1.4 Subspaces · Matrix Theory

Several of the vector spaces in Section 1.2 live inside a larger one and use its operations: $𝐶 [0, 1]$ inside $ℝ^{[0, 1]}$ , and ${𝐹 [𝑥]}_{\leq 𝑛}$ inside $𝐹 [𝑥]$ . So does every span. The span of a list in $𝑉$ is a subset of $𝑉$ , and by Lemma 1.3.9, sums and scalar multiples of linear combinations of the list are again linear combinations of it. This section names such subsets and gives a quick test for recognizing them. The lecture defines subspaces later (L11–12), alongside dimension. We meet them now because the span is their prototype, and because the kernel and the image of a linear map (Section 2.6) are subspaces the moment they are defined.

1.4.1The subspace criterion

Throughout, $𝑉$ is a vector space over a field $𝐹$ .

Definition 1.4.1 (Subspace). A subset

𝑊 \subseteq 𝑉

is a subspace of

𝑉

𝑊

is a vector space over

𝐹

under the operations of

𝑉

, restricted to

𝑊

. That is, for

𝐰, 𝐰^{'} \in 𝑊

and

𝑎 \in 𝐹

, the sum

𝐰 + 𝐰^{'}

and the scalar multiple

𝑎 𝐰

are computed in

𝑉

. A subspace

𝑊 \neq 𝑉

is a proper subspace.

So a subspace is a vector space in its own right, and its operations are not new: we add and scale elements of $𝑊$ exactly as we do in $𝑉$ . In particular, sums and scalar multiples of elements of $𝑊$ must lie in $𝑊$ , because closure is part of being a vector space (Remark 1.2.2).

Non-example 1.4.2 (A line that misses the origin). In

ℝ^{2}

, the line

𝐿 =

{(𝑐_{1}, 𝑐_{2}) | 𝑐_{1} + 𝑐_{2} = 1}

is not a subspace. The operations of

ℝ^{2}

are not even operations on

𝐿

: the vectors

(1, 0)

and

(0, 1)

lie on

𝐿

, but their sum

(1, 1)

does not, since

1 + 1 \neq 1

. Non-example 1.2.21 made the same observation for the line

𝑥 + 2 𝑦 = 2

. By contrast, the line

{(𝑟, 𝑟) | 𝑟 \in ℝ}

through the origin is a subspace (Example 1.4.8).

Checking all eight axioms for every candidate would be tedious, and most of the work is redundant.

Intuition (What has to be checked). Suppose

𝑊 \subseteq 𝑉

is closed under the operations of

𝑉

. Six of the axioms, (A1), (A2), and (A5)–(A8), are identities that hold for all vectors and scalars. An identity that holds for all vectors of

𝑉

holds in particular for all vectors of

𝑊

, so these six come for free. The other two axioms assert that something exists: (A3) a zero vector, and (A4) an additive inverse of each vector. These are where a subset can fail, because the vector that exists in

𝑉

might not lie in

𝑊

. Scalar multiplication produces both of them:

0 𝐰 = 𝟎

and

(- 1) 𝐰 = - 𝐰

(Proposition 1.2.32, Proposition 1.2.34). For the first we need some

𝐰 \in 𝑊

to start from, so

𝑊

must be nonempty. We used the same reasoning for subfields (Proposition 1.1.33) and for

𝐶 [0, 1]

(Example 1.2.12).

The lecture packages this as a criterion with two equivalent tests. The first speaks about linear combinations, which suits spans; the second is the one we check in practice.

Theorem 1.4.3 (Subspace criterion). Let

𝑊 \subseteq 𝑉

. The following are equivalent.

(I) $𝑊$ is a subspace of $𝑉$ .
(II) $𝑊 \neq \emptyset$ , and $𝑊$ is stable under linear combinations: for every $𝑘 \geq 1$ , all $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ , and all $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ , the vector $𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑘} 𝐰_{𝑘}$ lies in $𝑊$ .
(III) $𝟎 \in 𝑊$ , and $𝑊$ is closed under addition and scalar multiplication: $𝐰 + 𝐰^{'} \in 𝑊$ and $𝑎 𝐰 \in 𝑊$ for all $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ .

Here

𝟎

is the zero vector of

𝑉

Proof.routine

Uses. Definition 1.4.1 (Subspace), Definition 1.2.1 (Vector space), Proposition 1.2.32 (The scalar zero), Proposition 1.2.34 (Negatives and the scalar $- 1$ ), (1.3.1), induction on $𝑘$

Techniques. Proving TFAE (Technique 2.7.1)

Hint 1. Sort the axioms (A1)–(A8) into identities that hold “for all” vectors and scalars, and statements that something exists. Which kind can fail for a subset that is closed under the operations?

