

𝑈 \cap 𝑊 = {𝟎}

(Proposition 4.4.12). A basis of

𝑈

exists and extends (Proposition 4.1.21, Theorem 4.1.16). (b) If

𝑉 = 𝑈 \oplus 𝑊

, then

𝑊

is finite-dimensional, and a basis of

𝑈

followed by a basis of

𝑊

is a basis of

𝑉

(Theorem 4.4.13(b)), so

𝑊

arises as in (a), with

𝑟 = dim 𝑉 - 𝑘

(a) Let $𝑊 =$ $span (𝐰_{1}, \dots, 𝐰_{𝑟})$ . By Proposition 4.4.2(c), $𝑈 +$ $𝑊 =$ $span (𝐮_{1}, \dots, 𝐮_{𝑘}, 𝐰_{1}, \dots, 𝐰_{𝑟}) =$ $𝑉$ . The lists $𝐮_{1}, \dots, 𝐮_{𝑘}$ and $𝐰_{1}, \dots, 𝐰_{𝑟}$ are independent (the second as part of a basis), and their concatenation is independent, so $𝑈 \cap 𝑊 = {𝟎}$ by Proposition 4.4.12. Hence $𝑉 = 𝑈 \oplus 𝑊$ by Theorem 4.4.7.

For the existence of a complement: $𝑈$ is finite-dimensional (Proposition 4.1.21(a)), so it has a basis $𝐮_{1}, \dots, 𝐮_{𝑘}$ , which is independent in $𝑉$ and extends to a basis of $𝑉$ (Theorem 4.1.16). Apply the first part.

(b) Let $𝑊$ be a complement of $𝑈$ . It is finite-dimensional (Proposition 4.1.21(a)); let $𝐰_{1}, \dots, 𝐰_{𝑟}$ be a basis of $𝑊$ , and $𝐮_{1}, \dots, 𝐮_{𝑘}$ a basis of $𝑈$ . Since $𝑉 = 𝑈 \oplus 𝑊$ , Theorem 4.4.13(b) says that $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐰_{1},$ $\dots,$ $𝐰_{𝑟}$ is a basis of $𝑉$ , and $𝑊$ is the span of the added vectors, as in (a). Counting, $dim 𝑉 = 𝑘 + 𝑟$ , so $dim 𝑊 =$ $𝑟 =$ $dim 𝑉 -$ $dim 𝑈$ . □

Example 4.4.17 (Complements are not unique). In $ℝ^{2}$ , every line through $𝟎$ other than the $𝑥$ -axis is a complement of the $𝑥$ -axis: two distinct lines meet only in $𝟎$ , and their sum has dimension $1 + 1 = 2$ by Theorem 4.4.13(c), so it is $ℝ^{2}$ .

In $ℚ^{4}$ , Example 4.1.17 extended $𝐯_{1} = (1, 1, 0, 0)$ , $𝐯_{2} = (1, 1, 1, 1)$ by $𝐞_{1}, 𝐞_{3}$ to a basis of $ℚ^{4}$ . So $span (𝐞_{1}, 𝐞_{3})$ is a complement of $𝑈 = span (𝐯_{1}, 𝐯_{2})$ . So is $span (𝐞_{2}, 𝐞_{4})$ . A vector $𝑎 𝐯_{1} +$ $𝑏 𝐯_{2} =$ $(𝑎 + 𝑏, 𝑎 + 𝑏, 𝑏, 𝑏)$ of $𝑈$ lies in $span (𝐞_{2}, 𝐞_{4})$ only if its first and third coordinates vanish, that is, $𝑎 + 𝑏 = 0$ and $𝑏 = 0$ , so $𝑈 \cap$ $span (𝐞_{2}, 𝐞_{4}) =$ ${𝟎}$ . By Proposition 4.4.12, $𝐯_{1}, 𝐯_{2}, 𝐞_{2}, 𝐞_{4}$ is independent; having length $4$ , it is a basis (Theorem 4.1.10(b)), and (a) applies.

What a subspace does determine is the dimension of its complements, by (b). Exercise 4.4.46 shows that a subspace has only one complement exactly when it is ${𝟎}$ or $𝑉$ , and Exercise 4.4.49 counts the complements over a finite field.

Complements also show that nothing distinguishes kernels or images among subspaces.

Proposition 4.4.18 (Every subspace is a kernel and an image). Let $𝑉$ be finite-dimensional, and let $𝑈$ be a subspace of $𝑉$ with a complement $𝑊$ . There are $𝑆, 𝑇 \in L (𝑉)$ with

im 𝑆 = 𝑈, ker 𝑆 = 𝑊, ker 𝑇 = 𝑈, im 𝑇 = 𝑊 .

In particular, every subspace of $𝑉$ is the image of some operator on $𝑉$ and the kernel of another.

Proof.routine

Uses. Proposition 4.4.16 (Complements exist), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Extend a basis of a subspace (Technique 4.1.18), Construct a map by choosing it on a basis (Technique 2.2.4)

Hint 1. Concatenate bases of

𝑈

and

𝑊

to get a basis of

𝑉

, and define

𝑆

and

𝑇

on it.

Hint 2. Let

𝑆

fix the basis vectors of

𝑈

and kill those of

𝑊

; let

𝑇

do the opposite.

Sketch. With bases

𝐮_{1}, \dots, 𝐮_{𝑘}

𝑈

and

𝐰_{1}, \dots, 𝐰_{𝑟}

𝑊

, which together form a basis of

𝑉

, let

𝑆 (𝐮_{𝑖}) = 𝐮_{𝑖}

𝑆 (𝐰_{𝑗}) = 𝟎

𝑇 (𝐮_{𝑖}) = 𝟎

, and

𝑇 (𝐰_{𝑗}) = 𝐰_{𝑗}

. Then

𝑆 (\sum 𝑎_{𝑖} 𝐮_{𝑖} + \sum 𝑏_{𝑗} 𝐰_{𝑗}) =

\sum 𝑎_{𝑖} 𝐮_{𝑖}

, which is

𝟎

iff all

𝑎_{𝑖} = 0

; read off kernels and images. Complements exist by Proposition 4.4.16.

Let $𝐮_{1}, \dots, 𝐮_{𝑘}$ and $𝐰_{1}, \dots, 𝐰_{𝑟}$ be bases of $𝑈$ and $𝑊$ . Since $𝑉 = 𝑈 \oplus 𝑊$ , their concatenation is a basis of $𝑉$ (Theorem 4.4.13(b)). By Theorem 2.2.1 there are $𝑆, 𝑇 \in L (𝑉)$ with

𝑆 (𝐮_{𝑖}) = 𝐮_{𝑖}, 𝑆 (𝐰_{𝑗}) = 𝟎, 𝑇 (𝐮_{𝑖}) = 𝟎, 𝑇 (𝐰_{𝑗}) = 𝐰_{𝑗}

for all $𝑖$ and $𝑗$ . For $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ , we get $𝑆 (𝐯) =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}$ and $𝑇 (𝐯) =$ $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ .

The image of $𝑆$ is the span of $𝑆 (𝐮_{1}),$ $\dots,$ $𝑆 (𝐰_{𝑟})$ , which is $span (𝐮_{1}, \dots, 𝐮_{𝑘}, 𝟎, \dots, 𝟎) =$ $𝑈$ (Proposition 2.1.16). Since $𝐮_{1}, \dots, 𝐮_{𝑘}$ is independent, $𝑆 (𝐯) = 𝟎$ if and only if all $𝑎_{𝑖} = 0$ (Theorem 1.5.5), that is, if and only if $𝐯 \in 𝑊$ . So $ker 𝑆 = 𝑊$ . Exchanging the roles of $𝑈$ and $𝑊$ gives $ker 𝑇 = 𝑈$ and $im 𝑇 = 𝑊$ . Every subspace has a complement by Proposition 4.4.16, which proves the last sentence. □

For $𝑉 = 𝐹^{𝑛}$ this says that every subspace of $𝐹^{𝑛}$ is a column space (which is clear: put a spanning list in the columns) and a null space. The second statement is the useful one: every subspace of $𝐹^{𝑛}$ is the solution set of a homogeneous linear system. The operators $𝑆$ and $𝑇$ will reappear in Section 4.4.6 as the projections attached to $𝑉 = 𝑈 \oplus 𝑊$ .

Aside (Complements in infinite dimensions).Beyond 21-242 The proof of Proposition 4.4.16 extends a basis, which needs

𝑉

to be finite-dimensional. Some subspaces of infinite-dimensional spaces have evident complements; in

𝐹 [𝑥]

, the polynomials with

𝑝 (0) = 0

have the constants as a complement. In fact every subspace of every vector space has a complement, but in general the proof needs Zorn’s lemma (Section 15.1), and it produces no explicit complement.

4.4.5The dimension formula

Intuition (Inclusion–exclusion for subspaces). For finite sets,

| 𝐴 \cup

𝐵 | =

| 𝐴 | +

| 𝐵 | -

| 𝐴 \cap

𝐵 |

, because the elements of

𝐴 \cap 𝐵

are counted twice. For subspaces,

𝑈 + 𝑊

takes the place of the union (it is the smallest subspace containing it), and dimension takes the place of size. A basis of

𝑈

together with a basis of

𝑊

spans

𝑈 + 𝑊

, and we expect it to overcount by

dim (𝑈 \cap 𝑊)

directions. For arbitrary bases this is not visible: no entry of either basis need lie in

𝑈 \cap 𝑊

(Exercise 4.4.38). So the first proof chooses bases that are adapted to

𝑈 \cap 𝑊

, which is the extension technique again.

Theorem 4.4.19 (Dimension of a sum). Let $𝑈$ and $𝑊$ be finite-dimensional subspaces of $𝑉$ . Then $𝑈 + 𝑊$ and $𝑈 \cap 𝑊$ are finite-dimensional, and

dim (𝑈 + 𝑊) = dim 𝑈 + dim 𝑊 - dim (𝑈 \cap 𝑊) .

Proof (by extending a basis of $𝑈 \cap 𝑊$ ).standard

Uses. Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.1.16 (Extending to a basis), Proposition 1.4.14 (The span is the smallest subspace containing a list), Theorem 1.5.5 (Independence is decided at zero), Theorem 4.1.4 (All bases have the same length)

Techniques. Extend a basis of a subspace (Technique 4.1.18)

Hint 1. Start with a basis

𝐱_{1}, \dots, 𝐱_{𝑚}

𝑈 \cap 𝑊

and extend it twice: to a basis of

𝑈

and to a basis of

𝑊

Hint 2. Show that the

𝐱

’s, the new vectors of

𝑈

, and the new vectors of

𝑊

together form a basis of

𝑈 + 𝑊

. Spanning is easy.

Hint 3. For independence, take a relation and put the terms coming from

𝑊

on one side. That side lies in

𝑊

, and the other side lies in

𝑈

Sketch. Extend a basis

𝐱_{1}, \dots, 𝐱_{𝑚}

𝑈 \cap 𝑊

to bases

𝐱, 𝐮_{1}, \dots, 𝐮_{𝑗}

𝑈

and

𝐱, 𝐰_{1}, \dots, 𝐰_{𝑙}

𝑊

. The combined list spans

𝑈 + 𝑊

. In a relation

\sum 𝑎_{𝑖} 𝐱_{𝑖} +

\sum 𝑏_{𝑖} 𝐮_{𝑖} +

\sum 𝑐_{𝑖} 𝐰_{𝑖} =

𝟎

, the vector

\sum 𝑐_{𝑖} 𝐰_{𝑖}

equals

- (\sum 𝑎_{𝑖} 𝐱_{𝑖} + \sum 𝑏_{𝑖} 𝐮_{𝑖}) \in

𝑈

, so it lies in

𝑈 \cap 𝑊

and is a combination of the

𝐱

’s; independence of

𝐱, 𝐰

forces

𝑐 = 𝟎

, and then independence of

𝐱, 𝐮

forces

𝑎 = 𝑏 = 𝟎

. Count:

𝑚 +

𝑗 +

𝑙 =

(𝑚 + 𝑗) +

(𝑚 + 𝑙) -

𝑚

Since $𝑈 \cap 𝑊$ is a subspace of the finite-dimensional space $𝑈$ , it is finite-dimensional (Proposition 4.1.21(a)). Let $𝐱_{1}, \dots, 𝐱_{𝑚}$ be a basis of $𝑈 \cap 𝑊$ . It is an independent list in $𝑈$ , so it extends to a basis $𝐱_{1},$ $\dots,$ $𝐱_{𝑚},$ $𝐮_{1},$ $\dots,$ $𝐮_{𝑗}$ of $𝑈$ (Theorem 4.1.16, applied in $𝑈$ ), and likewise to a basis $𝐱_{1},$ $\dots,$ $𝐱_{𝑚},$ $𝐰_{1},$ $\dots,$ $𝐰_{𝑙}$ of $𝑊$ . Then $dim 𝑈 = 𝑚 + 𝑗$ and $dim 𝑊 = 𝑚 + 𝑙$ . We claim that

ℒ︀ = 𝐱_{1}, \dots, 𝐱_{𝑚}, 𝐮_{1}, \dots, 𝐮_{𝑗}, 𝐰_{1}, \dots, 𝐰_{𝑙}

is a basis of $𝑈 + 𝑊$ . Then $𝑈 + 𝑊$ is finite-dimensional, and

dim (𝑈 + 𝑊) = 𝑚 + 𝑗 + 𝑙 = (𝑚 + 𝑗) + (𝑚 + 𝑙) - 𝑚 = dim 𝑈 + dim 𝑊 - dim (𝑈 \cap 𝑊) .

$ℒ︀$ spans $𝑈 + 𝑊$ . Every entry of $ℒ︀$ lies in $𝑈$ or in $𝑊$ , hence in the subspace $𝑈 + 𝑊$ , so $span (ℒ︀) \subseteq 𝑈 + 𝑊$ (Proposition 1.4.14(b)). Conversely, let $𝐮 + 𝐰 \in 𝑈 + 𝑊$ with $𝐮 \in 𝑈$ and $𝐰 \in 𝑊$ . Then $𝐮$ is a combination of the $𝐱$ ’s and $𝐮$ ’s, and $𝐰$ is a combination of the $𝐱$ ’s and $𝐰$ ’s, so $𝐮 + 𝐰 \in span (ℒ︀)$ .

$ℒ︀$ is independent. Suppose

\sum_{𝑖 = 1}^{𝑚} 𝑎_{𝑖} 𝐱_{𝑖} + \sum_{𝑖 = 1}^{𝑗} 𝑏_{𝑖} 𝐮_{𝑖} + \sum_{𝑖 = 1}^{𝑙} 𝑐_{𝑖} 𝐰_{𝑖} = 𝟎 .

Then $𝐲 =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐰_{𝑖} =$ $- (\sum_{𝑖} 𝑎_{𝑖} 𝐱_{𝑖} + \sum_{𝑖} 𝑏_{𝑖} 𝐮_{𝑖})$ lies in $𝑊$ (by the first expression) and in $𝑈$ (by the second), so $𝐲 \in 𝑈 \cap 𝑊$ , and $𝐲 = \sum_{𝑖} 𝑑_{𝑖} 𝐱_{𝑖}$ for some scalars $𝑑_{𝑖}$ . Hence

\sum_{𝑖 = 1}^{𝑙} 𝑐_{𝑖} 𝐰_{𝑖} - \sum_{𝑖 = 1}^{𝑚} 𝑑_{𝑖} 𝐱_{𝑖} = 𝟎,

and since $𝐱_{1},$ $\dots,$ $𝐱_{𝑚},$ $𝐰_{1},$ $\dots,$ $𝐰_{𝑙}$ is independent, all $𝑐_{𝑖}$ (and $𝑑_{𝑖}$ ) are $0$ (Theorem 1.5.5). The original relation becomes $\sum_{𝑖} 𝑎_{𝑖} 𝐱_{𝑖} +$ $\sum_{𝑖} 𝑏_{𝑖} 𝐮_{𝑖} =$ $𝟎$ , and since $𝐱_{1},$ $\dots,$ $𝐱_{𝑚},$ $𝐮_{1},$ $\dots,$ $𝐮_{𝑗}$ is independent, all $𝑎_{𝑖}$ and $𝑏_{𝑖}$ are $0$ as well. □

The second proof reduces the formula to the direct case. Inside $𝑈$ , split off the part of $𝑈$ that overlaps $𝑊$ ; what remains of $𝑈$ meets $𝑊$ trivially.

Proof (via a complement inside $𝑈$ ).standard

Uses. Proposition 4.4.16 (Complements exist), Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.4.7 (Criteria for direct sums), Theorem 4.4.13 (Bases and dimension of a sum), Proposition 4.4.2 (Sums are the smallest containing subspaces)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Choose a complement

𝑋

𝑈 \cap 𝑊

inside the finite-dimensional space

𝑈

, so that

𝑈 = (𝑈 \cap 𝑊) \oplus 𝑋

. Show that

𝑈 + 𝑊 = 𝑋 \oplus 𝑊

Hint 2. Since

𝑋 \subseteq 𝑈

, a vector of

𝑋 \cap 𝑊

lies in

𝑈 \cap 𝑊

. For the sum,

𝑈 =

(𝑈 \cap 𝑊) +

𝑋 \subseteq

𝑊 +

𝑋

Sketch. Let

𝑋

be a complement of

𝑈 \cap 𝑊

𝑈

, so

dim 𝑋 =

dim 𝑈 -

dim (𝑈 \cap 𝑊)

. Since

𝑋 \subseteq 𝑈

𝑋 \cap

𝑊 =

𝑋 \cap

(𝑈 \cap 𝑊) =

{𝟎}

. Since

𝑈 =

(𝑈 \cap 𝑊) +

𝑋 \subseteq

𝑋 +

𝑊

𝑈 + 𝑊 = 𝑋 + 𝑊

. So

𝑈 + 𝑊 = 𝑋 \oplus 𝑊

, and Theorem 4.4.13(c) gives

dim (𝑈 + 𝑊) =

dim 𝑋 +

dim 𝑊

As in the first proof, $𝑈 \cap 𝑊$ is finite-dimensional. Apply Proposition 4.4.16 in the finite-dimensional space $𝑈$ to its subspace $𝑈 \cap 𝑊$ : there is a subspace $𝑋 \subseteq 𝑈$ with $𝑈 = (𝑈 \cap 𝑊) \oplus 𝑋$ , and $dim 𝑋 =$ $dim 𝑈 -$ $dim (𝑈 \cap 𝑊)$ .

First, $𝑋 \cap 𝑊 = {𝟎}$ : a vector of $𝑋 \cap 𝑊$ lies in $𝑈$ (as $𝑋 \subseteq 𝑈$ ) and in $𝑊$ , so it lies in $𝑋 \cap (𝑈 \cap 𝑊) = {𝟎}$ . Second, $𝑋 + 𝑊 = 𝑈 + 𝑊$ : since $𝑋 \subseteq 𝑈$ , we have $𝑋 + 𝑊 \subseteq 𝑈 + 𝑊$ ; and since $𝑈 \cap 𝑊 \subseteq 𝑊$ , we have $𝑈 =$ $(𝑈 \cap 𝑊) +$ $𝑋 \subseteq$ $𝑋 +$ $𝑊$ , so the subspace $𝑋 + 𝑊$ contains $𝑈$ and $𝑊$ and hence $𝑈 + 𝑊$ (Proposition 4.4.2(b)). By Theorem 4.4.7, $𝑈 + 𝑊 = 𝑋 \oplus 𝑊$ , and by Theorem 4.4.13, $𝑈 + 𝑊$ is finite-dimensional with

dim (𝑈 + 𝑊) = dim 𝑋 + dim 𝑊 = dim 𝑈 - dim (𝑈 \cap 𝑊) + dim 𝑊 .

□

The third proof counts a kernel and an image. It completes the remark after Proposition 4.2.26, whose first proof used the same map to bound $dim (𝑈 \cap 𝑊)$ .

Proof (by rank–nullity).standard

Uses. Theorem 4.2.11 (Rank–nullity), Proposition 4.2.6 (Injective maps preserve dimension), Theorem 1.5.5 (Independence is decided at zero), Proposition 4.2.26 (Subspaces that must meet), Proposition 4.1.21 (Dimensions of subspaces)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. With bases

𝐮_{1}, \dots, 𝐮_{𝑘}

𝑈

and

𝐰_{1}, \dots, 𝐰_{𝑙}

𝑊

, consider

𝐿 : 𝐹^{𝑘 + 𝑙} \to 𝑉

𝐿 (𝐚, 𝐛) =

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} -

\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}

. What is

im 𝐿

Hint 2. An element of

ker 𝐿

is a pair of coordinate vectors describing one vector of

𝑈 \cap 𝑊

. Show that

(𝐚, 𝐛) \mapsto

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}

is a bijection from

ker 𝐿

onto

𝑈 \cap 𝑊

Sketch.

im 𝐿 = 𝑈 + 𝑊

, since

𝐛

and

- 𝐛

range over the same set. The linear map

𝜑 (𝐚, 𝐛) =

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}

sends

ker 𝐿

into

𝑈 \cap 𝑊

and is injective, as in Proposition 4.2.26. It is onto

𝑈 \cap 𝑊

: if

𝐯 =

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} =

\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}

, then

(𝐚, 𝐛) \in ker 𝐿

and

𝜑 (𝐚, 𝐛) = 𝐯

. So

dim ker 𝐿 =

dim (𝑈 \cap 𝑊)

, and rank–nullity for

𝐿

reads

𝑘 +

𝑙 =

dim (𝑈 \cap 𝑊) +

dim (𝑈 + 𝑊)

Let $𝐮_{1}, \dots, 𝐮_{𝑘}$ and $𝐰_{1}, \dots, 𝐰_{𝑙}$ be bases of $𝑈$ and $𝑊$ . Write elements of $𝐹^{𝑘 + 𝑙}$ as $(𝐚, 𝐛)$ with $𝐚 \in 𝐹^{𝑘}$ and $𝐛 \in 𝐹^{𝑙}$ , and define the linear map

𝐿 : 𝐹^{𝑘 + 𝑙} \to 𝑉, 𝐿 (𝐚, 𝐛) = 𝑎_{1} 𝐮_{1} + \dots + 𝑎_{𝑘} 𝐮_{𝑘} - 𝑏_{1} 𝐰_{1} - \dots - 𝑏_{𝑙} 𝐰_{𝑙} .

Every value of $𝐿$ is a vector of $𝑈$ plus a vector of $𝑊$ , and every $𝐮 + 𝐰$ with $𝐮 = \sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}$ and $𝐰 = \sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ equals $𝐿 (𝐚, - 𝐛)$ . So $im 𝐿 = 𝑈 + 𝑊$ . By rank–nullity (Theorem 4.2.11(b)), $𝑈 + 𝑊$ is finite-dimensional and

𝑘 + 𝑙 = dim ker 𝐿 + dim (𝑈 + 𝑊) .

Define $𝜑 : ker 𝐿 \to 𝑉$ by $𝜑 (𝐚, 𝐛) =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}$ ; it is linear. For $(𝐚, 𝐛) \in ker 𝐿$ we have $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} =$ $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ , so $𝜑 (𝐚, 𝐛)$ lies in $𝑈$ and in $𝑊$ . Conversely, every $𝐯 \in 𝑈 \cap 𝑊$ can be written as $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} =$ $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ , and then $(𝐚, 𝐛) \in ker 𝐿$ with $𝜑 (𝐚, 𝐛) = 𝐯$ . So $im 𝜑 = 𝑈 \cap 𝑊$ . Finally, $𝜑$ is injective: if $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} = 𝟎$ , then $𝐚 = 𝟎$ , so $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗} = 𝟎$ and $𝐛 = 𝟎$ (Theorem 1.5.5). By Proposition 4.2.6(a), applied to $𝜑$ on the finite-dimensional space $ker 𝐿$ , $dim (𝑈 \cap 𝑊) =$ $dim ker 𝐿$ . Substituting gives $𝑘 +$ $𝑙 =$ $dim (𝑈 \cap 𝑊) +$ $dim (𝑈 + 𝑊)$ , as desired. □

The three proofs teach different things. The first builds an explicit basis of $𝑈 + 𝑊$ and is the one to adapt when a basis is wanted. The second shows that the formula is the direct case, $dim (𝑋 \oplus 𝑊) =$ $dim 𝑋 +$ $dim 𝑊$ , in disguise. The third is how one computes: for $𝑈, 𝑊 \subseteq 𝐹^{𝑛}$ , $ker 𝐿$ is the null space of the matrix whose columns are the $𝐮_{𝑖}$ and the $- 𝐰_{𝑗}$ , so one row reduction finds $dim (𝑈 + 𝑊)$ and a basis of $𝑈 \cap 𝑊$ at once. With the quotient spaces of Section 4.6, the formula can also be read as $(𝑈 + 𝑊) / 𝑊 ≅$ $𝑈 / (𝑈 \cap 𝑊)$ .

Example 4.4.20 (Computing a sum and an intersection). In $ℚ^{4}$ , let $𝑈 = span (𝐮_{1}, 𝐮_{2})$ and $𝑊 = span (𝐰_{1}, 𝐰_{2})$ , where

𝐮_{1} = (1, 0, 1, 0), 𝐮_{2} = (0, 1, 1, 1), 𝐰_{1} = (2, 0, - 1, 1), 𝐰_{2} = (1, 1, - 1, 2) .

Row reduce the matrix whose columns are $𝐮_{1}, 𝐮_{2}, 𝐰_{1}, 𝐰_{2}$ :

(\begin{matrix} 1 & 0 & 2 & 1 \\ 0 & 1 & 0 & 1 \\ 1 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & 2 \end{matrix}) ⇝ (\begin{matrix} 1 & 0 & 0 & - 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}) .

The first two columns are pivot columns, so $𝐮_{1}, 𝐮_{2}$ is independent and $dim 𝑈 = 2$ . The vectors $𝐰_{1}, 𝐰_{2}$ are not parallel (look at the second coordinates), so $dim 𝑊 = 2$ . By Proposition 4.4.2(c), $𝑈 + 𝑊$ is the column space, so $dim (𝑈 + 𝑊) = 3$ , the number of pivots (Theorem 4.2.7). The formula gives

dim (𝑈 \cap 𝑊) = 2 + 2 - 3 = 1 .

To find the line $𝑈 \cap 𝑊$ , read the free column: it says $𝐰_{2} = - 𝐮_{1} + 𝐮_{2} + 𝐰_{1}$ (Corollary 3.3.19(b)). Sorting the terms by subspace,

- 𝐮_{1} + 𝐮_{2} = - 𝐰_{1} + 𝐰_{2} = (- 1, 1, 0, 1) .

The left side lies in $𝑈$ and the right side in $𝑊$ , so this nonzero vector spans $𝑈 \cap 𝑊$ . In the language of the third proof, $(𝐚, 𝐛) =$ $((- 1, 1), (- 1, 1))$ spans $ker 𝐿$ .

Since $𝑈 + 𝑊 \subseteq 𝑉$ , the formula recovers Proposition 4.2.26: $dim (𝑈 \cap 𝑊) =$ $dim 𝑈 +$ $dim 𝑊 -$ $dim (𝑈 + 𝑊) \geq$ $dim 𝑈 +$ $dim 𝑊 -$ $dim 𝑉$ . It also says by how much that bound can miss, namely by $dim 𝑉 - dim (𝑈 + 𝑊)$ . The most frequent use is a test for complements.

Proposition 4.4.21 (Two out of three). Let $𝑉$ be finite-dimensional, and let $𝑈$ and $𝑊$ be subspaces of $𝑉$ . Consider the conditions

(i) $𝑈 \cap 𝑊 = {𝟎}$ ;
(ii) $𝑈 + 𝑊 = 𝑉$ ;
(iii) $dim 𝑈 + dim 𝑊 = dim 𝑉$ .

Any two of them imply the third, and then $𝑉 = 𝑈 \oplus 𝑊$ .

Proof.routine

Uses. Theorem 4.4.19 (Dimension of a sum), Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.4.7 (Criteria for direct sums)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Write the dimension formula as

dim (𝑈 + 𝑊) +

dim (𝑈 \cap 𝑊) =

dim 𝑈 +

dim 𝑊

and substitute each pair of conditions.

Hint 2. To get (ii) from a dimension count, use that a subspace of

𝑉

of dimension

dim 𝑉

𝑉

Sketch. By Theorem 4.4.19,

dim (𝑈 + 𝑊) +

dim (𝑈 \cap 𝑊) =

dim 𝑈 +

dim 𝑊

. (i) and (ii) make the left side

dim 𝑉

. (i) and (iii) give

dim (𝑈 + 𝑊) = dim 𝑉

, so

𝑈 + 𝑊 = 𝑉

. (ii) and (iii) give

dim (𝑈 \cap 𝑊) = 0

. Then

𝑉 = 𝑈 \oplus 𝑊

by Theorem 4.4.7.

By Theorem 4.4.19,

dim (𝑈 + 𝑊) + dim (𝑈 \cap 𝑊) = dim 𝑈 + dim 𝑊 .

If (i) and (ii) hold, the left side is $dim 𝑉 + 0$ , which gives (iii). If (i) and (iii) hold, then $dim (𝑈 + 𝑊) = dim 𝑉$ , so $𝑈 + 𝑊 = 𝑉$ by Proposition 4.1.21(b), which is (ii). If (ii) and (iii) hold, then $dim 𝑉 +$ $dim (𝑈 \cap 𝑊) =$ $dim 𝑉$ , so $dim (𝑈 \cap 𝑊) = 0$ and $𝑈 \cap 𝑊 = {𝟎}$ , which is (i). In each case (i) and (ii) hold, so $𝑉 = 𝑈 \oplus 𝑊$ by Theorem 4.4.7. □

This is the subspace version of the fact that a list of length $dim 𝑉$ is a basis as soon as it is independent or spanning (Theorem 4.1.10, Corollary 4.1.11): with the right dimensions, checking one of “meet trivially” and “add up to $𝑉$ ” gives the other for free.

Remark 4.4.22 (No formula for three subspaces). There is no formula of this kind for three subspaces in terms of the dimensions of their intersections. For the three lines

ℓ_{1}, ℓ_{2}, ℓ_{3}

of Non-example 4.4.6(b), all pairwise intersections and the triple intersection are

{𝟎}

, so the inclusion–exclusion expression

1 + 1 + 1 - 0 - 0 - 0 + 0

gives

3

, while

dim (ℓ_{1} + ℓ_{2} + ℓ_{3}) = 2

. The two-subspace proofs break down because

(𝑈 + 𝑊) \cap 𝑋

can be larger than

(𝑈 \cap 𝑋) + (𝑊 \cap 𝑋)

; Exercise 4.4.39 shows that this is the only obstruction.

Exercise 4.4.23.routine Let $𝑈 =$ ${𝐱 \in 𝐹^{4} | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} + 𝑥_{4} = 0}$ .

(a)Show that the complements of $𝑈$ in $𝐹^{4}$ are exactly the lines $span (𝐯)$ with $𝐯 \notin 𝑈$ .
(b)Is $span ((1, 1, 1, 1))$ a complement of $𝑈$ when $𝐹 = ℚ$ ? When $𝐹 = 𝔽_{2}$ ? When $𝐹 = 𝔽_{3}$ ?

Solution.

Uses. Example 4.2.25 (Kernels of linear functionals), Proposition 4.4.16 (Complements exist), Proposition 4.4.21 (Two out of three), Proposition 1.2.35 (Zero products)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1.

𝑈

is the kernel of a nonzero linear functional. What is

dim 𝑈

, and what must the dimension of a complement be?

Hint 2. For a nonzero

𝐯

, when is

𝑈 \cap span (𝐯) = {𝟎}

? Then apply Proposition 4.4.21.

Sketch.

dim 𝑈 = 3

(Example 4.2.25), so every complement is a line

span (𝐯)

𝐯 \neq 𝟎

(Proposition 4.4.16(b)). If

𝐯 \in 𝑈

, then

span (𝐯) \subseteq 𝑈

; if

𝐯 \notin 𝑈

, then

𝑎 𝐯 \in 𝑈

forces

𝑎 = 0

. So

span (𝐯)

is a complement iff

𝐯 \notin 𝑈

, by (i) and (iii) of Proposition 4.4.21. (b)

(1, 1, 1, 1) \in 𝑈

iff

1 + 1 + 1 + 1 = 0

: no over

ℚ

and

𝔽_{3}

, yes over

𝔽_{2}

(a) $𝑈$ is the kernel of the nonzero linear functional $𝐱 \mapsto$ $𝑥_{1} +$ $𝑥_{2} +$ $𝑥_{3} +$ $𝑥_{4}$ , so $dim 𝑈 = 3$ (Example 4.2.25). By Proposition 4.4.16(b), every complement of $𝑈$ has dimension $4 - 3 = 1$ , so it is $span (𝐯)$ for some $𝐯 \neq 𝟎$ . Conversely, let $𝐯 \neq 𝟎$ . If $𝐯 \in 𝑈$ , then $span (𝐯) \subseteq 𝑈$ meets $𝑈$ in a nonzero vector and is not a complement. If $𝐯 \notin 𝑈$ and $𝑎 𝐯 \in 𝑈$ with $𝑎 \neq 0$ , then $𝐯 = 𝑎^{- 1} (𝑎 𝐯) \in 𝑈$ , a contradiction; so $𝑈 \cap span (𝐯) = {𝟎}$ . Since also $dim 𝑈 +$ $dim span (𝐯) =$ $3 +$ $1 =$ $4$ , Proposition 4.4.21 gives $𝐹^{4} = 𝑈 \oplus span (𝐯)$ . So the complements of $𝑈$ are exactly the lines $span (𝐯)$ with $𝐯 \notin 𝑈$ (and then automatically $𝐯 \neq 𝟎$ ).

(b) By (a), $span ((1, 1, 1, 1))$ is a complement exactly when $(1, 1, 1, 1) \notin 𝑈$ , that is, when $1 + 1 + 1 + 1 \neq 0$ in $𝐹$ . In $ℚ$ , $1 + 1 + 1 + 1 = 4 \neq 0$ , so yes. In $𝔽_{2}$ , $1 + 1 + 1 + 1 = 0$ , so no: there $(1, 1, 1, 1) \in 𝑈$ . In $𝔽_{3}$ , $1 + 1 + 1 + 1 = 1 \neq 0$ , so yes.

4.4.6Projections

A decomposition $𝑉 = 𝑈 \oplus 𝑊$ gives every vector a well-defined $𝑈$ -component. Taking that component is a linear map, and the maps that arise in this way are exactly the operators $𝑃$ with $𝑃^{2} = 𝑃$ .

Definition 4.4.24 (Projection along a complement). Suppose $𝑉 = 𝑈 \oplus 𝑊$ . The projection onto $𝑈$ along $𝑊$ is the map

{Pr}_{𝑈, 𝑊} : 𝑉 \to 𝑉, 𝐮 + 𝐰 \mapsto 𝐮 (𝐮 \in 𝑈, 𝐰 \in 𝑊) .

It is well defined because every vector of $𝑉$ has exactly one decomposition $𝐮 + 𝐰$ . The order of the subscripts matters: ${Pr}_{𝑈, 𝑊}$ keeps the $𝑈$ -component and ${Pr}_{𝑊, 𝑈}$ keeps the $𝑊$ -component. Changing $𝑊$ changes the map, even though its image is still $𝑈$ . The orthogonal projection ${Pr}_{𝑊}$ of Section 6.4 is the special case ${Pr}_{𝑊, 𝑊^{⟂}}$ .

Example 4.4.25 (Projecting along a slanted line). In $𝐹^{2}$ , let $𝑈 = span (𝐞_{1})$ and $𝑊 = span ((1, 1))$ . Every vector decomposes as

(𝑥, 𝑦) = (𝑥 - 𝑦, 0) + (𝑦, 𝑦),

with $(𝑥 - 𝑦, 0) \in 𝑈$ and $(𝑦, 𝑦) \in 𝑊$ . So ${Pr}_{𝑈, 𝑊} (𝑥, 𝑦) =$ $(𝑥 - 𝑦, 0)$ and ${Pr}_{𝑊, 𝑈} (𝑥, 𝑦) =$ $(𝑦, 𝑦)$ , and their matrices with respect to the standard basis are

(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) and (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}) .

Each matrix squares to itself, the two add up to $𝐼_{2}$ , and their products in either order are $0$ . Over $ℝ$ , ${Pr}_{𝑈, 𝑊}$ slides every point parallel to $(1, 1)$ until it reaches the $𝑥$ -axis. It is not the perpendicular projection onto the $𝑥$ -axis, which is $(𝑥, 𝑦) \mapsto (𝑥, 0)$ , the projection onto $𝑈$ along the $𝑦$ -axis.

The projection

{Pr}_{𝑈, 𝑊}

of Example 4.4.25, with

𝑈

the

𝑥

-axis and

𝑊 = span ((1, 1))

. The plane collapses onto

𝑈

, every point moves parallel to

𝑊

, and the points of

𝑊

go to

𝟎

. In the static picture,

𝐯 = (1, 2)

goes to

{Pr}_{𝑈, 𝑊} (𝐯) =

(- 1, 0)

, and the dotted segment from

𝐯

to its image is parallel to

𝑊

. In the interactive version you can change the matrix; the result is a projection exactly when

𝐴^{2} = 𝐴

Proposition 4.4.26 (Properties of projections). Suppose $𝑉 = 𝑈 \oplus 𝑊$ , and let $𝑃 = {Pr}_{𝑈, 𝑊}$ and $𝑄 = {Pr}_{𝑊, 𝑈}$ .

(a) $𝑃$ is linear.
(b) $im 𝑃 = 𝑈$ and $ker 𝑃 = 𝑊$ , and $𝑃 (𝐮) = 𝐮$ for every $𝐮 \in 𝑈$ .
(c) $𝑃^{2} = 𝑃$ .
(d) $𝑃 + 𝑄 = {id}_{𝑉}$ and $𝑃 𝑄 = 𝑄 𝑃 = 0$ .

Proof.routine

Uses. Definition 4.4.24 (Projection along a complement), Definition 4.4.4 (Direct sum), Definition 2.1.1 (Linear map)

Techniques. Compare components (Technique 4.4.8)

Hint 1. For (a), if

𝐯 = 𝐮 + 𝐰

and

𝐯^{'} = 𝐮^{'} + 𝐰^{'}

, write down a decomposition of

𝐯 + 𝐯^{'}

and of

𝑎 𝐯

. Uniqueness does the rest.

Hint 2. For (b), the decomposition of a vector

𝐮 \in 𝑈

𝐮 + 𝟎

. For (d),

𝐯 = 𝑃 (𝐯) + 𝑄 (𝐯)

by definition.

Sketch. (a)

(𝐮 + 𝐮^{'}) + (𝐰 + 𝐰^{'})

and

𝑎 𝐮 + 𝑎 𝐰

are decompositions of

𝐯 + 𝐯^{'}

and

𝑎 𝐯

, hence the decompositions, so

𝑃 (𝐯 + 𝐯^{'}) = 𝐮 + 𝐮^{'}

and

𝑃 (𝑎 𝐯) = 𝑎 𝐮

. (b)

𝐮 = 𝐮 + 𝟎

gives

𝑃 (𝐮) = 𝐮

, so

im 𝑃 = 𝑈

; and

𝑃 (𝐮 + 𝐰) = 𝟎

iff

𝐮 = 𝟎

. (c)

𝑃 (𝐯) \in 𝑈

is fixed. (d)

𝐯 =

𝐮 +

𝐰 =

𝑃 (𝐯) +

𝑄 (𝐯)

, and

𝑃 (𝑄 (𝐯)) =

𝑃 (𝐰) =

𝟎

since

𝐰 \in 𝑊 = ker 𝑃

; symmetrically

𝑄 𝑃 = 0

Throughout, $𝐯 = 𝐮 + 𝐰$ and $𝐯^{'} = 𝐮^{'} + 𝐰^{'}$ with $𝐮, 𝐮^{'} \in 𝑈$ and $𝐰, 𝐰^{'} \in 𝑊$ .

(a) Since $𝑈$ and $𝑊$ are subspaces, $𝐯 +$ $𝐯^{'} =$ $(𝐮 + 𝐮^{'}) +$ $(𝐰 + 𝐰^{'})$ and $𝑎 𝐯 = 𝑎 𝐮 + 𝑎 𝐰$ are decompositions of $𝐯 + 𝐯^{'}$ and $𝑎 𝐯$ along $𝑈 \oplus 𝑊$ , so they are the decompositions. Hence $𝑃 (𝐯 + 𝐯^{'}) =$ $𝐮 +$ $𝐮^{'} =$ $𝑃 (𝐯) +$ $𝑃 (𝐯^{'})$ and $𝑃 (𝑎 𝐯) =$ $𝑎 𝐮 =$ $𝑎 𝑃 (𝐯)$ .

(b) Every value of $𝑃$ lies in $𝑈$ . For $𝐮 \in 𝑈$ , the decomposition of $𝐮$ is $𝐮 + 𝟎$ , so $𝑃 (𝐮) = 𝐮$ ; hence $𝑈 \subseteq im 𝑃$ , and $im 𝑃 = 𝑈$ . Next, $𝑃 (𝐯) = 𝐮$ is $𝟎$ if and only if $𝐯 = 𝐰 \in 𝑊$ , so $ker 𝑃 = 𝑊$ .

(c) For every $𝐯$ , $𝑃 (𝐯) \in 𝑈$ , so $𝑃 (𝑃 (𝐯)) = 𝑃 (𝐯)$ by (b).

(d) By definition $𝑃 (𝐯) +$ $𝑄 (𝐯) =$ $𝐮 +$ $𝐰 =$ $𝐯$ , so $𝑃 + 𝑄 = {id}_{𝑉}$ . Since $𝑄 (𝐯) =$ $𝐰 \in$ $𝑊 =$ $ker 𝑃$ , we get $𝑃 𝑄 (𝐯) = 𝟎$ . Exchanging the roles of $𝑈$ and $𝑊$ (so that $𝑄 = {Pr}_{𝑊, 𝑈}$ has kernel $𝑈$ ) gives $𝑄 𝑃 = 0$ . □

An operator $𝑃$ with $𝑃^{2} = 𝑃$ is called idempotent. The converse of (c) is the main result of this subsection: every idempotent operator is a projection.

Theorem 4.4.27 (Idempotents are projections). Let $𝑃 \in L (𝑉)$ satisfy $𝑃^{2} = 𝑃$ . Then

𝑉 = im 𝑃 \oplus ker 𝑃,

and $𝑃 = {Pr}_{im 𝑃, ker 𝑃}$ . Moreover, $im 𝑃 =$ ${𝐯 \in 𝑉 | 𝑃 (𝐯) = 𝐯}$ .

Together with Proposition 4.4.26, the theorem gives a bijection between the ways of writing $𝑉 = 𝑈 \oplus 𝑊$ (with $𝑈$ and $𝑊$ in this order) and the idempotent operators on $𝑉$ : a decomposition gives ${Pr}_{𝑈, 𝑊}$ , and an idempotent $𝑃$ gives the decomposition $𝑉 = im 𝑃 \oplus ker 𝑃$ .

Proof (by writing down the decomposition).routine

Uses. Theorem 4.4.7 (Criteria for direct sums), Definition 4.4.24 (Projection along a complement), Proposition 2.6.8 (Kernels and images are subspaces)

Techniques. Compare components (Technique 4.4.8)

Hint 1. Every

𝐯

can be written as

𝑃 (𝐯) +

(𝐯 - 𝑃 (𝐯))

. Where do the two pieces lie?

Hint 2. If

𝐯 \in im 𝑃 \cap ker 𝑃

, write

𝐯 = 𝑃 (𝐱)

and apply

𝑃

Sketch.

𝑃 (𝐯 - 𝑃 (𝐯)) =

𝑃 (𝐯) -

𝑃^{2} (𝐯) =

𝟎

, so

𝐯 =

𝑃 (𝐯) +

(𝐯 - 𝑃 (𝐯)) \in

im 𝑃 +

ker 𝑃

. If

𝐯 = 𝑃 (𝐱)

and

𝑃 (𝐯) = 𝟎

, then

𝐯 =

𝑃^{2} (𝐱) =

𝑃 (𝐯) =

𝟎

. So

𝑉 = im 𝑃 \oplus ker 𝑃

, and the

im 𝑃

-component of

𝐯

𝑃 (𝐯)

. Vectors

𝑃 (𝐱)

im 𝑃

are fixed, as

𝑃 (𝑃 (𝐱)) = 𝑃 (𝐱)

Let $𝐯 \in 𝑉$ . Then $𝑃 (𝐯) \in im 𝑃$ , and

𝑃 (𝐯 - 𝑃 (𝐯)) = 𝑃 (𝐯) - 𝑃^{2} (𝐯) = 𝑃 (𝐯) - 𝑃 (𝐯) = 𝟎,

so $𝐯 - 𝑃 (𝐯) \in ker 𝑃$ . Hence

𝐯 = 𝑃 (𝐯) + (𝐯 - 𝑃 (𝐯))

is a decomposition of $𝐯$ along $im 𝑃$ and $ker 𝑃$ , and $𝑉 = im 𝑃 + ker 𝑃$ . Next, let $𝐯 \in im 𝑃 \cap ker 𝑃$ , say $𝐯 = 𝑃 (𝐱)$ . Then $𝐯 =$ $𝑃 (𝐱) =$ $𝑃^{2} (𝐱) =$ $𝑃 (𝐯) =$ $𝟎$ . By Theorem 4.4.7, $𝑉 = im 𝑃 \oplus ker 𝑃$ , and by the displayed decomposition, the $im 𝑃$ -component of $𝐯$ is $𝑃 (𝐯)$ , that is, $𝑃 = {Pr}_{im 𝑃, ker 𝑃}$ .

Finally, if $𝑃 (𝐯) = 𝐯$ , then $𝐯 \in im 𝑃$ ; conversely, if $𝐯 = 𝑃 (𝐱)$ , then $𝑃 (𝐯) =$ $𝑃^{2} (𝐱) =$ $𝑃 (𝐱) =$ $𝐯$ . □

In finite dimensions there is a second proof, which never writes down the decomposition $𝐯 =$ $𝑃 (𝐯) +$ $(𝐯 - 𝑃 (𝐯))$ and lets a dimension count produce the sum instead.

Proof (in finite dimensions, by counting).routine

Uses. Theorem 4.2.11 (Rank–nullity), Proposition 4.4.21 (Two out of three), Definition 4.4.24 (Projection along a complement)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Show

im 𝑃 \cap ker 𝑃 = {𝟎}

as in the first proof.

Hint 2. Rank–nullity gives

dim im 𝑃 +

dim ker 𝑃 =

dim 𝑉

. Which result turns this and a trivial intersection into a direct sum?

Sketch. If

𝐯 = 𝑃 (𝐱)

and

𝑃 (𝐯) = 𝟎

, then

𝐯 = 𝑃^{2} (𝐱) = 𝟎

. Rank–nullity gives

dim im 𝑃 +

dim ker 𝑃 =

dim 𝑉

, so

𝑉 = im 𝑃 \oplus ker 𝑃

by Proposition 4.4.21. For

𝐯 = 𝑃 (𝐱) + 𝐰

with

𝐰 \in ker 𝑃

𝑃 (𝐯) =

𝑃^{2} (𝐱) =

𝑃 (𝐱)

, the

im 𝑃

-component.

Assume that

𝑉

is finite-dimensional. As in the first proof, if

𝐯 = 𝑃 (𝐱)

lies in

ker 𝑃

, then

𝐯 =

𝑃^{2} (𝐱) =

𝑃 (𝐯) =

𝟎

, so

im 𝑃 \cap ker 𝑃 = {𝟎}

. By Theorem 4.2.11,

dim im 𝑃 +

dim ker 𝑃 =

dim 𝑉

. Conditions (i) and (iii) of Proposition 4.4.21 hold, so

𝑉 = im 𝑃 \oplus ker 𝑃

. To identify

𝑃

, let

𝐯 = 𝑃 (𝐱) + 𝐰

with

𝐰 \in ker 𝑃

. Then

𝑃 (𝐯) =

𝑃^{2} (𝐱) +

𝑃 (𝐰) =

𝑃 (𝐱)

, which is the

im 𝑃

-component of

𝐯

. So

𝑃 = {Pr}_{im 𝑃, ker 𝑃}

. □

The counting proof does not reach infinite-dimensional spaces, and even in finite dimensions the first proof is more informative: it says what the components of $𝐯$ are.

Technique 4.4.28 (Split the space with an idempotent). To decompose $𝑉$ as a direct sum, look for an operator $𝑃$ with $𝑃^{2} = 𝑃$ . Then $𝑉 = im 𝑃 \oplus ker 𝑃$ , every vector splits as $𝐯 =$ $𝑃 (𝐯) +$ $(𝐯 - 𝑃 (𝐯))$ , and $im 𝑃$ is the set of vectors that $𝑃$ fixes (Theorem 4.4.27). In the other direction, every decomposition $𝑉 = 𝑈 \oplus 𝑊$ comes with its projections, and statements about decompositions become algebra of operators: $𝑃^{2} = 𝑃$ , $𝑃 + 𝑄 = {id}_{𝑉}$ , $𝑃 𝑄 = 𝑄 𝑃 = 0$ .

Reach for it when a formula suggests “the part of $𝐯$ that …”, such as the symmetric part $(𝐴 + 𝐴^{𝖳}) / 2$ of a matrix, or more generally the average of $𝐯$ and $𝑇 (𝐯)$ for an operator with $𝑇^{2} = {id}_{𝑉}$ (Exercise 4.4.43). Each is $𝑃 (𝐯)$ for an idempotent $𝑃$ . The canonical example is Theorem 4.4.27. Later, idempotent matrices over $ℚ$ , $ℝ$ , $ℂ$ have trace equal to rank (Exercise 4.3.43), and orthogonal projections are the idempotents that are self-adjoint (Section 6.6).

Used in 4 places and 1 practice exercise

Exercise 4.4.30, Exercise 4.4.43, Exercise 4.4.51, Problem 4.4; 1 practice exercise.

A direct sum of $𝑘$ subspaces comes with $𝑘$ projections, one for each summand. They are the algebraic form of the decomposition, and they are how Chapters 7 and 8 talk about splitting $𝑉$ into eigenspaces.

Proposition 4.4.29 (Projections of a direct sum).

(a)Suppose $𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘}$ , and let $𝑃_{𝑖} : 𝑉 \to 𝑉$ send each vector to its component in $𝑈_{𝑖}$ . Then each $𝑃_{𝑖}$ is linear with $im 𝑃_{𝑖} = 𝑈_{𝑖}$ , and
$𝑃_{1} + \dots + 𝑃_{𝑘} = {id}_{𝑉}, 𝑃_{𝑖}^{2} = 𝑃_{𝑖}, 𝑃_{𝑖} 𝑃_{𝑗} = 0 (𝑖 \neq 𝑗) .$
(b)Conversely, let $𝑃_{1}, \dots, 𝑃_{𝑘} \in L (𝑉)$ satisfy $𝑃_{1} + \dots + 𝑃_{𝑘} = {id}_{𝑉}$ and $𝑃_{𝑖} 𝑃_{𝑗} = 0$ for $𝑖 \neq 𝑗$ . Then each $𝑃_{𝑖}$ is idempotent, $𝑉 = im 𝑃_{1} \oplus \dots \oplus im 𝑃_{𝑘}$ , and $𝑃_{𝑖}$ sends each vector to its component in $im 𝑃_{𝑖}$ .

Proof.standard

Uses. Theorem 4.4.7 (Criteria for direct sums), Proposition 4.4.26 (Properties of projections), Definition 4.4.24 (Projection along a complement), Proposition 4.4.2 (Sums are the smallest containing subspaces)

Techniques. Apply an operator to a relation to kill terms (Technique 4.1.25), Compare components (Technique 4.4.8)

Hint 1. For (a), let

𝑊_{𝑖}

be the sum of the summands other than

𝑈_{𝑖}

. Show

𝑉 = 𝑈_{𝑖} \oplus 𝑊_{𝑖}

and

𝑃_{𝑖} = {Pr}_{𝑈_{𝑖}, 𝑊_{𝑖}}

Hint 2. For (b), multiply

{id}_{𝑉} = \sum_{𝑗} 𝑃_{𝑗}

on the left by

𝑃_{𝑖}

to get idempotence.

Hint 3. For directness in (b), take

𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎

with

𝐮_{𝑖} = 𝑃_{𝑖} (𝐱_{𝑖})

and apply

𝑃_{𝑗}

: every term but one is killed.

Sketch. (a) With

𝑊_{𝑖} = \sum_{𝑗 \neq 𝑖} 𝑈_{𝑗}

, criterion (III) gives

𝑉 = 𝑈_{𝑖} \oplus 𝑊_{𝑖}

, and

𝑃_{𝑖} = {Pr}_{𝑈_{𝑖}, 𝑊_{𝑖}}

, which is linear and idempotent with image

𝑈_{𝑖}

and kernel

𝑊_{𝑖}

. For

𝑗 \neq 𝑖

𝑃_{𝑗} (𝐯) \in

𝑈_{𝑗} \subseteq

𝑊_{𝑖}

, so

𝑃_{𝑖} 𝑃_{𝑗} = 0

; and

𝐯 = \sum_{𝑖} 𝑃_{𝑖} (𝐯)

. (b)

𝑃_{𝑖} =

𝑃_{𝑖} \sum_{𝑗} 𝑃_{𝑗} =

𝑃_{𝑖}^{2}

. Since

𝐯 = \sum_{𝑖} 𝑃_{𝑖} (𝐯)

, the images add up to

𝑉

. If

\sum_{𝑖} 𝑃_{𝑖} (𝐱_{𝑖}) = 𝟎

, applying

𝑃_{𝑗}

leaves

𝑃_{𝑗}^{2} (𝐱_{𝑗}) =

𝑃_{𝑗} (𝐱_{𝑗}) =

𝟎

, so the sum is direct, and

𝐯 = \sum_{𝑖} 𝑃_{𝑖} (𝐯)

is the decomposition of

𝐯

(a) Fix $𝑖$ and let $𝑊_{𝑖} = \sum_{𝑗 \neq 𝑖} 𝑈_{𝑗}$ . Every $𝐯 = 𝐮_{1} + \dots + 𝐮_{𝑘}$ (with $𝐮_{𝑗} \in 𝑈_{𝑗}$ ) is $𝐮_{𝑖} +$ $\sum_{𝑗 \neq 𝑖} 𝐮_{𝑗} \in$ $𝑈_{𝑖} +$ $𝑊_{𝑖}$ , and $𝑈_{𝑖} \cap 𝑊_{𝑖} = {𝟎}$ by (III) of Theorem 4.4.7. So $𝑉 = 𝑈_{𝑖} \oplus 𝑊_{𝑖}$ , and the $𝑈_{𝑖}$ -component of $𝐯$ in this decomposition is again $𝐮_{𝑖}$ . That is, $𝑃_{𝑖} = {Pr}_{𝑈_{𝑖}, 𝑊_{𝑖}}$ , which is linear and idempotent with image $𝑈_{𝑖}$ and kernel $𝑊_{𝑖}$ by Proposition 4.4.26. For $𝑗 \neq 𝑖$ and $𝐯 \in 𝑉$ , $𝑃_{𝑗} (𝐯) \in$ $𝑈_{𝑗} \subseteq$ $𝑊_{𝑖} =$ $ker 𝑃_{𝑖}$ , so $𝑃_{𝑖} 𝑃_{𝑗} = 0$ . Finally, $𝐯 =$ $𝐮_{1} +$ $\dots +$ $𝐮_{𝑘} =$ $𝑃_{1} (𝐯) +$ $\dots +$ $𝑃_{𝑘} (𝐯)$ , so $𝑃_{1} + \dots + 𝑃_{𝑘} = {id}_{𝑉}$ .

(b) For each $𝑖$ ,

𝑃_{𝑖} = 𝑃_{𝑖} {id}_{𝑉} = 𝑃_{𝑖} (𝑃_{1} + \dots + 𝑃_{𝑘}) = 𝑃_{𝑖} 𝑃_{1} + \dots + 𝑃_{𝑖} 𝑃_{𝑘} = 𝑃_{𝑖}^{2},

since every term $𝑃_{𝑖} 𝑃_{𝑗}$ with $𝑗 \neq 𝑖$ is $0$ . For $𝐯 \in 𝑉$ , $𝐯 =$ $𝑃_{1} (𝐯) +$ $\dots +$ $𝑃_{𝑘} (𝐯)$ , so $𝑉 = im 𝑃_{1} + \dots + im 𝑃_{𝑘}$ . For directness we check (II) of Theorem 4.4.7. Let $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ with $𝐮_{𝑖} =$ $𝑃_{𝑖} (𝐱_{𝑖}) \in$ $im 𝑃_{𝑖}$ , and fix $𝑗$ . Applying $𝑃_{𝑗}$ kills every term with $𝑖 \neq 𝑗$ , because $𝑃_{𝑗} (𝐮_{𝑖}) =$ $𝑃_{𝑗} 𝑃_{𝑖} (𝐱_{𝑖}) =$ $𝟎$ , and leaves $𝑃_{𝑗} (𝐮_{𝑗}) =$ $𝑃_{𝑗}^{2} (𝐱_{𝑗}) =$ $𝑃_{𝑗} (𝐱_{𝑗}) =$ $𝐮_{𝑗}$ . So $𝐮_{𝑗} = 𝑃_{𝑗} (𝟎) = 𝟎$ . Hence $𝑉 = im 𝑃_{1} \oplus \dots \oplus im 𝑃_{𝑘}$ , and since $𝐯 = \sum_{𝑖} 𝑃_{𝑖} (𝐯)$ with $𝑃_{𝑖} (𝐯) \in im 𝑃_{𝑖}$ , the component of $𝐯$ in $im 𝑃_{𝑖}$ is $𝑃_{𝑖} (𝐯)$ . □

Exercise 4.4.30.routine In $ℚ^{3}$ , let $𝑈 =$ ${𝐱 | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0}$ and $𝑊 = span ((1, 1, 1))$ .

(a)Show that $ℚ^{3} = 𝑈 \oplus 𝑊$ , and find the matrices of ${Pr}_{𝑈, 𝑊}$ and ${Pr}_{𝑊, 𝑈}$ with respect to the standard basis.
(b)Show that over $𝔽_{3}$ , $𝑈 + 𝑊 \neq 𝔽_{3}^{3}$ . Which step of (a) fails?

Solution.

Uses. Definition 4.4.24 (Projection along a complement), Theorem 4.4.7 (Criteria for direct sums), Example 4.2.25 (Kernels of linear functionals)

Techniques. Split the space with an idempotent (Technique 4.4.28)

Hint 1. Subtract from

𝐱

the multiple of

(1, 1, 1)

that makes the coordinates sum to

0

Hint 2. Over

𝔽_{3}

, is

(1, 1, 1)

𝑈

Sketch. With

𝑚 =

(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) / 3

𝐱 =

(𝐱 - 𝑚 (1, 1, 1)) +

𝑚 (1, 1, 1)

is a decomposition, and

𝑡 (1, 1, 1) \in 𝑈

forces

3 𝑡 = 0

. So

{Pr}_{𝑊, 𝑈}

has all entries

1 / 3

and

{Pr}_{𝑈, 𝑊} =

𝐼_{3} -

{Pr}_{𝑊, 𝑈}

. Over

𝔽_{3}

1 + 1 + 1 = 0

puts

𝑊

inside

𝑈

; the division by

3

is impossible.

(a) For $𝐱 \in ℚ^{3}$ , let $𝑚 =$ $(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) / 3$ . The coordinates of $𝐱 - 𝑚 (1, 1, 1)$ sum to $(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) -$ $3 𝑚 =$ $0$ , so

𝐱 = (𝐱 - 𝑚 (1, 1, 1)) + 𝑚 (1, 1, 1)

is a decomposition of $𝐱$ along $𝑈$ and $𝑊$ . If $𝑡 (1, 1, 1) \in 𝑈$ , then $3 𝑡 = 0$ and $𝑡 = 0$ , so $𝑈 \cap 𝑊 = {𝟎}$ . By Theorem 4.4.7, $ℚ^{3} = 𝑈 \oplus 𝑊$ , with ${Pr}_{𝑊, 𝑈} (𝐱) =$ $𝑚 (1, 1, 1)$ and ${Pr}_{𝑈, 𝑊} (𝐱) =$ $𝐱 -$ $𝑚 (1, 1, 1)$ . Their matrices are

{Pr}_{𝑊, 𝑈} : \frac{1}{3} (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}), {Pr}_{𝑈, 𝑊} : \frac{1}{3} (\begin{matrix} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{matrix}) .

(b) In $𝔽_{3}$ , $1 + 1 + 1 = 0$ , so $(1, 1, 1) \in 𝑈$ and $𝑊 \subseteq 𝑈$ . Then $𝑈 + 𝑊 = 𝑈$ , which has dimension $2$ (Example 4.2.25), so $𝑈 + 𝑊 \neq 𝔽_{3}^{3}$ . The step that fails is the division by $3$ in the definition of $𝑚$ , since $3 = 0$ in $𝔽_{3}$ . Correspondingly, the uniqueness argument fails too: $3 𝑡 = 0$ holds for every $𝑡 \in 𝔽_{3}$ .

4.4.7Linear maps on direct sums

A linear map is determined by its values on a basis, and those values can be prescribed freely (Theorem 2.2.1). A direct sum decomposition gives the same freedom, one summand at a time.

Proposition 4.4.31 (Defining maps summand by summand). Suppose $𝑉 = 𝑈_{1} \oplus \dots \oplus 𝑈_{𝑘}$ , and let $𝑆_{𝑖} \in L (𝑈_{𝑖}, 𝑊)$ for each $𝑖$ . There is exactly one $𝑇 \in L (𝑉, 𝑊)$ with $𝑇 (𝐮) = 𝑆_{𝑖} (𝐮)$ for every $𝑖$ and every $𝐮 \in 𝑈_{𝑖}$ , namely

𝑇 (𝐮_{1} + \dots + 𝐮_{𝑘}) = 𝑆_{1} (𝐮_{1}) + \dots + 𝑆_{𝑘} (𝐮_{𝑘}) (𝐮_{𝑖} \in 𝑈_{𝑖}) .

Proof.routine

Uses. Definition 4.4.4 (Direct sum), Proposition 4.4.29 (Projections of a direct sum), Proposition 2.1.17 (Compositions of linear maps are linear), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Compare components (Technique 4.4.8)

Hint 1. Uniqueness: a linear

𝑇

that agrees with each

𝑆_{𝑖}

𝑈_{𝑖}

must satisfy the displayed formula.

Hint 2. Existence: the formula is well defined because decompositions are unique. For linearity, write

𝑇

in terms of the projections of Proposition 4.4.29.

Sketch. If

𝑇

is linear and agrees with

𝑆_{𝑖}

𝑈_{𝑖}

, then

𝑇 (\sum_{𝑖} 𝐮_{𝑖}) =

\sum_{𝑖} 𝑇 (𝐮_{𝑖}) =

\sum_{𝑖} 𝑆_{𝑖} (𝐮_{𝑖})

, so

𝑇

is unique. The formula defines a function because each vector has one decomposition, and it equals

\sum_{𝑖} 𝑆_{𝑖} \circ 𝑃_{𝑖}^{'}

, where

𝑃_{𝑖}^{'} : 𝑉 \to 𝑈_{𝑖}

takes the

𝑈_{𝑖}

-component; this is linear. On

𝐮 \in 𝑈_{𝑖}

the decomposition is

𝐮

in slot

𝑖

and

𝟎

elsewhere, so

𝑇 (𝐮) = 𝑆_{𝑖} (𝐮)

Uniqueness. If $𝑇 \in L (𝑉, 𝑊)$ agrees with $𝑆_{𝑖}$ on $𝑈_{𝑖}$ for each $𝑖$ , then for $𝐮_{𝑖} \in 𝑈_{𝑖}$ ,

𝑇 (𝐮_{1} + \dots + 𝐮_{𝑘}) = 𝑇 (𝐮_{1}) + \dots + 𝑇 (𝐮_{𝑘}) = 𝑆_{1} (𝐮_{1}) + \dots + 𝑆_{𝑘} (𝐮_{𝑘}) .

Every vector of $𝑉$ is of this form, so $𝑇$ is determined.

Existence. Since every $𝐯 \in 𝑉$ has exactly one decomposition, the displayed formula defines a function $𝑇 : 𝑉 \to 𝑊$ . Let $𝑃_{𝑖}^{'} : 𝑉 \to 𝑈_{𝑖}$ send $𝐯$ to its component in $𝑈_{𝑖}$ ; it is the projection $𝑃_{𝑖}$ of Proposition 4.4.29(a) with its target shrunk to $𝑈_{𝑖}$ , so it is linear. Then $𝑇 =$ $𝑆_{1} \circ$ $𝑃_{1}^{'} +$ $\dots +$ $𝑆_{𝑘} \circ$ $𝑃_{𝑘}^{'}$ , which is linear by Proposition 2.1.17 and Proposition 2.1.23. Finally, for $𝐮 \in 𝑈_{𝑖}$ , the decomposition of $𝐮$ has component $𝐮$ in $𝑈_{𝑖}$ and $𝟎$ in every other summand, so $𝑇 (𝐮) =$ $𝑆_{𝑖} (𝐮) +$ $\sum_{𝑗 \neq 𝑖} 𝑆_{𝑗} (𝟎) =$ $𝑆_{𝑖} (𝐮)$ . □

When $𝐯_{1}, \dots, 𝐯_{𝑛}$ is a basis and $𝑈_{𝑖} = span (𝐯_{𝑖})$ , a linear map $𝑆_{𝑖} : 𝑈_{𝑖} \to 𝑊$ is determined by $𝑆_{𝑖} (𝐯_{𝑖})$ , which can be any vector of $𝑊$ , and the proposition becomes Theorem 2.2.1.

Remark 4.4.32 (Block diagonal matrices). Suppose $𝑉 = 𝑈 \oplus 𝑊$ is finite-dimensional, and $𝑇 \in L (𝑉)$ maps $𝑈$ into $𝑈$ and $𝑊$ into $𝑊$ . Let $𝑇_{𝑈} \in L (𝑈)$ and $𝑇_{𝑊} \in L (𝑊)$ be the operators $𝐮 \mapsto 𝑇 (𝐮)$ and $𝐰 \mapsto 𝑇 (𝐰)$ . Glue a basis ${ℬ︀}_{𝑈}$ of $𝑈$ and a basis ${ℬ︀}_{𝑊}$ of $𝑊$ into a basis $ℬ︀$ of $𝑉$ (Technique 4.4.14). The image of a vector of ${ℬ︀}_{𝑈}$ is a combination of ${ℬ︀}_{𝑈}$ alone, and similarly for ${ℬ︀}_{𝑊}$ , so the matrix of $𝑇$ is block diagonal (Definition 2.4.33):

{[𝑇]}_{ℬ︀ ℬ︀} = (\begin{matrix} {[𝑇_{𝑈}]}_{{ℬ︀}_{𝑈} {ℬ︀}_{𝑈}} & 0 \\ 0 & {[𝑇_{𝑊}]}_{{ℬ︀}_{𝑊} {ℬ︀}_{𝑊}} \end{matrix}) .

By Proposition 4.4.31, $𝑇$ is determined by $𝑇_{𝑈}$ and $𝑇_{𝑊}$ , so $𝑇$ can be studied through the two smaller operators. This is the strategy of the eigenvalue chapters. An operator is diagonalizable when $𝑉$ is a direct sum of subspaces on each of which $𝑇$ acts as multiplication by a scalar (Section 7.4), and the Jordan form refines such a splitting (Section 10.3). Exercise 4.4.44 computes the kernel and image of such a $𝑇$ .

Exercises

Exercise 4.4.33.routine In $𝐹^{4}$ , let $𝑈 =$ $span ((1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1))$ and

𝑊 = {𝐱 \in 𝐹^{4} | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} + 𝑥_{4} = 0 and 𝑥_{1} = 𝑥_{3}} .

(a)For $𝐹 = ℚ$ , find $dim 𝑈$ , $dim 𝑊$ , a basis of $𝑈 \cap 𝑊$ , and $dim (𝑈 + 𝑊)$ .
(b)Repeat (a) for $𝐹 = 𝔽_{2}$ .

Solution.

Uses. Theorem 4.4.19 (Dimension of a sum), Theorem 4.2.11 (Rank–nullity), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Theorem 1.5.5 (Independence is decided at zero)

Hint 1. Write a general vector of

𝑈

(𝑎, 𝑎 + 𝑏, 𝑏 + 𝑐, 𝑐)

and substitute it into the two equations that define

𝑊

Hint 2. Over

ℚ

the first equation becomes

2 (𝑎 + 𝑏 + 𝑐) = 0

. What does it become over

𝔽_{2}

Sketch. Over every field, the spanning vectors of

𝑈

are independent, so

dim 𝑈 = 3

, and

𝑊

is cut out by two independent equations, so

dim 𝑊 = 2

. A vector

(𝑎, 𝑎 + 𝑏, 𝑏 + 𝑐, 𝑐)

𝑈

lies in

𝑊

iff

2 (𝑎 + 𝑏 + 𝑐) = 0

and

𝑎 = 𝑏 + 𝑐

. Over

ℚ

this forces

𝑎 = 0

and

𝑐 = - 𝑏

𝑈 \cap

𝑊 =

span ((0, 1, 0, - 1))

, and

dim (𝑈 + 𝑊) = 4

. Over

𝔽_{2}

the first condition is empty,

dim (𝑈 \cap 𝑊) = 2

, so

𝑊 \subseteq 𝑈

and

dim (𝑈 + 𝑊) = 3

(a) If $𝑎 (1, 1, 0, 0) +$ $𝑏 (0, 1, 1, 0) +$ $𝑐 (0, 0, 1, 1) =$ $(𝑎, 𝑎 + 𝑏, 𝑏 + 𝑐, 𝑐) =$ $𝟎$ , then $𝑎 = 𝑐 = 0$ from the first and last coordinates, and then $𝑏 = 0$ . So the spanning list of $𝑈$ is independent and $dim 𝑈 = 3$ ; this works over every field. The subspace $𝑊$ is the kernel of the linear map $𝐹^{4} \to 𝐹^{2}$ with matrix

(\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 0 & - 1 & 0 \end{matrix}) .

Its first two columns $(1, 1)$ and $(1, 0)$ are independent, so it has rank $2$ (Theorem 4.2.7), and $dim 𝑊 = 4 - 2 = 2$ by Theorem 4.2.11.

A general vector of $𝑈$ is $(𝑎, 𝑎 + 𝑏, 𝑏 + 𝑐, 𝑐)$ . It lies in $𝑊$ if and only if

2 𝑎 + 2 𝑏 + 2 𝑐 = 0 and 𝑎 = 𝑏 + 𝑐 .

Over $ℚ$ , the first condition says $𝑎 + 𝑏 + 𝑐 = 0$ , and together with $𝑎 = 𝑏 + 𝑐$ it gives $2 𝑎 = 0$ , so $𝑎 = 0$ and $𝑐 = - 𝑏$ . Hence $𝑈 \cap$ $𝑊 =$ ${(0, 𝑏, 0, - 𝑏) | 𝑏 \in ℚ} =$ $span ((0, 1, 0, - 1))$ , of dimension $1$ . By Theorem 4.4.19, $dim (𝑈 + 𝑊) =$ $3 +$ $2 -$ $1 =$ $4$ , so $𝑈 + 𝑊 = ℚ^{4}$ .

(b) Over $𝔽_{2}$ the same arguments give $dim 𝑈 = 3$ and $dim 𝑊 = 2$ (the matrix is now $(\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 \end{matrix})$ , with the same first two columns). But $2 = 0$ in $𝔽_{2}$ , so the first condition holds for every vector of $𝑈$ , and $𝑈 \cap 𝑊$ consists of the vectors with $𝑎 = 𝑏 + 𝑐$ :

(𝑏 + 𝑐, 𝑐, 𝑏 + 𝑐, 𝑐) = 𝑏 (1, 0, 1, 0) + 𝑐 (1, 1, 1, 1), 𝑏, 𝑐 \in 𝔽_{2} .

So $dim (𝑈 \cap 𝑊) =$ $2 =$ $dim 𝑊$ , which means $𝑊 \subseteq 𝑈$ , and $dim (𝑈 + 𝑊) =$ $3 +$ $2 -$ $2 =$ $3$ , so $𝑈 + 𝑊 = 𝑈$ . This is visible directly: over $𝔽_{2}$ , each spanning vector of $𝑈$ has an even number of entries $1$ , so $𝑈$ is the $3$ -dimensional subspace $𝑥_{1} + 𝑥_{2} + 𝑥_{3} + 𝑥_{4} = 0$ , and $𝑊$ lies in it by definition. (Over $ℚ$ , $𝑈$ is the subspace $𝑥_{1} - 𝑥_{2} + 𝑥_{3} - 𝑥_{4} = 0$ .)

Exercise 4.4.34.routine Let $𝑈$ , $𝑊$ , $𝑋$ be subspaces of a finite-dimensional space $𝑉$ . Prove or disprove:

(a)If $𝑈 \cap$ $𝑊 =$ $𝑈 \cap$ $𝑋 =$ $𝑊 \cap$ $𝑋 =$ ${𝟎}$ , then $𝑈 + 𝑊 + 𝑋$ is direct.
(b)If $𝑈 + 𝑊 + 𝑋$ is direct, then $𝑈 + 𝑊$ is direct.
(c)If $𝑉 = 𝑈 \oplus 𝑊$ and $𝑉 = 𝑈 \oplus 𝑋$ , then $𝑊 = 𝑋$ .
(d)If $𝑉 = 𝑈 \oplus 𝑊$ and $𝑉 = 𝑈 \oplus 𝑋$ , then $dim 𝑊 = dim 𝑋$ .
(e)If $𝑉 = 𝑈 \oplus 𝑊$ , then $𝑋 =$ $(𝑋 \cap 𝑈) \oplus$ $(𝑋 \cap 𝑊)$ .
(f)If $𝑈 + 𝑊 = 𝑉$ and $dim 𝑈 + dim 𝑊 = dim 𝑉$ , then $𝑉 = 𝑈 \oplus 𝑊$ .

Solution.

Uses. Non-example 4.4.6 (Sums that are not direct), Proposition 4.4.9 (Grouping summands), Example 4.4.17 (Complements are not unique), Proposition 4.4.16 (Complements exist), Proposition 4.4.21 (Two out of three)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Dimension counting (Technique 4.2.24)

Hint 1. Three of the statements are false, and lines through

𝟎

ℝ^{2}

give counterexamples to all three.

Hint 2. For the true ones, use Proposition 4.4.9, Proposition 4.4.16(b), and Proposition 4.4.21.

Sketch. (a) False: three lines in

ℝ^{2}

. (b) True: a selection of independent subspaces is independent. (c) False: the

𝑦

-axis and the line

𝑦 = 𝑥

are both complements of the

𝑥

-axis. (d) True: both dimensions are

dim 𝑉 - dim 𝑈

. (e) False:

𝑈

𝑊

the axes of

ℝ^{2}

and

𝑋

the line

𝑦 = 𝑥

. (f) True: two out of three.

(a) False. In $ℝ^{2}$ , the lines $span (𝐞_{1})$ , $span (𝐞_{2})$ , $span (𝐞_{1} + 𝐞_{2})$ meet pairwise in ${𝟎}$ , but their sum is not direct (Non-example 4.4.6(b)).

(b) True, by the last part of Proposition 4.4.9: a selection of independent subspaces is independent.

(c) False. In $ℝ^{2}$ , with $𝑈 = span (𝐞_{1})$ , both $𝑊 = span (𝐞_{2})$ and $𝑋 = span ((1, 1))$ are complements of $𝑈$ (Example 4.4.17), and $𝑊 \neq 𝑋$ .

(d) True. By Proposition 4.4.16(b), $dim 𝑊 =$ $dim 𝑉 -$ $dim 𝑈 =$ $dim 𝑋$ .

(e) False. In $ℝ^{2}$ , let $𝑈 = span (𝐞_{1})$ , $𝑊 = span (𝐞_{2})$ , and $𝑋 = span ((1, 1))$ . Then $ℝ^{2} = 𝑈 \oplus 𝑊$ , but $𝑋 \cap 𝑈 = 𝑋 \cap 𝑊 = {𝟎}$ , so $(𝑋 \cap 𝑈) \oplus$ $(𝑋 \cap 𝑊) =$ ${𝟎} \neq$ $𝑋$ . The components of a vector of $𝑋$ lie in $𝑈$ and $𝑊$ , but not in $𝑋$ .

(f) True. Conditions (ii) and (iii) of Proposition 4.4.21 hold, so $𝑉 = 𝑈 \oplus 𝑊$ .

Exercise 4.4.35.routine Let $𝑇 \in L (𝑉, 𝑊)$ , and let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ .

(a)Show that $im (𝑇 ↾_{𝑈_{1} + \dots + 𝑈_{𝑘}}) =$ $im (𝑇 ↾_{𝑈_{1}}) +$ $\dots +$ $im (𝑇 ↾_{𝑈_{𝑘}})$ .
(b)Show that if $𝑇$ is injective and $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent, then $im (𝑇 ↾_{𝑈_{1}}),$ $\dots,$ $im (𝑇 ↾_{𝑈_{𝑘}})$ are independent.
(c)Show by an example that (b) can fail if $𝑇$ is not injective.

Solution.

Uses. Definition 4.2.1 (Restriction), Theorem 4.4.7 (Criteria for direct sums), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Test injectivity at $𝟎$ (Technique 2.6.12)

Hint 1. For (a),

𝑇 (𝐮_{1} + \dots + 𝐮_{𝑘}) =

𝑇 (𝐮_{1}) +

\dots +

𝑇 (𝐮_{𝑘})

Hint 2. For (b), use criterion (II): pull a decomposition of

𝟎

back through

𝑇

Sketch. (a) Both sides consist of the vectors

𝑇 (𝐮_{1}) +

\dots +

𝑇 (𝐮_{𝑘})

with

𝐮_{𝑖} \in 𝑈_{𝑖}

. (b) If

𝑇 (𝐮_{1}) +

\dots +

𝑇 (𝐮_{𝑘}) =

𝟎

, then

𝑇 (\sum_{𝑖} 𝐮_{𝑖}) = 𝟎

, so

\sum_{𝑖} 𝐮_{𝑖} = 𝟎

, so every

𝐮_{𝑖} = 𝟎

and every

𝑇 (𝐮_{𝑖}) = 𝟎

. (c)

𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦

maps both axes of

𝐹^{2}

onto

𝐹

(a) An element of the left side is $𝑇 (𝐮_{1} + \dots + 𝐮_{𝑘})$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ , and by linearity this equals $𝑇 (𝐮_{1}) +$ $\dots +$ $𝑇 (𝐮_{𝑘})$ , a typical element of the right side. Reading the same equation backwards gives the other containment.

(b) We check (II) of Theorem 4.4.7. Let $𝑇 (𝐮_{1}) +$ $\dots +$ $𝑇 (𝐮_{𝑘}) =$ $𝟎$ with $𝐮_{𝑖} \in 𝑈_{𝑖}$ . Then $𝑇 (𝐮_{1} + \dots + 𝐮_{𝑘}) = 𝟎$ , so $𝐮_{1} + \dots + 𝐮_{𝑘} = 𝟎$ since $𝑇$ is injective (Theorem 2.6.11). As $𝑈_{1}, \dots, 𝑈_{𝑘}$ are independent, every $𝐮_{𝑖} = 𝟎$ , so every $𝑇 (𝐮_{𝑖}) = 𝟎$ .

(c) Let $𝑇 : 𝐹^{2} \to 𝐹$ , $𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦$ , and $𝑈_{1} = span (𝐞_{1})$ , $𝑈_{2} = span (𝐞_{2})$ , which are independent. Both images are all of $𝐹$ , and $𝐹 + 𝐹$ is not direct, since $𝐹 \cap 𝐹 = 𝐹 \neq {𝟎}$ .

Exercise 4.4.36.routine In $ℂ^{2}$ , let $𝑈 = span ((1, 𝑖))$ and $𝑊 = span ((1, - 𝑖))$ .

(a)Show that $ℂ^{2} = 𝑈 \oplus 𝑊$ . Find the matrix $𝑃$ of ${Pr}_{𝑈, 𝑊}$ with respect to the standard basis, and check that $𝑃^{2} = 𝑃$ .
(b)Let $𝑅 (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ , the rotation by a quarter turn. Find scalars $𝜆$ and $𝜇$ with $𝑅 (𝐮) = 𝜆 𝐮$ for all $𝐮 \in 𝑈$ and $𝑅 (𝐰) = 𝜇 𝐰$ for all $𝐰 \in 𝑊$ .

Solution.

Uses. Example 4.4.5 (Direct sums), Definition 4.4.24 (Projection along a complement), Theorem 4.1.10 (Independent lists are short)

Techniques. Compare components (Technique 4.4.8)

Hint 1. Solve

(𝑥, 𝑦) =

𝑎 (1, 𝑖) +

𝑏 (1, - 𝑖)

for

𝑎

and

𝑏

Hint 2. Compute

𝑅 (1, 𝑖)

and

𝑅 (1, - 𝑖)

Sketch. The vectors

(1, 𝑖)

and

(1, - 𝑖)

are not proportional, so they form a basis and

ℂ^{2} = 𝑈 \oplus 𝑊

. Solving gives

𝑎 = (𝑥 - 𝑖 𝑦) / 2

, so

{Pr}_{𝑈, 𝑊} (𝑥, 𝑦) =

((𝑥 - 𝑖 𝑦) / 2, (𝑖 𝑥 + 𝑦) / 2)

and

𝑃 = \frac{1}{2} (\begin{matrix} 1 & - 𝑖 \\ 𝑖 & 1 \end{matrix})

. (b)

𝑅 (1, 𝑖) =

(- 𝑖, 1) =

- 𝑖 (1, 𝑖)

and

𝑅 (1, - 𝑖) =

(𝑖, 1) =

𝑖 (1, - 𝑖)

, so

𝜆 = - 𝑖

and

𝜇 = 𝑖

(a) If $(1, - 𝑖) = 𝑐 (1, 𝑖)$ , then $𝑐 = 1$ and $- 𝑖 = 𝑖$ , which is false; so $(1, 𝑖), (1, - 𝑖)$ is independent, hence a basis of $ℂ^{2}$ (Theorem 4.1.10(b)), and $ℂ^{2} = 𝑈 \oplus 𝑊$ by Example 4.4.5(a). To decompose $(𝑥, 𝑦)$ , solve $𝑎 (1, 𝑖) +$ $𝑏 (1, - 𝑖) =$ $(𝑥, 𝑦)$ , that is, $𝑎 + 𝑏 = 𝑥$ and $𝑖 (𝑎 - 𝑏) = 𝑦$ . Then $𝑎 - 𝑏 = - 𝑖 𝑦$ , so

𝑎 = \frac{𝑥 - 𝑖 𝑦}{2}, 𝑏 = \frac{𝑥 + 𝑖 𝑦}{2} .

Hence ${Pr}_{𝑈, 𝑊} (𝑥, 𝑦) =$ $𝑎 (1, 𝑖) =$ $((𝑥 - 𝑖 𝑦) / 2, (𝑖 𝑥 + 𝑦) / 2)$ , and

𝑃 = \frac{1}{2} (\begin{matrix} 1 & - 𝑖 \\ 𝑖 & 1 \end{matrix}), 𝑃^{2} = \frac{1}{4} (\begin{matrix} 1 + (- 𝑖) 𝑖 & - 𝑖 - 𝑖 \\ 𝑖 + 𝑖 & 𝑖 (- 𝑖) + 1 \end{matrix}) = \frac{1}{4} (\begin{matrix} 2 & - 2 𝑖 \\ 2 𝑖 & 2 \end{matrix}) = 𝑃 .

(b) $𝑅 (1, 𝑖) =$ $(- 𝑖, 1) =$ $- 𝑖 (1, 𝑖)$ , since $(- 𝑖) 𝑖 = 1$ ; and $𝑅 (1, - 𝑖) =$ $(𝑖, 1) =$ $𝑖 (1, - 𝑖)$ , since $𝑖 (- 𝑖) = 1$ . By linearity, $𝑅 (𝐮) = - 𝑖 𝐮$ for $𝐮 \in 𝑈$ and $𝑅 (𝐰) = 𝑖 𝐰$ for $𝐰 \in 𝑊$ , so $𝜆 = - 𝑖$ and $𝜇 = 𝑖$ .

Over $ℝ$ , the quarter turn maps no line through $𝟎$ into itself. Over $ℂ$ , it splits $ℂ^{2}$ into two lines on which it acts as multiplication by a scalar. This is diagonalization (Section 7.4) in its simplest form.

Exercise 4.4.37.standard Find the error in the following argument.

Claim. Let $𝑈_{1}$ , $𝑈_{2}$ , $𝑈_{3}$ be finite-dimensional subspaces of $𝑉$ with $𝑈_{𝑖} \cap 𝑈_{𝑗} = {𝟎}$ whenever $𝑖 \neq 𝑗$ . Then $dim (𝑈_{1} + 𝑈_{2} + 𝑈_{3}) =$ $dim 𝑈_{1} +$ $dim 𝑈_{2} +$ $dim 𝑈_{3}$ .

Proof. Let ${ℬ︀}_{𝑖}$ be a basis of $𝑈_{𝑖}$ , and let $ℬ︀$ be the concatenation ${ℬ︀}_{1}, {ℬ︀}_{2}, {ℬ︀}_{3}$ , which spans $𝑈_{1} + 𝑈_{2} + 𝑈_{3}$ . Suppose $ℬ︀$ is dependent. By the linear dependence lemma, some entry $𝐛$ of $ℬ︀$ is a combination of the entries before it. Say $𝐛$ is in ${ℬ︀}_{3}$ ; the other cases are similar. Move the terms that come from ${ℬ︀}_{3}$ to the left side. The left side is now $𝐛$ minus a combination of earlier entries of ${ℬ︀}_{3}$ , so it lies in $𝑈_{3}$ . The right side is a combination of entries of ${ℬ︀}_{1}$ and ${ℬ︀}_{2}$ , so it lies in $𝑈_{1}$ or in $𝑈_{2}$ . Hence both sides lie in $𝑈_{3} \cap 𝑈_{1}$ or in $𝑈_{3} \cap 𝑈_{2}$ , which are ${𝟎}$ . So $𝐛$ is a combination of earlier entries of ${ℬ︀}_{3}$ , contradicting the independence of ${ℬ︀}_{3}$ . Therefore $ℬ︀$ is a basis, and the claim follows.

Solution.

Uses. Non-example 4.4.6 (Sums that are not direct), Theorem 4.4.7 (Criteria for direct sums), Theorem 4.4.13 (Bases and dimension of a sum)

Techniques. Induction on the number of summands (Technique 4.4.10)

Hint 1. Test the claim on three lines in

ℝ^{2}

before looking for the error.

Hint 2. Where does a combination of entries of

{ℬ︀}_{1}

and

{ℬ︀}_{2}

lie?

Sketch. The claim is false: three lines in

ℝ^{2}

that meet pairwise in

𝟎

have dimensions adding up to

3 > 2

. The error is “it lies in

𝑈_{1}

or in

𝑈_{2}

”: a combination of entries of

{ℬ︀}_{1}

and

{ℬ︀}_{2}

lies in

𝑈_{1} + 𝑈_{2}

, which is usually much larger than

𝑈_{1} \cup 𝑈_{2}

. The argument is correct under criterion (IV),

𝑈_{3} \cap

(𝑈_{1} + 𝑈_{2}) =

{𝟎}

The claim is false. The lines $span (𝐞_{1})$ , $span (𝐞_{2})$ , and $span (𝐞_{1} + 𝐞_{2})$ in $ℝ^{2}$ meet pairwise in ${𝟎}$ , their dimensions add up to $3$ , and their sum is $ℝ^{2}$ , of dimension $2$ .

The error is the sentence “it lies in $𝑈_{1}$ or in $𝑈_{2}$ ”. A combination of entries of ${ℬ︀}_{1}$ and ${ℬ︀}_{2}$ is a vector of $𝑈_{1}$ plus a vector of $𝑈_{2}$ , so it lies in $𝑈_{1} + 𝑈_{2}$ , not in $𝑈_{1} \cup 𝑈_{2}$ . To conclude that both sides vanish, the argument needs $𝑈_{3} \cap$ $(𝑈_{1} + 𝑈_{2}) =$ ${𝟎}$ . In the example, $𝐞_{1} + 𝐞_{2}$ lies in $𝑈_{3}$ and also in $𝑈_{1} + 𝑈_{2}$ , and the argument breaks exactly there.

With the hypotheses $𝑈_{2} \cap 𝑈_{1} = {𝟎}$ and $𝑈_{3} \cap$ $(𝑈_{1} + 𝑈_{2}) =$ ${𝟎}$ instead, the argument is valid. (For an entry of ${ℬ︀}_{2}$ , the right side is a combination of entries of ${ℬ︀}_{1}$ and lies in $𝑈_{1}$ , so the pairwise hypothesis is enough there.) These are condition (IV) of Theorem 4.4.7, and the corrected argument is a proof of (IV) $\Rightarrow$ (I) combined with Theorem 4.4.13(c).

Exercise 4.4.38.standard Find the error in the following proof of Theorem 4.4.19.

Proof. Let $ℬ︀$ be a basis of $𝑈$ and $𝒞︀$ a basis of $𝑊$ . The concatenation $ℬ︀, 𝒞︀$ spans $𝑈 + 𝑊$ . The vectors that occur in both $ℬ︀$ and $𝒞︀$ lie in $𝑈 \cap 𝑊$ , and they form a basis of $𝑈 \cap 𝑊$ . Deleting one copy of each of them from $ℬ︀, 𝒞︀$ leaves a basis of $𝑈 + 𝑊$ with $dim 𝑈 +$ $dim 𝑊 -$ $dim (𝑈 \cap 𝑊)$ entries.

Solution.

Uses. Theorem 4.4.19 (Dimension of a sum)

Techniques. Extend a basis of a subspace (Technique 4.1.18)

Hint 1. Do a basis of

𝑈

and a basis of

𝑊

have to share any vectors at all?

Hint 2. Try

𝑈 = span (𝐞_{1}, 𝐞_{2})

with basis

𝐞_{1}, 𝐞_{2}

, and

𝑊 =

span (𝐞_{1} + 𝐞_{2} + 𝐞_{3}, 𝐞_{3})

with basis

𝐞_{1} + 𝐞_{2} + 𝐞_{3}, 𝐞_{3}

, in

ℝ^{3}

Sketch. The claim that the shared vectors form a basis of

𝑈 \cap 𝑊

is unjustified and usually false. With

ℬ︀ = 𝐞_{1}, 𝐞_{2}

and

𝒞︀ =

𝐞_{1} +

𝐞_{2} +

𝐞_{3},

𝐞_{3}

, no vector is shared and no entry of either basis lies in

𝑈 \cap

𝑊 =

span (𝐞_{1} + 𝐞_{2})

. The repair is to choose the bases: extend one basis of

𝑈 \cap 𝑊

to both, as in the first proof of Theorem 4.4.19.

The error is the claim that the vectors occurring in both $ℬ︀$ and $𝒞︀$ form a basis of $𝑈 \cap 𝑊$ . Arbitrary bases need not share any vector, and the entries of $ℬ︀$ that happen to lie in $𝑈 \cap 𝑊$ need not span it.

For example, in $ℝ^{3}$ let $𝑈 = span (𝐞_{1}, 𝐞_{2})$ with $ℬ︀ = 𝐞_{1}, 𝐞_{2}$ , and $𝑊 =$ $span (𝐞_{1} + 𝐞_{2} + 𝐞_{3}, 𝐞_{3})$ with $𝒞︀ =$ $𝐞_{1} +$ $𝐞_{2} +$ $𝐞_{3},$ $𝐞_{3}$ . The bases share no vector, so the procedure deletes nothing and returns four vectors in $ℝ^{3}$ , which cannot be a basis. Here $𝑈 \cap$ $𝑊 =$ $span (𝐞_{1} + 𝐞_{2})$ , since a vector $𝑎 (𝐞_{1} + 𝐞_{2} + 𝐞_{3}) +$ $𝑏 𝐞_{3} =$ $(𝑎, 𝑎, 𝑎 + 𝑏)$ of $𝑊$ lies in $𝑈$ exactly when $𝑏 = - 𝑎$ ; and none of $𝐞_{1}$ , $𝐞_{2}$ , $𝐞_{1} + 𝐞_{2} + 𝐞_{3}$ , $𝐞_{3}$ lies in this line.

The argument becomes correct if the bases are chosen with care: start from a basis of $𝑈 \cap 𝑊$ and extend it to a basis of $𝑈$ and to a basis of $𝑊$ . Then the shared vectors are exactly the chosen basis of $𝑈 \cap 𝑊$ , and what remains to be shown is that the merged list is independent. That is the first proof of Theorem 4.4.19.

Exercise 4.4.39.standard For finite-dimensional subspaces $𝑈$ , $𝑊$ , $𝑋$ of $𝑉$ , let

𝑒 (𝑈, 𝑊, 𝑋) = dim 𝑈 + dim 𝑊 + dim 𝑋 - dim (𝑈 \cap 𝑊) - dim (𝑈 \cap 𝑋) - dim (𝑊 \cap 𝑋) + dim (𝑈 \cap 𝑊 \cap 𝑋) .

(a)Find subspaces of $ℝ^{2}$ with $dim (𝑈 + 𝑊 + 𝑋) \neq$ $𝑒 (𝑈, 𝑊, 𝑋)$ .
(b)Show that $(𝑈 \cap 𝑋) +$ $(𝑊 \cap 𝑋) \subseteq$ $(𝑈 + 𝑊) \cap$ $𝑋$ always.
(c)Show that $dim (𝑈 + 𝑊 + 𝑋) =$ $𝑒 (𝑈, 𝑊, 𝑋)$ whenever $(𝑈 + 𝑊) \cap$ $𝑋 =$ $(𝑈 \cap 𝑋) +$ $(𝑊 \cap 𝑋)$ .

Solution.

Uses. Remark 4.4.22 (No formula for three subspaces), Proposition 1.4.16 (Intersections of subspaces), Proposition 4.4.2 (Sums are the smallest containing subspaces), Theorem 4.4.19 (Dimension of a sum)

Hint 1. For (a), see Remark 4.4.22.

Hint 2. For (c), apply the dimension formula three times: to

𝑈 + 𝑊

and

𝑋

, to

𝑈

and

𝑊

, and to

𝑈 \cap 𝑋

and

𝑊 \cap 𝑋

Sketch. (a) Three distinct lines:

dim (𝑈 + 𝑊 + 𝑋) = 2

but

𝑒 = 3

. (b)

𝑈 \cap 𝑋

and

𝑊 \cap 𝑋

lie in the subspace

(𝑈 + 𝑊) \cap 𝑋

. (c)

dim (𝑈 + 𝑊 + 𝑋) =

dim (𝑈 + 𝑊) +

dim 𝑋 -

dim ((𝑈 \cap 𝑋) + (𝑊 \cap 𝑋))

; expand

dim (𝑈 + 𝑊)

and

dim ((𝑈 \cap 𝑋) + (𝑊 \cap 𝑋))

by the formula, using

(𝑈 \cap 𝑋) \cap

(𝑊 \cap 𝑋) =

𝑈 \cap

𝑊 \cap

𝑋

(a) Let $𝑈 = span (𝐞_{1})$ , $𝑊 = span (𝐞_{2})$ , and $𝑋 = span (𝐞_{1} + 𝐞_{2})$ . All pairwise intersections and the triple intersection are ${𝟎}$ , so $𝑒 (𝑈, 𝑊, 𝑋) = 3$ , but $dim (𝑈 + 𝑊 + 𝑋) =$ $dim ℝ^{2} =$ $2$ (Remark 4.4.22). Here $(𝑈 + 𝑊) \cap 𝑋 = 𝑋$ , while $(𝑈 \cap 𝑋) +$ $(𝑊 \cap 𝑋) =$ ${𝟎}$ .

(b) $𝑈 \cap 𝑋 \subseteq 𝑈 \subseteq 𝑈 + 𝑊$ and $𝑈 \cap 𝑋 \subseteq 𝑋$ , so $𝑈 \cap 𝑋 \subseteq (𝑈 + 𝑊) \cap 𝑋$ ; likewise $𝑊 \cap 𝑋 \subseteq (𝑈 + 𝑊) \cap 𝑋$ . Since $(𝑈 + 𝑊) \cap 𝑋$ is a subspace (Proposition 1.4.16), it contains $(𝑈 \cap 𝑋) + (𝑊 \cap 𝑋)$ by Proposition 4.4.2(b).

(c) All the subspaces involved are finite-dimensional, being contained in $𝑈 + 𝑊 + 𝑋$ (Theorem 4.4.13(a)). By Proposition 4.4.2(a), $𝑈 +$ $𝑊 +$ $𝑋 =$ $(𝑈 + 𝑊) +$ $𝑋$ , so Theorem 4.4.19 and the hypothesis give

dim (𝑈 + 𝑊 + 𝑋) = dim (𝑈 + 𝑊) + dim 𝑋 - dim ((𝑈 \cap 𝑋) + (𝑊 \cap 𝑋)) .

By Theorem 4.4.19 again, $dim (𝑈 + 𝑊) =$ $dim 𝑈 +$ $dim 𝑊 -$ $dim (𝑈 \cap 𝑊)$ , and since $(𝑈 \cap 𝑋) \cap$ $(𝑊 \cap 𝑋) =$ $𝑈 \cap$ $𝑊 \cap$ $𝑋$ ,

dim ((𝑈 \cap 𝑋) + (𝑊 \cap 𝑋)) = dim (𝑈 \cap 𝑋) + dim (𝑊 \cap 𝑋) - dim (𝑈 \cap 𝑊 \cap 𝑋) .

Substituting both gives $dim (𝑈 + 𝑊 + 𝑋) =$ $𝑒 (𝑈, 𝑊, 𝑋)$ .

For instance, the hypothesis holds when $𝑋 \subseteq 𝑈$ , since then both sides equal $𝑋$ .

Exercise 4.4.40.standard Let $𝑉$ have dimension $𝑛$ , and let $𝑈_{1}, \dots, 𝑈_{𝑘}$ be subspaces of $𝑉$ .

(a)Show that $dim (𝑈_{1} \cap \dots \cap 𝑈_{𝑘}) \geq$ $dim 𝑈_{1} +$ $\dots +$ $dim 𝑈_{𝑘} -$ $(𝑘 - 1) 𝑛$ .
(b)Let $𝑘 \leq 𝑛$ . Show that the bound in (a) is attained by the subspaces $𝑈_{𝑖} =$ ${𝐱 \in 𝐹^{𝑛} | 𝑥_{𝑖} = 0}$ , $𝑖 = 1, \dots, 𝑘$ .
(c)Deduce that if $𝑛 \geq 2$ , any $𝑛 - 1$ subspaces of $𝑉$ of dimension $𝑛 - 1$ have a nonzero vector in common.

Solution.

Uses. Theorem 4.4.19 (Dimension of a sum), Proposition 4.1.21 (Dimensions of subspaces), Example 4.2.25 (Kernels of linear functionals)

Techniques. Induction on the number of summands (Technique 4.4.10), Dimension counting (Technique 4.2.24)

Hint 1. Induct on

𝑘

. In the inductive step, apply the dimension formula to

𝑋 = 𝑈_{1} \cap \dots \cap 𝑈_{𝑘 - 1}

and

𝑈_{𝑘}

Hint 2.

dim (𝑋 + 𝑈_{𝑘}) \leq 𝑛

Sketch. (a) For

𝑘 = 1

it is an equality. With

𝑋 = 𝑈_{1} \cap \dots \cap 𝑈_{𝑘 - 1}

dim (𝑋 \cap 𝑈_{𝑘}) =

dim 𝑋 +

dim 𝑈_{𝑘} -

dim (𝑋 + 𝑈_{𝑘}) \geq

dim 𝑋 +

dim 𝑈_{𝑘} -

𝑛

, and the inductive hypothesis bounds

dim 𝑋

. (b) The intersection is

span (𝐞_{𝑘 + 1}, \dots, 𝐞_{𝑛})

, of dimension

𝑛 -

𝑘 =

𝑘 (𝑛 - 1) -

(𝑘 - 1) 𝑛

. (c) For

𝑛 - 1

subspaces of dimension

𝑛 - 1

, the bound is

{(𝑛 - 1)}^{2} -

(𝑛 - 2) 𝑛 =

1

(a) We induct on $𝑘$ . For $𝑘 = 1$ both sides equal $dim 𝑈_{1}$ . Let $𝑘 \geq 2$ , and put $𝑋 = 𝑈_{1} \cap \dots \cap 𝑈_{𝑘 - 1}$ ; by the inductive hypothesis, $dim 𝑋 \geq$ $dim 𝑈_{1} +$ $\dots +$ $dim 𝑈_{𝑘 - 1} -$ $(𝑘 - 2) 𝑛$ . By Theorem 4.4.19 and $dim (𝑋 + 𝑈_{𝑘}) \leq 𝑛$ (Proposition 4.1.21(a)),

\begin{aligned} dim (𝑋 \cap 𝑈_{𝑘}) & = dim 𝑋 + dim 𝑈_{𝑘} - dim (𝑋 + 𝑈_{𝑘}) \\ \geq dim 𝑈_{1} + \dots + dim 𝑈_{𝑘 - 1} - (𝑘 - 2) 𝑛 + dim 𝑈_{𝑘} - 𝑛 \\ = dim 𝑈_{1} + \dots + dim 𝑈_{𝑘} - (𝑘 - 1) 𝑛 . \end{aligned}

Since $𝑋 \cap$ $𝑈_{𝑘} =$ $𝑈_{1} \cap$ $\dots \cap$ $𝑈_{𝑘}$ , this is the claim.

(b) Each $𝑈_{𝑖}$ is the kernel of the nonzero functional $𝐱 \mapsto 𝑥_{𝑖}$ , so $dim 𝑈_{𝑖} = 𝑛 - 1$ (Example 4.2.25), and the bound in (a) is $𝑘 (𝑛 - 1) -$ $(𝑘 - 1) 𝑛 =$ $𝑛 -$ $𝑘$ . The intersection consists of the vectors with $𝑥_{1} = \dots = 𝑥_{𝑘} = 0$ , which is $span (𝐞_{𝑘 + 1}, \dots, 𝐞_{𝑛})$ , of dimension $𝑛 - 𝑘$ .

(c) For $𝑘 = 𝑛 - 1$ subspaces of dimension $𝑛 - 1$ , (a) gives

dim (𝑈_{1} \cap \dots \cap 𝑈_{𝑛 - 1}) \geq {(𝑛 - 1)}^{2} - (𝑛 - 2) 𝑛 = 1 .

Exercise 4.4.41.standard Let $𝐴, 𝐵 \in 𝐹^{𝑚 \times 𝑛}$ , and let $(\begin{matrix} 𝐴 & 𝐵 \end{matrix}) \in 𝐹^{𝑚 \times 2 𝑛}$ be the matrix whose columns are those of $𝐴$ followed by those of $𝐵$ .

(a)Show that $rank (\begin{matrix} 𝐴 & 𝐵 \end{matrix}) =$ $rank 𝐴 +$ $rank 𝐵 -$ $dim (im 𝐴 \cap im 𝐵)$ .
(b)Deduce that $rank (𝐴 + 𝐵) \leq$ $rank 𝐴 +$ $rank 𝐵 -$ $dim (im 𝐴 \cap im 𝐵)$ . This sharpens Proposition 4.3.20.
(c)Find $𝐴, 𝐵 \in 𝐹^{2 \times 2}$ with $im 𝐴 \cap im 𝐵 = {𝟎}$ but $rank (𝐴 + 𝐵) <$ $rank 𝐴 +$ $rank 𝐵$ .

Solution.

Uses. Proposition 2.6.18 (The column space is spanned by the columns), Proposition 4.4.2 (Sums are the smallest containing subspaces), Theorem 4.4.19 (Dimension of a sum), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Proposition 4.1.21 (Dimensions of subspaces)

Hint 1. The column space of

(\begin{matrix} 𝐴 & 𝐵 \end{matrix})

is spanned by the columns of

𝐴

together with those of

𝐵

Hint 2.

(𝐴 + 𝐵) 𝐱 =

𝐴 𝐱 +

𝐵 𝐱

Sketch. (a)

im (\begin{matrix} 𝐴 & 𝐵 \end{matrix}) =

im 𝐴 +

im 𝐵

by Proposition 2.6.18 and Proposition 4.4.2(c), so (a) is the dimension formula. (b)

im (𝐴 + 𝐵) \subseteq

im 𝐴 +

im 𝐵

. (c)

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

and

𝐵 = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})

have images

span (𝐞_{1})

and

span (𝐞_{2})

, but

𝐴 + 𝐵 = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix})

has rank

1

(a) By Proposition 2.6.18, $im 𝐴$ , $im 𝐵$ , and $im (\begin{matrix} 𝐴 & 𝐵 \end{matrix})$ are the spans of the columns of $𝐴$ , of $𝐵$ , and of both together. So $im (\begin{matrix} 𝐴 & 𝐵 \end{matrix}) =$ $im 𝐴 +$ $im 𝐵$ by Proposition 4.4.2(c), and since rank is the dimension of the image (Theorem 4.2.7), Theorem 4.4.19 gives

rank (\begin{matrix} 𝐴 & 𝐵 \end{matrix}) = dim (im 𝐴 + im 𝐵) = rank 𝐴 + rank 𝐵 - dim (im 𝐴 \cap im 𝐵) .

(b) For every $𝐱 \in 𝐹^{𝑛}$ , $(𝐴 + 𝐵) 𝐱 =$ $𝐴 𝐱 +$ $𝐵 𝐱 \in$ $im 𝐴 +$ $im 𝐵$ . So $im (𝐴 + 𝐵) \subseteq$ $im 𝐴 +$ $im 𝐵$ , and by Proposition 4.1.21(a) and (a),

rank (𝐴 + 𝐵) \leq dim (im 𝐴 + im 𝐵) = rank 𝐴 + rank 𝐵 - dim (im 𝐴 \cap im 𝐵) .

(c) Let $𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ . Then $im 𝐴 = span (𝐞_{1})$ and $im 𝐵 = span (𝐞_{2})$ meet only in $𝟎$ , and $rank 𝐴 + rank 𝐵 = 2$ . But $𝐴 + 𝐵 = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix})$ has rank $1$ . The loss comes from the other side: $𝐴$ and $𝐵$ both kill $𝐞_{2}$ , so $𝐴 + 𝐵$ does too.

Exercise 4.4.42.standard Let $𝑉$ be finite-dimensional, and let $𝑈$ and $𝑊$ be subspaces of $𝑉$ .

(a)Show that there is $𝑇 \in L (𝑉)$ with $ker 𝑇 = 𝑈$ and $im 𝑇 = 𝑊$ if and only if $dim 𝑈 + dim 𝑊 = dim 𝑉$ .
(b)Show that $𝑇$ in (a) can be chosen with $𝑇^{2} = 𝑇$ if and only if $𝑉 = 𝑈 \oplus 𝑊$ .
(c)In $ℝ^{2}$ , let $𝑈 = 𝑊 = span (𝐞_{1})$ . Find $𝑇$ as in (a), and explain directly why no such $𝑇$ satisfies $𝑇^{2} = 𝑇$ .

Solution.

Uses. Theorem 4.2.11 (Rank–nullity), Proposition 4.4.16 (Complements exist), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Theorem 4.4.27 (Idempotents are projections), Proposition 4.4.26 (Properties of projections)

Techniques. Extend a basis of a subspace (Technique 4.1.18), Construct a map by choosing it on a basis (Technique 2.2.4)

Hint 1. One direction of (a) is rank–nullity. For the other, a complement

𝑋

𝑈

has the same dimension as

𝑊

Hint 2. Define

𝑇

to kill a basis of

𝑈

and to send a basis of

𝑋

to a basis of

𝑊

. For (b), use Theorem 4.4.27.

Sketch. (a) (

\Rightarrow

) is rank–nullity. (

\leq

) Let

𝑋

be a complement of

𝑈

, so

dim 𝑋 =

dim 𝑉 -

dim 𝑈 =

dim 𝑊

. Glue bases

𝐮_{𝑖}

𝑈

and

𝐱_{𝑗}

𝑋

into a basis of

𝑉

, and let

𝑇 (𝐮_{𝑖}) = 𝟎

and

𝑇 (𝐱_{𝑗}) = 𝐰_{𝑗}

, a basis of

𝑊

. Then

im 𝑇 = 𝑊

and

ker 𝑇 = 𝑈

. (b) An idempotent has

𝑉 =

im 𝑇 \oplus

ker 𝑇 =

𝑊 \oplus

𝑈

; conversely,

{Pr}_{𝑊, 𝑈}

works. (c)

𝑇 (𝑥, 𝑦) = (𝑦, 0)

. An idempotent fixes its image and kills its kernel, so these meet only in

𝟎

(a) If $ker 𝑇 = 𝑈$ and $im 𝑇 = 𝑊$ , then $dim 𝑈 + dim 𝑊 = dim 𝑉$ by Theorem 4.2.11. Conversely, suppose $dim 𝑈 + dim 𝑊 = dim 𝑉$ . Let $𝑋$ be a complement of $𝑈$ (Proposition 4.4.16), so $dim 𝑋 =$ $dim 𝑉 -$ $dim 𝑈 =$ $dim 𝑊$ ; call this number $𝑟$ . Let $𝐮_{1}, \dots, 𝐮_{𝑘}$ , $𝐱_{1}, \dots, 𝐱_{𝑟}$ , and $𝐰_{1}, \dots, 𝐰_{𝑟}$ be bases of $𝑈$ , $𝑋$ , and $𝑊$ . Since $𝑉 = 𝑈 \oplus 𝑋$ , the list $𝐮_{1},$ $\dots,$ $𝐮_{𝑘},$ $𝐱_{1},$ $\dots,$ $𝐱_{𝑟}$ is a basis of $𝑉$ (Theorem 4.4.13(b)), so there is $𝑇 \in L (𝑉)$ with $𝑇 (𝐮_{𝑖}) = 𝟎$ and $𝑇 (𝐱_{𝑗}) = 𝐰_{𝑗}$ for all $𝑖$ , $𝑗$ (Theorem 2.2.1). Its image is $span (𝐰_{1}, \dots, 𝐰_{𝑟}) =$ $𝑊$ (Proposition 2.1.16). For $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗}$ , we get $𝑇 (𝐯) =$ $\sum_{𝑗} 𝑏_{𝑗} 𝐰_{𝑗}$ , which is $𝟎$ if and only if all $𝑏_{𝑗} = 0$ , that is, if and only if $𝐯 \in 𝑈$ . So $ker 𝑇 = 𝑈$ .

(b) If $𝑇^{2} = 𝑇$ , then $𝑉 =$ $im 𝑇 \oplus$ $ker 𝑇 =$ $𝑊 \oplus$ $𝑈$ by Theorem 4.4.27. Conversely, if $𝑉 = 𝑈 \oplus 𝑊$ , then $𝑇 = {Pr}_{𝑊, 𝑈}$ satisfies $𝑇^{2} = 𝑇$ , $im 𝑇 = 𝑊$ , and $ker 𝑇 = 𝑈$ by Proposition 4.4.26.

(c) Here $dim 𝑈 + dim 𝑊 = 2$ , and $𝑇 (𝑥, 𝑦) = (𝑦, 0)$ has $ker 𝑇 =$ $span (𝐞_{1}) =$ $im 𝑇$ . Suppose some $𝑇$ with $𝑇^{2} = 𝑇$ had $ker 𝑇 =$ $im 𝑇 =$ $span (𝐞_{1})$ . Then $𝐞_{1} = 𝑇 (𝐱)$ for some $𝐱$ , so $𝑇 (𝐞_{1}) =$ $𝑇^{2} (𝐱) =$ $𝑇 (𝐱) =$ $𝐞_{1}$ ; but $𝐞_{1} \in ker 𝑇$ , so $𝑇 (𝐞_{1}) = 𝟎$ , a contradiction.

Exercise 4.4.43.standard Suppose $1 + 1 \neq 0$ in $𝐹$ , and let $𝑇 \in L (𝑉)$ satisfy $𝑇^{2} = {id}_{𝑉}$ . Let

𝑉_{+} = {𝐯 \in 𝑉 | 𝑇 (𝐯) = 𝐯}, 𝑉_{-} = {𝐯 \in 𝑉 | 𝑇 (𝐯) = - 𝐯} .

(a)Show that $𝑉 = 𝑉_{+} \oplus 𝑉_{-}$ .
(b)Deduce that $𝐹^{𝑛 \times 𝑛}$ is the direct sum of the subspace of symmetric matrices ( $𝐴^{𝖳} = 𝐴$ ) and the subspace of skew-symmetric matrices ( $𝐴^{𝖳} = - 𝐴$ ), and find the dimensions of both.
(c)Over $𝔽_{2}$ , find $𝑇 \in L (𝔽_{2}^{2})$ with $𝑇^{2} = id$ but $𝑉_{+} + 𝑉_{-} \neq 𝔽_{2}^{2}$ .

Solution.

Uses. Proposition 2.6.8 (Kernels and images are subspaces), Theorem 4.4.7 (Criteria for direct sums), Proposition 1.2.35 (Zero products), Theorem 4.4.27 (Idempotents are projections), Definition 2.4.21 (Transpose), Lemma 2.4.24 (Matrix units), Theorem 4.4.13 (Bases and dimension of a sum)

Techniques. Split the space with an idempotent (Technique 4.4.28)

Hint 1. Write

2 = 1 + 1

, which is invertible in

𝐹

. Split

𝐯 =

2^{- 1} (𝐯 + 𝑇 (𝐯)) +

2^{- 1} (𝐯 - 𝑇 (𝐯))

and check where each piece lies.

Hint 2. Equivalently,

𝑃 = 2^{- 1} ({id}_{𝑉} + 𝑇)

is idempotent. For (b), apply (a) to

𝐴 \mapsto 𝐴^{𝖳}

Sketch. (a)

𝑇 (𝐯 + 𝑇 𝐯) =

𝐯 +

𝑇 𝐯

and

𝑇 (𝐯 - 𝑇 𝐯) =

- (𝐯 - 𝑇 𝐯)

, so the split shows

𝑉 = 𝑉_{+} + 𝑉_{-}

; if

𝑇 𝐯 = 𝐯 = - 𝐯

, then

2 𝐯 = 𝟎

and

𝐯 = 𝟎

. (b) Transposition is linear with square the identity, and

𝑉_{+}

𝑉_{-}

are the symmetric and skew-symmetric matrices. The symmetric ones have basis

𝑈_{𝑖 𝑖}

and

𝑈_{𝑖 𝑗} + 𝑈_{𝑗 𝑖}

(

𝑖 < 𝑗

), so dimension

𝑛 (𝑛 + 1) / 2

, and the skew-symmetric ones have dimension

𝑛^{2} -

𝑛 (𝑛 + 1) / 2 =

𝑛 (𝑛 - 1) / 2

. (c)

𝑇 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})

𝑉_{+} = 𝑉_{-} = span (𝐞_{1})

(a) $𝑉_{+} = ker (𝑇 - {id}_{𝑉})$ and $𝑉_{-} = ker (𝑇 + {id}_{𝑉})$ are subspaces (Proposition 2.6.8). Write $2 = 1 + 1$ , which is nonzero by hypothesis. For $𝐯 \in 𝑉$ , put $𝐯_{+} =$ $2^{- 1} (𝐯 + 𝑇 (𝐯))$ and $𝐯_{-} =$ $2^{- 1} (𝐯 - 𝑇 (𝐯))$ . Then $𝐯_{+} +$ $𝐯_{-} =$ $2^{- 1} (2 𝐯) =$ $𝐯$ , and since $𝑇^{2} = {id}_{𝑉}$ ,

𝑇 (𝐯_{+}) = 2^{- 1} (𝑇 (𝐯) + 𝐯) = 𝐯_{+}, 𝑇 (𝐯_{-}) = 2^{- 1} (𝑇 (𝐯) - 𝐯) = - 𝐯_{-} .

So $𝐯_{+} \in 𝑉_{+}$ and $𝐯_{-} \in 𝑉_{-}$ , and $𝑉 = 𝑉_{+} + 𝑉_{-}$ . If $𝐯 \in 𝑉_{+} \cap 𝑉_{-}$ , then $𝐯 = 𝑇 (𝐯) = - 𝐯$ , so $2 𝐯 = 𝟎$ and $𝐯 = 𝟎$ (Proposition 1.2.35). By Theorem 4.4.7, $𝑉 = 𝑉_{+} \oplus 𝑉_{-}$ .

In the language of Theorem 4.4.27, $𝑃 = 2^{- 1} ({id}_{𝑉} + 𝑇)$ satisfies $𝑃^{2} =$ $4^{- 1} ({id}_{𝑉} + 2 𝑇 + 𝑇^{2}) =$ $2^{- 1} ({id}_{𝑉} + 𝑇) =$ $𝑃$ , with $im 𝑃 = 𝑉_{+}$ and $ker 𝑃 = 𝑉_{-}$ .

(b) The map $𝐴 \mapsto 𝐴^{𝖳}$ on $𝐹^{𝑛 \times 𝑛}$ is linear, and ${(𝐴^{𝖳})}^{𝖳} = 𝐴$ (Definition 2.4.21). Its $𝑉_{+}$ consists of the symmetric matrices and its $𝑉_{-}$ of the skew-symmetric ones, so (a) gives the direct sum, with $𝐴 =$ $2^{- 1} (𝐴 + 𝐴^{𝖳}) +$ $2^{- 1} (𝐴 - 𝐴^{𝖳})$ . Write $𝑈_{𝑖 𝑗} = 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳}$ . A matrix $𝐴 =$ $\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝑈_{𝑖 𝑗}$ is symmetric if and only if $𝑎_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ for all $𝑖, 𝑗$ , that is, if and only if

𝐴 = \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖 𝑖} 𝑈_{𝑖 𝑖} + \sum_{𝑖 < 𝑗} 𝑎_{𝑖 𝑗} (𝑈_{𝑖 𝑗} + 𝑈_{𝑗 𝑖}) .

So the $𝑛 + 𝑛 (𝑛 - 1) / 2$ matrices $𝑈_{𝑖 𝑖}$ and $𝑈_{𝑖 𝑗} + 𝑈_{𝑗 𝑖}$ ( $𝑖 < 𝑗$ ) span the symmetric matrices, and they are independent, since a combination of them has entry $𝑎_{𝑖 𝑗}$ in position $(𝑖, 𝑗)$ for $𝑖 \leq 𝑗$ , and the matrix units are independent (Lemma 2.4.24). Hence the symmetric matrices have dimension $𝑛 (𝑛 + 1) / 2$ , and by Theorem 4.4.13(c) the skew-symmetric matrices have dimension $𝑛^{2} -$ $𝑛 (𝑛 + 1) / 2 =$ $𝑛 (𝑛 - 1) / 2$ .

(c) Over $𝔽_{2}$ , let $𝑇$ have matrix $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ . Then $𝑇^{2}$ has matrix $(\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) = 𝐼_{2}$ . $𝑇 (𝑥, 𝑦) =$ $(𝑥 + 𝑦, 𝑦)$ equals $(𝑥, 𝑦)$ if and only if $𝑦 = 0$ , so $𝑉_{+} = span (𝐞_{1})$ , and since $- 1 = 1$ in $𝔽_{2}$ , $𝑉_{-} = 𝑉_{+}$ . So $𝑉_{+} +$ $𝑉_{-} =$ $span (𝐞_{1}) \neq$ $𝔽_{2}^{2}$ . (Likewise, over $𝔽_{2}$ the skew-symmetric matrices are exactly the symmetric ones.)

Exercise 4.4.44.standard Suppose $𝑉 = 𝑈 \oplus 𝑊$ , and let $𝑇 \in L (𝑉)$ satisfy $𝑇 (𝐮) \in 𝑈$ for all $𝐮 \in 𝑈$ and $𝑇 (𝐰) \in 𝑊$ for all $𝐰 \in 𝑊$ .

(a)Show that $ker 𝑇 =$ $(𝑈 \cap ker 𝑇) \oplus$ $(𝑊 \cap ker 𝑇)$ and $im 𝑇 =$ $im (𝑇 ↾_{𝑈}) \oplus$ $im (𝑇 ↾_{𝑊})$ .
(b)Deduce that if $𝑉$ is finite-dimensional, then $rank 𝑇 =$ $rank (𝑇 ↾_{𝑈}) +$ $rank (𝑇 ↾_{𝑊})$ .
(c)Show that (a) can fail without the hypothesis on $𝑇$ : find $𝑇 \in L (ℝ^{2})$ with $ker 𝑇 \neq$ $(𝑈 \cap ker 𝑇) +$ $(𝑊 \cap ker 𝑇)$ , where $𝑈 = span (𝐞_{1})$ and $𝑊 = span (𝐞_{2})$ .

Solution.

Uses. Theorem 4.4.7 (Criteria for direct sums), Theorem 4.4.13 (Bases and dimension of a sum), Definition 4.2.14 (Rank and nullity of a linear map), Proposition 4.4.2 (Sums are the smallest containing subspaces)

Techniques. Compare components (Technique 4.4.8)

Hint 1. If

𝑇 (𝐮 + 𝐰) = 𝟎

, then

𝑇 (𝐮) + 𝑇 (𝐰) = 𝟎

with

𝑇 (𝐮) \in 𝑈

and

𝑇 (𝐰) \in 𝑊

. Compare components.

Hint 2. Subspaces of independent subspaces are independent:

(𝑈 \cap ker 𝑇) \cap

(𝑊 \cap ker 𝑇) \subseteq

𝑈 \cap

𝑊

Sketch. (a) If

𝐮 + 𝐰 \in ker 𝑇

, then

𝑇 𝐮 + 𝑇 𝐰 = 𝟎

is a decomposition of

𝟎

along

𝑈 \oplus 𝑊

, so

𝑇 𝐮 = 𝑇 𝐰 = 𝟎

. The image:

𝑇 (𝐮 + 𝐰) =

𝑇 𝐮 +

𝑇 𝐰

. Both sums are direct because their summands lie in

𝑈

and

𝑊

. (b) Dimensions add in a direct sum. (c)

𝑇 (𝑥, 𝑦) =

(𝑥 + 𝑦, 0)

has kernel

span ((1, - 1))

, which meets both axes trivially.

(a) Let $𝐯 = 𝐮 + 𝐰 \in ker 𝑇$ with $𝐮 \in 𝑈$ and $𝐰 \in 𝑊$ . Then $𝟎 = 𝑇 (𝐮) + 𝑇 (𝐰)$ with $𝑇 (𝐮) \in 𝑈$ and $𝑇 (𝐰) \in 𝑊$ , a decomposition of $𝟎$ along $𝑈 \oplus 𝑊$ , so $𝑇 (𝐮) = 𝑇 (𝐰) = 𝟎$ by (II) of Theorem 4.4.7. Hence $𝐮 \in 𝑈 \cap ker 𝑇$ and $𝐰 \in 𝑊 \cap ker 𝑇$ , so $ker 𝑇 \subseteq$ $(𝑈 \cap ker 𝑇) +$ $(𝑊 \cap ker 𝑇)$ . The reverse containment holds because $ker 𝑇$ is a subspace containing both summands (Proposition 4.4.2(b)). The sum is direct, since $(𝑈 \cap ker 𝑇) \cap$ $(𝑊 \cap ker 𝑇) \subseteq$ $𝑈 \cap$ $𝑊 =$ ${𝟎}$ .

For the image, every $𝑇 (𝐮 + 𝐰) =$ $𝑇 (𝐮) +$ $𝑇 (𝐰)$ lies in $im (𝑇 ↾_{𝑈}) +$ $im (𝑇 ↾_{𝑊})$ , and conversely $𝑇 (𝐮) +$ $𝑇 (𝐰) =$ $𝑇 (𝐮 + 𝐰) \in$ $im 𝑇$ . By hypothesis $im (𝑇 ↾_{𝑈}) \subseteq 𝑈$ and $im (𝑇 ↾_{𝑊}) \subseteq 𝑊$ , so their intersection lies in $𝑈 \cap 𝑊 = {𝟎}$ , and the sum is direct.

(b) By (a) and Theorem 4.4.13(c), $dim im 𝑇 =$ $dim im (𝑇 ↾_{𝑈}) +$ $dim im (𝑇 ↾_{𝑊})$ , which is the claim (Definition 4.2.14).

(c) Let $𝑇 (𝑥, 𝑦) =$ $(𝑥 + 𝑦, 0)$ . Then $ker 𝑇 =$ $span ((1, - 1))$ , and a vector $(𝑡, - 𝑡)$ lies on an axis only if $𝑡 = 0$ . So $(𝑈 \cap ker 𝑇) +$ $(𝑊 \cap ker 𝑇) =$ ${𝟎} \neq$ $ker 𝑇$ . Here $𝑇 (𝐞_{2}) = 𝐞_{1} \notin 𝑊$ , so the hypothesis fails.

Exercise 4.4.45.standard Let

𝐴 \in 𝐹^{𝑚 \times 𝑛}

. Prove that

𝐹^{𝑛} = ker 𝐴 \oplus im 𝐴^{𝖳}

when

𝐹 = ℝ

, and show by examples that this can fail when

𝐹 = ℂ

and when

𝐹 = 𝔽_{2}

Solution.

Uses. Theorem 4.2.11 (Rank–nullity), Theorem 4.3.8 (Row rank equals column rank), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Proposition 4.4.21 (Two out of three), Proposition 2.4.23 (Transpose of a product)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Over every field the dimensions add up to

𝑛

, so by Proposition 4.4.21 only the intersection matters.

Hint 2. If

𝐱 = 𝐴^{𝖳} 𝐲

and

𝐴 𝐱 = 𝟎

, compute

𝐱^{𝖳} 𝐱

Sketch.

dim ker 𝐴 +

dim im 𝐴^{𝖳} =

(𝑛 - rank 𝐴) +

rank 𝐴 =

𝑛

. Over

ℝ

, if

𝐱 = 𝐴^{𝖳} 𝐲 \in ker 𝐴

, then

𝐱^{𝖳} 𝐱 = 𝐲^{𝖳} 𝐴 𝐱 = 0

, and

𝑥_{1}^{2} + \dots + 𝑥_{𝑛}^{2} = 0

forces

𝐱 = 𝟎

. Over

ℂ

𝐴 = (\begin{matrix} 1 & 𝑖 \end{matrix})

has

ker 𝐴 =

span ((- 𝑖, 1)) =

im 𝐴^{𝖳}

. Over

𝔽_{2}

𝐴 = (\begin{matrix} 1 & 1 \end{matrix})

has

ker 𝐴 =

span ((1, 1)) =

im 𝐴^{𝖳}

Over every field, $dim ker 𝐴 =$ $𝑛 -$ $rank 𝐴$ by Theorem 4.2.11, and $dim im 𝐴^{𝖳} =$ $rank 𝐴^{𝖳} =$ $rank 𝐴$ by Theorem 4.2.7 and Theorem 4.3.8. So condition (iii) of Proposition 4.4.21 always holds, and $𝐹^{𝑛} = ker 𝐴 \oplus im 𝐴^{𝖳}$ if and only if $ker 𝐴 \cap im 𝐴^{𝖳} = {𝟎}$ .

Over $ℝ$ . Let $𝐱 \in ker 𝐴 \cap im 𝐴^{𝖳}$ , say $𝐱 = 𝐴^{𝖳} 𝐲$ . By Proposition 2.4.23, $𝐱^{𝖳} = 𝐲^{𝖳} 𝐴$ , so

𝑥_{1}^{2} + \dots + 𝑥_{𝑛}^{2} = 𝐱^{𝖳} 𝐱 = 𝐲^{𝖳} (𝐴 𝐱) = 𝐲^{𝖳} 𝟎 = 0 .

A sum of squares of real numbers vanishes only if each term does, so $𝐱 = 𝟎$ .

Over $ℂ$ . Let $𝐴 = (\begin{matrix} 1 & 𝑖 \end{matrix}) \in ℂ^{1 \times 2}$ . Then $𝐴 𝐱 = 𝑥_{1} + 𝑖 𝑥_{2} = 0$ if and only if $𝑥_{1} = - 𝑖 𝑥_{2}$ , so $ker 𝐴 =$ $span ((- 𝑖, 1))$ . Also $im 𝐴^{𝖳} =$ $span ((1, 𝑖))$ , and $(- 𝑖, 1) = - 𝑖 (1, 𝑖)$ . So $ker 𝐴 = im 𝐴^{𝖳}$ is a line, and the sum is not direct. The real argument breaks because ${(1, 𝑖)}^{𝖳} (1, 𝑖) =$ $1 +$ $𝑖^{2} =$ $0$ . Over $ℂ$ the correct statement uses the conjugate transpose (Section 6.6).

Over $𝔽_{2}$ . Let $𝐴 = (\begin{matrix} 1 & 1 \end{matrix}) \in 𝔽_{2}^{1 \times 2}$ . Then $ker 𝐴 =$ ${𝐱 | 𝑥_{1} + 𝑥_{2} = 0} =$ $span ((1, 1)) =$ $im 𝐴^{𝖳}$ , and again the sum is not direct; here ${(1, 1)}^{𝖳} (1, 1) =$ $1 +$ $1 =$ $0$ .

Exercise 4.4.46.standard Let

𝑉

be finite-dimensional, and let

𝑈

be a subspace of

𝑉

. Show that

𝑈

has exactly one complement if and only if

𝑈 = {𝟎}

𝑈 = 𝑉

Solution.

Uses. Proposition 4.4.16 (Complements exist), Proposition 4.4.21 (Two out of three), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. If

𝑈 = {𝟎}

𝑈 = 𝑉

, what must a complement be?

Hint 2. Otherwise, take a complement

𝑊

with basis

𝐰_{1}, \dots, 𝐰_{𝑟}

and a nonzero

𝐮 \in 𝑈

. Tilt

𝐰_{1}

towards

𝐮

: try

span (𝐰_{1} + 𝐮, 𝐰_{2}, \dots, 𝐰_{𝑟})

Sketch. A complement

𝑊

{𝟎}

satisfies

𝑊 = {𝟎} + 𝑊 = 𝑉

, and a complement of

𝑉

satisfies

𝑊 = 𝑉 \cap 𝑊 = {𝟎}

. If

{𝟎} \neq 𝑈 \neq 𝑉

, take a complement

𝑊

with basis

𝐰_{1}, \dots, 𝐰_{𝑟}

(

𝑟 \geq 1

) and

𝐮 \in 𝑈

nonzero. If

𝑎 (𝐰_{1} + 𝐮) +

\sum_{𝑗 \geq 2} 𝑎_{𝑗} 𝐰_{𝑗} \in

𝑈

, then

𝑎 𝐰_{1} +

\sum_{𝑗 \geq 2} 𝑎_{𝑗} 𝐰_{𝑗} \in

𝑈 \cap

𝑊

, so all coefficients vanish. So

𝑊^{'} =

span (𝐰_{1} + 𝐮, 𝐰_{2}, \dots, 𝐰_{𝑟})

has dimension

𝑟

and meets

𝑈

trivially, hence is a complement, and

𝐰_{1} + 𝐮 \notin 𝑊

If $𝑈 = {𝟎}$ , then $𝑉 = {𝟎} \oplus 𝑉$ , and any complement $𝑊$ satisfies $𝑊 = {𝟎} + 𝑊 = 𝑉$ . If $𝑈 = 𝑉$ , then $𝑉 = 𝑉 \oplus {𝟎}$ , and any complement $𝑊$ satisfies $𝑊 = 𝑉 \cap 𝑊 = {𝟎}$ . In both cases the complement is unique.

Now suppose ${𝟎} \neq 𝑈 \neq 𝑉$ . By Proposition 4.4.16, $𝑈$ has a complement $𝑊$ , of dimension $𝑟 = dim 𝑉 - dim 𝑈 \geq 1$ ; let $𝐰_{1}, \dots, 𝐰_{𝑟}$ be a basis of $𝑊$ , and let $𝐮 \in 𝑈$ be nonzero. Let $𝑊^{'} =$ $span (𝐰_{1} + 𝐮, 𝐰_{2}, \dots, 𝐰_{𝑟})$ . Suppose

𝐲 = 𝑎 (𝐰_{1} + 𝐮) + 𝑎_{2} 𝐰_{2} + \dots + 𝑎_{𝑟} 𝐰_{𝑟} \in 𝑈 .

Then $𝑎 𝐰_{1} +$ $𝑎_{2} 𝐰_{2} +$ $\dots +$ $𝑎_{𝑟} 𝐰_{𝑟} =$ $𝐲 -$ $𝑎 𝐮$ lies in $𝑊$ and in $𝑈$ , hence is $𝟎$ , and since $𝐰_{1}, \dots, 𝐰_{𝑟}$ is independent, $𝑎 = 𝑎_{2} = \dots = 𝑎_{𝑟} = 0$ (Theorem 1.5.5), so $𝐲 = 𝟎$ . This shows $𝑈 \cap 𝑊^{'} = {𝟎}$ , and, taking $𝐲 = 𝟎$ , that the spanning list of $𝑊^{'}$ is independent, so $dim 𝑊^{'} = 𝑟$ . By Proposition 4.4.21, $𝑉 = 𝑈 \oplus 𝑊^{'}$ . Finally $𝑊^{'} \neq 𝑊$ : if $𝐰_{1} + 𝐮 \in 𝑊$ , then $𝐮 =$ $(𝐰_{1} + 𝐮) -$ $𝐰_{1} \in$ $𝑈 \cap$ $𝑊 =$ ${𝟎}$ , a contradiction. So $𝑈$ has at least two complements.

Exercise 4.4.47.standard Let $0 \leq 𝑘 \leq 𝑛$ , let $𝑐_{0}, \dots, 𝑐_{𝑘} \in 𝐹$ be distinct, and let

𝑍 = {𝑝 \in {𝐹 [𝑥]}_{\leq 𝑛} | 𝑝 (𝑐_{0}) = 𝑝 (𝑐_{1}) = \dots = 𝑝 (𝑐_{𝑘}) = 0} .

Show that ${𝐹 [𝑥]}_{\leq 𝑛} =$ $𝑍 \oplus$ ${𝐹 [𝑥]}_{\leq 𝑘}$ , and compute $dim 𝑍$ in two ways.

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Lemma 3.4.23 (Too many roots), Theorem 3.4.24 (Polynomial interpolation), Theorem 4.4.7 (Criteria for direct sums), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 4.2.11 (Rank–nullity)

Hint 1. A polynomial of degree at most

𝑘

that vanishes at

𝑘 + 1

distinct points is

𝟎

Hint 2. Given

𝑝

, interpolate its values at

𝑐_{0}, \dots, 𝑐_{𝑘}

by a polynomial

𝑞

of degree at most

𝑘

. Then

𝑝 - 𝑞 \in 𝑍

Hint 3. For the second count,

𝑍

is the kernel of

𝑝 \mapsto

(𝑝 (𝑐_{0}), \dots, 𝑝 (𝑐_{𝑘}))

Sketch.

𝑍 \cap {𝐹 [𝑥]}_{\leq 𝑘} = {𝟎}

by Lemma 3.4.23. For

𝑝

, let

𝑞 \in {𝐹 [𝑥]}_{\leq 𝑘}

interpolate the values

𝑝 (𝑐_{𝑖})

(Theorem 3.4.24); then

𝑝 = (𝑝 - 𝑞) + 𝑞

with

𝑝 - 𝑞 \in 𝑍

. So

dim 𝑍 =

(𝑛 + 1) -

(𝑘 + 1) =

𝑛 -

𝑘

by Theorem 4.4.13(c). Alternatively, evaluation

{𝐹 [𝑥]}_{\leq 𝑛} \to 𝐹^{𝑘 + 1}

is onto by interpolation and has kernel

𝑍

; rank–nullity gives the same answer.

The map $ev : {𝐹 [𝑥]}_{\leq 𝑛} \to$ $𝐹^{𝑘 + 1}$ , $𝑝 \mapsto$ $(𝑝 (𝑐_{0}), \dots, 𝑝 (𝑐_{𝑘}))$ , is linear (Example 2.1.9), and $𝑍 = ker ev$ is a subspace.

If $𝑝 \in 𝑍 \cap {𝐹 [𝑥]}_{\leq 𝑘}$ , then $𝑝$ has degree at most $𝑘$ and vanishes at the $𝑘 + 1$ distinct points $𝑐_{0}, \dots, 𝑐_{𝑘}$ , so $𝑝 = 𝟎$ by Lemma 3.4.23. Next, let $𝑝 \in {𝐹 [𝑥]}_{\leq 𝑛}$ . By Theorem 3.4.24 there is $𝑞 \in {𝐹 [𝑥]}_{\leq 𝑘}$ with $𝑞 (𝑐_{𝑖}) = 𝑝 (𝑐_{𝑖})$ for $𝑖 = 0, \dots, 𝑘$ . Since $𝑘 \leq 𝑛$ , $𝑝 - 𝑞 \in {𝐹 [𝑥]}_{\leq 𝑛}$ , and it vanishes at every $𝑐_{𝑖}$ , so $𝑝 - 𝑞 \in 𝑍$ and $𝑝 =$ $(𝑝 - 𝑞) +$ $𝑞 \in$ $𝑍 +$ ${𝐹 [𝑥]}_{\leq 𝑘}$ . By Theorem 4.4.7, ${𝐹 [𝑥]}_{\leq 𝑛} =$ $𝑍 \oplus$ ${𝐹 [𝑥]}_{\leq 𝑘}$ .

First count. By Theorem 4.4.13(c), $𝑛 + 1 = dim 𝑍 + (𝑘 + 1)$ , so $dim 𝑍 = 𝑛 - 𝑘$ .

Second count. The map $ev$ is surjective: for $(𝑑_{0}, \dots, 𝑑_{𝑘}) \in$ $𝐹^{𝑘 + 1}$ , the interpolating polynomial of Theorem 3.4.24 lies in ${𝐹 [𝑥]}_{\leq 𝑘} \subseteq$ ${𝐹 [𝑥]}_{\leq 𝑛}$ and has these values. By Theorem 4.2.11, $dim 𝑍 =$ $dim ker ev =$ $(𝑛 + 1) -$ $(𝑘 + 1) =$ $𝑛 -$ $𝑘$ .

The decomposition $𝑝 = (𝑝 - 𝑞) + 𝑞$ says that the projection onto ${𝐹 [𝑥]}_{\leq 𝑘}$ along $𝑍$ sends $𝑝$ to its interpolating polynomial at $𝑐_{0}, \dots, 𝑐_{𝑘}$ . Note that the hypothesis requires $𝐹$ to have at least $𝑘 + 1$ elements.

Exercise 4.4.48.standard Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ satisfy $𝐴 𝐵 = 0$ , and suppose that $𝐴 + 𝐵$ is invertible.

(a)Show that $rank 𝐴 + rank 𝐵 = 𝑛$ and $𝐹^{𝑛} = im 𝐴 \oplus im 𝐵$ .
(b)Show that $im 𝐵 = ker 𝐴$ .
(c)Show that if also $𝐵 𝐴 = 0$ , then $im 𝐴 = ker 𝐵$ . Show by an example that $im 𝐴 = ker 𝐵$ can fail without the hypothesis $𝐵 𝐴 = 0$ .

Solution.

Uses. Theorem 4.4.13 (Bases and dimension of a sum), Theorem 4.2.11 (Rank–nullity), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Proposition 4.1.21 (Dimensions of subspaces), Proposition 4.4.21 (Two out of three)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Since

𝐴 + 𝐵

is invertible, every

𝐱

equals

(𝐴 + 𝐵) 𝐲 =

𝐴 𝐲 +

𝐵 𝐲

for some

𝐲

. What does that say about

im 𝐴 + im 𝐵

Hint 2.

𝐴 𝐵 = 0

means

im 𝐵 \subseteq ker 𝐴

. Combine this with rank–nullity for

𝐴

Sketch. (a) Every

𝐱 = (𝐴 + 𝐵) 𝐲

lies in

im 𝐴 + im 𝐵

, so

im 𝐴 + im 𝐵 = 𝐹^{𝑛}

and

rank 𝐴 + rank 𝐵 \geq 𝑛

. Also

im 𝐵 \subseteq ker 𝐴

, so

rank 𝐵 \leq 𝑛 - rank 𝐴

. Hence

rank 𝐴 + rank 𝐵 = 𝑛

, and two out of three gives the direct sum. (b)

im 𝐵 \subseteq ker 𝐴

, with equal dimensions. (c) Symmetrically

im 𝐴 = ker 𝐵

. Without

𝐵 𝐴 = 0

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

𝐵 = (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})

(a) For $𝐱 \in 𝐹^{𝑛}$ , invertibility of $𝐴 + 𝐵$ gives $𝐲$ with $𝐱 =$ $(𝐴 + 𝐵) 𝐲 =$ $𝐴 𝐲 +$ $𝐵 𝐲$ , so $im 𝐴 + im 𝐵 = 𝐹^{𝑛}$ . By Theorem 4.4.13(a) and Theorem 4.2.7, $𝑛 \leq rank 𝐴 + rank 𝐵$ . On the other hand, $𝐴 (𝐵 𝐲) = 𝟎$ for every $𝐲$ , so $im 𝐵 \subseteq ker 𝐴$ , and by Proposition 4.1.21(a) and Theorem 4.2.11, $rank 𝐵 \leq$ $dim ker 𝐴 =$ $𝑛 -$ $rank 𝐴$ . Hence $rank 𝐴 + rank 𝐵 = 𝑛$ . Conditions (ii) and (iii) of Proposition 4.4.21 hold for $im 𝐴$ and $im 𝐵$ , so $𝐹^{𝑛} = im 𝐴 \oplus im 𝐵$ .

(b) $im 𝐵 \subseteq ker 𝐴$ , and $dim im 𝐵 =$ $rank 𝐵 =$ $𝑛 -$ $rank 𝐴 =$ $dim ker 𝐴$ , so $im 𝐵 = ker 𝐴$ by Proposition 4.1.21(b).

(c) If $𝐵 𝐴 = 0$ , the hypotheses are symmetric in $𝐴$ and $𝐵$ (note $𝐵 + 𝐴 = 𝐴 + 𝐵$ ), so (b) with the roles exchanged gives $im 𝐴 = ker 𝐵$ .

Without $𝐵 𝐴 = 0$ , let $𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})$ . Then $𝐴 𝐵 = 0$ and $𝐴 + 𝐵 = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ is invertible, with inverse $(\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix})$ , but $𝐵 𝐴 = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}) \neq 0$ , and $im 𝐴 = span (𝐞_{1})$ while $ker 𝐵 =$ $span ((1, - 1))$ .

The kernels are complementary in any case: $ker 𝐴 \cap$ $ker 𝐵 \subseteq$ $ker (𝐴 + 𝐵) =$ ${𝟎}$ , and by (a), $dim ker 𝐴 +$ $dim ker 𝐵 =$ $(𝑛 - rank 𝐴) +$ $(𝑛 - rank 𝐵) =$ $𝑛$ , so $𝐹^{𝑛} = ker 𝐴 \oplus ker 𝐵$ by Proposition 4.4.21, whether or not $𝐵 𝐴 = 0$ . In the example, $ker 𝐴 = span (𝐞_{2})$ and $ker 𝐵 =$ $span ((1, - 1))$ .

Exercise 4.4.49.hard Let $𝑉$ be finite-dimensional with $𝑉 = 𝑈 \oplus 𝑊_{0}$ .

(a)For $𝑆 \in L (𝑊_{0}, 𝑈)$ , let $Γ_{𝑆} =$ ${𝐰 + 𝑆 (𝐰) | 𝐰 \in 𝑊_{0}}$ . Show that $Γ_{𝑆}$ is a complement of $𝑈$ .
(b)Show that every complement of $𝑈$ equals $Γ_{𝑆}$ for exactly one $𝑆 \in L (𝑊_{0}, 𝑈)$ .
(c)Deduce that if $𝐹$ has $𝑞$ elements, $dim 𝑉 = 𝑛$ , and $dim 𝑈 = 𝑘$ , then $𝑈$ has exactly $𝑞^{𝑘 (𝑛 - 𝑘)}$ complements in $𝑉$ .
(d)List the complements of $𝑈 = span ((1, 1, 0))$ in $𝔽_{2}^{3}$ .

Solution.

Uses. Proposition 4.2.6 (Injective maps preserve dimension), Proposition 4.4.21 (Two out of three), Proposition 4.4.16 (Complements exist), Definition 4.4.24 (Projection along a complement), Proposition 4.4.26 (Properties of projections), Proposition 4.1.21 (Dimensions of subspaces), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 4.1.34 (Size of a vector space over a finite field)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. (a)

Γ_{𝑆}

is the image of the linear map

𝐰 \mapsto 𝐰 + 𝑆 (𝐰)

. Show that this map is injective and that

Γ_{𝑆} \cap 𝑈 = {𝟎}

, and count dimensions.

Hint 2. (b) Given a complement

𝑊

, decompose each

𝐰 \in 𝑊_{0}

along

𝑉 = 𝑈 \oplus 𝑊

. Its

𝑊

-component should be

𝐰 + 𝑆 (𝐰)

Hint 3. (c) A linear map

𝑊_{0} \to 𝑈

is determined by the images of a basis of

𝑊_{0}

, which can be any vectors of

𝑈

Sketch. (a) If

𝐰 + 𝑆 𝐰 = 𝟎

𝐰 + 𝑆 𝐰 \in 𝑈

, then

𝐰 \in 𝑈 \cap 𝑊_{0} = {𝟎}

; so

dim Γ_{𝑆} =

dim 𝑊_{0} =

𝑛 -

𝑘

and two out of three applies. (b) For a complement

𝑊

, let

𝑆 = - {Pr}_{𝑈, 𝑊} ↾_{𝑊_{0}}

, with values in

𝑈

. Then

𝐰 +

𝑆 (𝐰) =

{Pr}_{𝑊, 𝑈} (𝐰) \in

𝑊

, so

Γ_{𝑆} \subseteq 𝑊

, with equal dimensions. If

Γ_{𝑆} = Γ_{𝑆^{'}}

, compare components in

𝑊_{0} \oplus 𝑈

. (c)

| L (𝑊_{0}, 𝑈) | =

{(𝑞^{𝑘})}^{𝑛 - 𝑘}

. (d) The four planes

𝑥_{1} = 0

𝑥_{2} = 0

𝑥_{1} + 𝑥_{3} = 0

𝑥_{2} + 𝑥_{3} = 0

Let $𝑛 = dim 𝑉$ and $𝑘 = dim 𝑈$ , so $dim 𝑊_{0} = 𝑛 - 𝑘$ (Proposition 4.4.16(b)).

(a) The map $𝑔 : 𝑊_{0} \to 𝑉$ , $𝑔 (𝐰) = 𝐰 + 𝑆 (𝐰)$ , is linear with image $Γ_{𝑆}$ , so $Γ_{𝑆}$ is a subspace. If $𝑔 (𝐰) \in 𝑈$ , then $𝐰 =$ $𝑔 (𝐰) -$ $𝑆 (𝐰) \in$ $𝑈$ , as $𝑆 (𝐰) \in 𝑈$ ; so $𝐰 \in 𝑈 \cap 𝑊_{0} = {𝟎}$ . Taking $𝑔 (𝐰) = 𝟎$ shows that $𝑔$ is injective, so $dim Γ_{𝑆} = 𝑛 - 𝑘$ (Proposition 4.2.6(a)); and in general it shows $𝑈 \cap Γ_{𝑆} = {𝟎}$ . By Proposition 4.4.21, $𝑉 = 𝑈 \oplus Γ_{𝑆}$ .

(b) Existence. Let $𝑊$ be a complement of $𝑈$ . Define $𝑆 : 𝑊_{0} \to 𝑈$ by $𝑆 (𝐰) =$ $- {Pr}_{𝑈, 𝑊} (𝐰)$ ; it is linear with values in $𝑈$ (Proposition 4.4.26). For $𝐰 \in 𝑊_{0}$ ,

𝐰 + 𝑆 (𝐰) = 𝐰 - {Pr}_{𝑈, 𝑊} (𝐰) = {Pr}_{𝑊, 𝑈} (𝐰) \in 𝑊,

by Proposition 4.4.26(d). So $Γ_{𝑆} \subseteq 𝑊$ . Both have dimension $𝑛 - 𝑘$ , by (a) and Proposition 4.4.16(b), so $Γ_{𝑆} = 𝑊$ (Proposition 4.1.21(b)).

Uniqueness. Suppose $Γ_{𝑆} = Γ_{𝑆^{'}}$ , and let $𝐰 \in 𝑊_{0}$ . Then $𝐰 + 𝑆 (𝐰) \in Γ_{𝑆^{'}}$ , so $𝐰 +$ $𝑆 (𝐰) =$ $𝐰^{'} +$ $𝑆^{'} (𝐰^{'})$ for some $𝐰^{'} \in 𝑊_{0}$ . Both sides are decompositions along $𝑉 = 𝑊_{0} \oplus 𝑈$ , so comparing components gives $𝐰 = 𝐰^{'}$ and $𝑆 (𝐰) =$ $𝑆^{'} (𝐰^{'}) =$ $𝑆^{'} (𝐰)$ . Hence $𝑆 = 𝑆^{'}$ .

(c) By (a) and (b), $𝑆 \mapsto Γ_{𝑆}$ is a bijection from $L (𝑊_{0}, 𝑈)$ onto the set of complements of $𝑈$ . Fix a basis $𝐰_{1}, \dots, 𝐰_{𝑛 - 𝑘}$ of $𝑊_{0}$ . By Theorem 2.2.1, $𝑆 \mapsto$ $(𝑆 (𝐰_{1}), \dots, 𝑆 (𝐰_{𝑛 - 𝑘}))$ is a bijection from $L (𝑊_{0}, 𝑈)$ onto the set of $(𝑛 - 𝑘)$ -tuples of vectors of $𝑈$ . Since $| 𝑈 | = 𝑞^{𝑘}$ (Proposition 4.1.34), there are ${(𝑞^{𝑘})}^{𝑛 - 𝑘} =$ $𝑞^{𝑘 (𝑛 - 𝑘)}$ complements.

(d) Here $𝑛 = 3$ and $𝑘 = 1$ , so there are $2^{2} = 4$ complements. Take $𝑊_{0} =$ $span (𝐞_{2}, 𝐞_{3}) =$ ${𝐱 | 𝑥_{1} = 0}$ . Since $(1, 1, 0) \notin 𝑊_{0}$ , $𝑈 \cap 𝑊_{0} = {𝟎}$ , and $1 + 2 = 3$ , so $𝑊_{0}$ is a complement of $𝑈$ (Proposition 4.4.21). A map $𝑆 : 𝑊_{0} \to 𝑈$ is given by $𝑆 (𝐞_{2}) = 𝑎 (1, 1, 0)$ and $𝑆 (𝐞_{3}) = 𝑏 (1, 1, 0)$ with $𝑎, 𝑏 \in 𝔽_{2}$ , and $Γ_{𝑆} =$ $span (𝐞_{2} + 𝑎 (1, 1, 0), 𝐞_{3} + 𝑏 (1, 1, 0))$ . The four choices give

\begin{aligned} (𝑎, 𝑏) = (0, 0) & : span ((0, 1, 0), (0, 0, 1)) = {𝑥_{1} = 0}, \\ (𝑎, 𝑏) = (1, 0) & : span ((1, 0, 0), (0, 0, 1)) = {𝑥_{2} = 0}, \\ (𝑎, 𝑏) = (0, 1) & : span ((0, 1, 0), (1, 1, 1)) = {𝑥_{1} + 𝑥_{3} = 0}, \\ (𝑎, 𝑏) = (1, 1) & : span ((1, 0, 0), (1, 1, 1)) = {𝑥_{2} + 𝑥_{3} = 0} . \end{aligned}

These are exactly the planes of $𝔽_{2}^{3}$ that do not contain $(1, 1, 0)$ .

Exercise 4.4.50.hard Let $𝑉$ be finite-dimensional.

(a)Let $𝑈$ and $𝑊$ be subspaces of $𝑉$ with $dim 𝑈 = dim 𝑊$ . Show that $𝑈$ and $𝑊$ have a common complement, that is, a subspace $𝑋$ with $𝑉 = 𝑈 \oplus 𝑋 = 𝑊 \oplus 𝑋$ .
(b)Show that the three lines through $𝟎$ in $𝔽_{2}^{2}$ have no common complement.

Solution.

Uses. Exercise 1.4.21, Proposition 4.1.21 (Dimensions of subspaces), Theorem 4.4.13 (Bases and dimension of a sum), Proposition 4.4.21 (Two out of three), Proposition 4.4.16 (Complements exist)

Techniques. Dimension counting (Technique 4.2.24), Extend a basis of a subspace (Technique 4.1.18)

Hint 1. Induct on

𝑐 = dim 𝑉 - dim 𝑈

. If

𝑐 \geq 1

, find a vector

𝐯

that lies in neither

𝑈

nor

𝑊

Hint 2. Two proper subspaces never cover

𝑉

(Exercise 1.4.21). Enlarge

𝑈

and

𝑊

𝐯

, apply the inductive hypothesis, and add

𝐯

back to the common complement.

Hint 3. For (b), a complement of a line in

𝔽_{2}^{2}

is another line. How many lines are there?

Sketch. (a) Induct on

𝑐

. If

𝑐 = 0

, take

𝑋 = {𝟎}

. Otherwise

𝑈

𝑊

are proper, so some

𝐯

lies in neither.

𝑈^{'} = 𝑈 + span (𝐯)

and

𝑊^{'} = 𝑊 + span (𝐯)

have dimension

dim 𝑈 + 1

; take a common complement

𝑋^{'}

and set

𝑋 = 𝑋^{'} + span (𝐯)

. Then

dim 𝑋 = 𝑐

, and if

𝐱^{'} + 𝑎 𝐯 \in 𝑈

with

𝐱^{'} \in 𝑋^{'}

, then

𝐱^{'} \in 𝑈^{'} \cap 𝑋^{'} = {𝟎}

and

𝑎 𝐯 \in 𝑈

forces

𝑎 = 0

. Two out of three. (b) There are only three lines, and a common complement would be a fourth.

(a) Let $𝑛 = dim 𝑉$ and $𝑑 = dim 𝑈 = dim 𝑊$ . We prove the claim by induction on $𝑐 = 𝑛 - 𝑑$ , for all pairs of subspaces of $𝑉$ of equal dimension. If $𝑐 = 0$ , then $𝑈 = 𝑊 = 𝑉$ (Proposition 4.1.21(b)) and $𝑋 = {𝟎}$ works.

Let $𝑐 \geq 1$ . Then $𝑈$ and $𝑊$ are proper subspaces, so $𝑈 \cup 𝑊 \neq 𝑉$ by Exercise 1.4.21; choose $𝐯 \in 𝑉$ with $𝐯 \notin 𝑈$ and $𝐯 \notin 𝑊$ . If $𝑎 𝐯 \in 𝑈$ with $𝑎 \neq 0$ , then $𝐯 = 𝑎^{- 1} (𝑎 𝐯) \in 𝑈$ ; so $𝑈 \cap span (𝐯) = {𝟎}$ , and $𝑈^{'} = 𝑈 + span (𝐯)$ has dimension $𝑑 + 1$ by Theorem 4.4.13(c). Likewise $𝑊^{'} = 𝑊 + span (𝐯)$ has dimension $𝑑 + 1$ . By the inductive hypothesis there is $𝑋^{'}$ with $𝑉 =$ $𝑈^{'} \oplus$ $𝑋^{'} =$ $𝑊^{'} \oplus$ $𝑋^{'}$ , and $dim 𝑋^{'} = 𝑛 - 𝑑 - 1$ . Let $𝑋 = 𝑋^{'} + span (𝐯)$ . Since $𝐯 \in 𝑈^{'}$ , we have $𝑋^{'} \cap$ $span (𝐯) \subseteq$ $𝑋^{'} \cap$ $𝑈^{'} =$ ${𝟎}$ , so $dim 𝑋 = 𝑛 - 𝑑$ .

We show $𝑈 \cap 𝑋 = {𝟎}$ . Let $𝐲 = 𝐱^{'} + 𝑎 𝐯 \in 𝑈$ with $𝐱^{'} \in 𝑋^{'}$ and $𝑎 \in 𝐹$ . Then $𝐱^{'} = 𝐲 - 𝑎 𝐯$ lies in $𝑈^{'}$ , since $𝐲 \in 𝑈 \subseteq 𝑈^{'}$ and $𝐯 \in 𝑈^{'}$ . So $𝐱^{'} \in 𝑈^{'} \cap 𝑋^{'} = {𝟎}$ , and $𝑎 𝐯 = 𝐲 \in 𝑈$ , which forces $𝑎 = 0$ since $𝐯 \notin 𝑈$ . Hence $𝐲 = 𝟎$ . As $dim 𝑈 + dim 𝑋 = 𝑛$ , Proposition 4.4.21 gives $𝑉 = 𝑈 \oplus 𝑋$ . The same argument with $𝑊$ and $𝑊^{'}$ in place of $𝑈$ and $𝑈^{'}$ gives $𝑉 = 𝑊 \oplus 𝑋$ .

(b) Each line through $𝟎$ in $𝔽_{2}^{2}$ contains exactly one nonzero vector, so there are exactly three lines: $span (𝐞_{1})$ , $span (𝐞_{2})$ , $span (𝐞_{1} + 𝐞_{2})$ . A complement of a line has dimension $1$ (Proposition 4.4.16(b)) and is a line different from it. A common complement of all three would be a fourth line, which does not exist.

So (a) does not extend to three subspaces over every field. Over an infinite field it does extend to any finite number of subspaces of equal dimension, by the same induction together with Exercise 1.4.43.

Exercise 4.4.51.hard Let $𝐹$ be $ℚ$ , $ℝ$ , or $ℂ$ , and let $𝑃_{1},$ $\dots,$ $𝑃_{𝑘} \in$ $𝐹^{𝑛 \times 𝑛}$ satisfy $𝑃_{𝑖}^{2} = 𝑃_{𝑖}$ for each $𝑖$ and $𝑃_{1} + \dots + 𝑃_{𝑘} = 𝐼_{𝑛}$ .

(a)Show that $𝐹^{𝑛} =$ $im 𝑃_{1} \oplus$ $\dots \oplus$ $im 𝑃_{𝑘}$ .
(b)Deduce that $𝑃_{𝑖} 𝑃_{𝑗} = 0$ whenever $𝑖 \neq 𝑗$ .
(c)Show that the conclusion of (b) can fail over $𝔽_{2}$ .

Solution.

Uses. Exercise 4.3.43, Proposition 2.4.27 (The trace is linear), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Theorem 4.4.13 (Bases and dimension of a sum), Proposition 4.4.29 (Projections of a direct sum)

Techniques. Compare components (Technique 4.4.8), Split the space with an idempotent (Technique 4.4.28)

Hint 1. Every

𝐱

equals

\sum_{𝑖} 𝑃_{𝑖} 𝐱

, so the images add up to

𝐹^{𝑛}

. Directness is then a dimension count. How is

rank 𝑃_{𝑖}

related to

tr 𝑃_{𝑖}

over these fields?

Hint 2. For (b), decompose

𝑃_{𝑗} 𝐱

in two ways: as itself, a vector of

im 𝑃_{𝑗}

, and as

\sum_{𝑖} 𝑃_{𝑖} (𝑃_{𝑗} 𝐱)

Sketch. (a)

𝐱 = \sum_{𝑖} 𝑃_{𝑖} 𝐱

, so

\sum_{𝑖} im 𝑃_{𝑖} = 𝐹^{𝑛}

. By Exercise 4.3.43,

rank 𝑃_{𝑖} = tr 𝑃_{𝑖}

, so

\sum_{𝑖} dim im 𝑃_{𝑖} =

tr (\sum_{𝑖} 𝑃_{𝑖}) =

tr 𝐼_{𝑛} =

𝑛

, and Theorem 4.4.13(c) makes the sum direct. (b)

𝑃_{𝑗} 𝐱 =

\sum_{𝑖} 𝑃_{𝑖} 𝑃_{𝑗} 𝐱

has components

𝑃_{𝑖} 𝑃_{𝑗} 𝐱 \in im 𝑃_{𝑖}

, while

𝑃_{𝑗} 𝐱 \in im 𝑃_{𝑗}

also has the decomposition with

𝑃_{𝑗} 𝐱

in slot

𝑗

and

𝟎

elsewhere; compare components. (c) Over

𝔽_{2}

, take

𝑃_{1} = 𝑃_{2} = 𝑃_{3} = 𝐼_{1}

(a) For $𝐱 \in 𝐹^{𝑛}$ , $𝐱 =$ $𝐼_{𝑛} 𝐱 =$ $𝑃_{1} 𝐱 +$ $\dots +$ $𝑃_{𝑘} 𝐱$ , so $im 𝑃_{1} +$ $\dots +$ $im 𝑃_{𝑘} =$ $𝐹^{𝑛}$ . Over $ℚ$ , $ℝ$ , and $ℂ$ , an idempotent matrix has rank equal to its trace (Exercise 4.3.43(c)). So, using Theorem 4.2.7 and Proposition 2.4.27,

dim im 𝑃_{1} + \dots + dim im 𝑃_{𝑘} = tr 𝑃_{1} + \dots + tr 𝑃_{𝑘} = tr (𝑃_{1} + \dots + 𝑃_{𝑘}) = tr 𝐼_{𝑛} = 𝑛 .

This is the dimension of the sum, so the sum is direct by Theorem 4.4.13(c).

(b) Fix $𝑗$ and $𝐱 \in 𝐹^{𝑛}$ , and let $𝐲 = 𝑃_{𝑗} 𝐱 \in im 𝑃_{𝑗}$ . Since $\sum_{𝑖} 𝑃_{𝑖} = 𝐼_{𝑛}$ ,

𝐲 = 𝑃_{1} 𝐲 + \dots + 𝑃_{𝑘} 𝐲,

a decomposition of $𝐲$ along $im 𝑃_{1}, \dots, im 𝑃_{𝑘}$ . Another decomposition of $𝐲$ has component $𝐲$ in $im 𝑃_{𝑗}$ and $𝟎$ in every other $im 𝑃_{𝑖}$ . By (a) the two agree, so $𝑃_{𝑖} 𝐲 =$ $𝑃_{𝑖} 𝑃_{𝑗} 𝐱 =$ $𝟎$ for $𝑖 \neq 𝑗$ . As $𝐱$ was arbitrary, $𝑃_{𝑖} 𝑃_{𝑗} = 0$ .

By (b) and Proposition 4.4.29(b), the $𝑃_{𝑖}$ are the projections of the decomposition in (a).

(c) Over $𝔽_{2}$ , let $𝑛 = 1$ , $𝑘 = 3$ , and $𝑃_{1} =$ $𝑃_{2} =$ $𝑃_{3} =$ $𝐼_{1} =$ $(\begin{matrix} 1 \end{matrix})$ . Each is idempotent, and $𝑃_{1} +$ $𝑃_{2} +$ $𝑃_{3} =$ $(\begin{matrix} 1 + 1 + 1 \end{matrix}) =$ $𝐼_{1}$ , but $𝑃_{1} 𝑃_{2} = 𝐼_{1} \neq 0$ . What fails is (a): the ranks add up to $3 > 1$ , while the trace only records them modulo $2$ .

Exercise 4.4.52.challenge Let $𝐹$ be a field with $𝑞$ elements. Call subspaces $𝑈_{1}, \dots, 𝑈_{𝑚}$ of $𝐹^{4}$ a spread if each has dimension $2$ and any two of them are complements of each other.

(a)Show that a spread has at most $𝑞^{2} + 1$ members.
(b)For $𝐹 = 𝔽_{2}$ , construct a spread with $5$ members. (For $𝑀 \in 𝔽_{2}^{2 \times 2}$ , consider the subspace of vectors $(𝑥_{1}, 𝑥_{2}, 𝑦_{1}, 𝑦_{2})$ with $(𝑦_{1}, 𝑦_{2}) =$ $𝑀 (𝑥_{1}, 𝑥_{2})$ .)

Solution.

Uses. Proposition 4.1.34 (Size of a vector space over a finite field), Theorem 4.4.7 (Criteria for direct sums), Proposition 4.4.21 (Two out of three), Proposition 4.2.6 (Injective maps preserve dimension)

Techniques. Double counting (Technique 4.1.36)

Hint 1. Two members of a spread share only

𝟎

. Count nonzero vectors.

Hint 2. The subspace

Γ_{𝑀}

of vectors

(𝐱, 𝑀 𝐱)

meets

Γ_{𝑀^{'}}

only in

𝟎

when

𝑀 - 𝑀^{'}

is invertible. Add the subspace of vectors

(𝟎, 𝐲)

Hint 3. Over

𝔽_{2}

𝐶 = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})

satisfies

𝐶^{2} = 𝐶 + 𝐼_{2}

and

𝐶^{3} = 𝐼_{2}

. Try

𝑀 \in {0, 𝐼_{2}, 𝐶, 𝐶^{2}}

Sketch. (a) Each member has

𝑞^{2} - 1

nonzero vectors, no nonzero vector lies in two members, and

𝐹^{4}

has

𝑞^{4} - 1

nonzero vectors; so

𝑚 (𝑞^{2} - 1) \leq 𝑞^{4} - 1

, i.e.

𝑚 \leq 𝑞^{2} + 1

. (b) Take

Γ_{𝑀}

for

𝑀 \in {0, 𝐼_{2}, 𝐶, 𝐶^{2}}

and

Γ_{\infty} = {(𝟎, 𝐲)}

. Every difference of two of the four matrices is

𝐼_{2}

𝐶

, or

𝐶^{2}

, all invertible, so

Γ_{𝑀} \cap Γ_{𝑀^{'}} = {𝟎}

; and

Γ_{𝑀} \cap Γ_{\infty} = {𝟎}

. Two-dimensional subspaces of

𝔽_{2}^{4}

meeting trivially are complements.

Write vectors of $𝐹^{4}$ as $(𝐱, 𝐲)$ with $𝐱, 𝐲 \in 𝐹^{2}$ .

(a) Let $𝑈_{1}, \dots, 𝑈_{𝑚}$ be a spread. For $𝑖 \neq 𝑗$ , $𝑈_{𝑖} \cap 𝑈_{𝑗} = {𝟎}$ , so the sets $𝑈_{𝑖} ∖ {𝟎}$ are pairwise disjoint subsets of $𝐹^{4} ∖ {𝟎}$ . Each $𝑈_{𝑖}$ has $𝑞^{2}$ elements (Proposition 4.1.34), so

𝑚 (𝑞^{2} - 1) \leq 𝑞^{4} - 1 = (𝑞^{2} - 1) (𝑞^{2} + 1),

and $𝑚 \leq 𝑞^{2} + 1$ .

(b) For $𝑀 \in 𝔽_{2}^{2 \times 2}$ let $Γ_{𝑀} =$ ${(𝐱, 𝑀 𝐱) | 𝐱 \in 𝔽_{2}^{2}}$ , the image of the injective linear map $𝐱 \mapsto (𝐱, 𝑀 𝐱)$ , so $dim Γ_{𝑀} = 2$ (Proposition 4.2.6(a)). Let $Γ_{\infty} =$ ${(𝟎, 𝐲) | 𝐲 \in 𝔽_{2}^{2}}$ , also of dimension $2$ . Let

𝐶 = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}), 𝐶^{2} = (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) = 𝐶 + 𝐼_{2}, 𝐶^{3} = 𝐶 𝐶^{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = 𝐼_{2} .

So $𝐶$ and $𝐶^{2}$ are invertible, with $𝐶^{- 1} = 𝐶^{2}$ . We claim that $Γ_{0},$ $Γ_{𝐼_{2}},$ $Γ_{𝐶},$ $Γ_{𝐶^{2}},$ $Γ_{\infty}$ is a spread.

Since $- 1 = 1$ in $𝔽_{2}$ , the differences of two distinct matrices among $0, 𝐼_{2}, 𝐶, 𝐶^{2}$ are

𝐼_{2} - 0 = 𝐼_{2}, 𝐶 - 0 = 𝐶, 𝐶^{2} - 0 = 𝐶^{2}, 𝐶 - 𝐼_{2} = 𝐶^{2}, 𝐶^{2} - 𝐼_{2} = 𝐶, 𝐶^{2} - 𝐶 = 𝐼_{2},

all invertible. If $(𝐱, 𝑀 𝐱) =$ $(𝐱^{'}, 𝑀^{'} 𝐱^{'})$ with $𝑀 \neq 𝑀^{'}$ among them, then $𝐱 = 𝐱^{'}$ and $(𝑀 - 𝑀^{'}) 𝐱 = 𝟎$ , so $𝐱 =$ ${(𝑀 - 𝑀^{'})}^{- 1} 𝟎 =$ $𝟎$ . Hence $Γ_{𝑀} \cap Γ_{𝑀^{'}} = {𝟎}$ . Also, $(𝐱, 𝑀 𝐱) = (𝟎, 𝐲)$ forces $𝐱 = 𝟎$ and then $𝐲 = 𝟎$ , so $Γ_{𝑀} \cap Γ_{\infty} = {𝟎}$ . Any two of the five subspaces have dimensions adding up to $4$ and meet only in $𝟎$ , so they are complements of each other by Proposition 4.4.21. This spread has $2^{2} + 1$ members, so the bound in (a) is attained.

The four matrices $0, 𝐼_{2}, 𝐶, 𝐶^{2}$ , with matrix addition and multiplication, form the field with four elements (Section 14.3). The same construction, with $𝔽_{𝑞^{2}}$ realized as $𝑞^{2}$ matrices in $𝐹^{2 \times 2}$ , gives a spread of size $𝑞^{2} + 1$ for every $𝑞$ .

More practice

The practice bank has 29 exercises on this section’s topics.