Hint 2. For (I)

\Rightarrow

(III), the subspace

𝑊

has a zero vector of its own, but (III) asks for the zero vector of

𝑉

. Multiply the zero vector of

𝑊

by the scalar

0

Hint 3. For (III)

\Rightarrow

(II), induct on

𝑘

and split off the last term. For (II)

\Rightarrow

(I), produce the zero vector and negatives inside

𝑊

0 𝐰_{0}

and

(- 1) 𝐰

Sketch. (I)

\Rightarrow

(III): the operations of

𝑊

land in

𝑊

, and

𝟎 = 0 \cdot 𝟎_{𝑊} \in 𝑊

. (III)

\Rightarrow

(II): induction on

𝑘

, writing a combination of length

𝑘 + 1

as one of length

𝑘

plus a single term. (II)

\Rightarrow

(I): sums and multiples are combinations of length

2

and

1

, so the operations restrict to

𝑊

; the identities (A1), (A2), (A5)–(A8) are inherited from

𝑉

; and

𝟎 = 0 𝐰_{0}

and

- 𝐰 = (- 1) 𝐰

lie in

𝑊

and supply (A3) and (A4).

We prove (I) $\Rightarrow$ (III) $\Rightarrow$ (II) $\Rightarrow$ (I).

(I) $\Rightarrow$ (III). Suppose $𝑊$ is a subspace. Since $𝑊$ is a vector space under the operations of $𝑉$ , these operations map $𝑊 \times 𝑊$ and $𝐹 \times 𝑊$ into $𝑊$ ; that is, $𝑊$ is closed under addition and scalar multiplication. By (A3) for the vector space $𝑊$ , there is a zero vector $𝟎_{𝑊} \in 𝑊$ . By closure under scalar multiplication, $0 \cdot 𝟎_{𝑊} \in 𝑊$ , and $0 \cdot 𝟎_{𝑊} = 𝟎$ by Proposition 1.2.32, applied in $𝑉$ . Hence $𝟎 \in 𝑊$ .

(III) $\Rightarrow$ (II). Since $𝟎 \in 𝑊$ , $𝑊 \neq \emptyset$ . We show by induction on $𝑘 \geq 1$ that $𝑎_{1} 𝐰_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐰_{𝑘} \in$ $𝑊$ for all $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ and $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ . For $𝑘 = 1$ this is closure under scalar multiplication. Suppose it holds for some $𝑘 \geq 1$ , and let $𝐰_{1}, \dots, 𝐰_{𝑘 + 1} \in 𝑊$ and $𝑎_{1}, \dots, 𝑎_{𝑘 + 1} \in 𝐹$ . By (1.3.1),

\sum_{𝑖 = 1}^{𝑘 + 1} 𝑎_{𝑖} 𝐰_{𝑖} = (\sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} 𝐰_{𝑖}) + 𝑎_{𝑘 + 1} 𝐰_{𝑘 + 1} .

The first summand lies in $𝑊$ by the inductive hypothesis, and the second by closure under scalar multiplication. So their sum lies in $𝑊$ , by closure under addition.

(II) $\Rightarrow$ (I). Let $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ . Then $𝐰 + 𝐰^{'} = 1 𝐰 + 1 𝐰^{'}$ by (A5), and $𝑎 𝐰$ are linear combinations of elements of $𝑊$ , of lengths $2$ and $1$ , so they lie in $𝑊$ by (II). Hence the operations of $𝑉$ restrict to functions $𝑊 \times 𝑊 \to 𝑊$ and $𝐹 \times 𝑊 \to 𝑊$ . Axioms (A1), (A2), and (A5)–(A8) are identities that hold for all vectors of $𝑉$ and all scalars, so in particular for all vectors of $𝑊$ . It remains to check (A3) and (A4).

(A3). Since $𝑊 \neq \emptyset$ , we may choose some $𝐰_{0} \in 𝑊$ . By (II), $0 𝐰_{0} \in 𝑊$ , and $0 𝐰_{0} = 𝟎$ by Proposition 1.2.32. So $𝟎 \in 𝑊$ , and $𝐰 + 𝟎 = 𝐰$ for every $𝐰 \in 𝑊$ , because this holds in $𝑉$ .

(A4). Let $𝐰 \in 𝑊$ . By (II), $(- 1) 𝐰 \in 𝑊$ , and $𝐰 +$ $(- 1) 𝐰 =$ $𝐰 +$ $(- 𝐰) =$ $𝟎$ by Proposition 1.2.34, where $𝟎$ is the zero vector of $𝑊$ just found.

So $𝑊$ satisfies all eight axioms, and $𝑊$ is a subspace of $𝑉$ . □

Remark 1.4.4 (The zero vector of a subspace). Let

𝑊

be a subspace of

𝑉

. The proof of (I)

\Rightarrow

(III) showed that

𝟎 \in 𝑊

. Since

𝐰 + 𝟎 = 𝐰

for every

𝐰 \in 𝑊

, the vector

𝟎

is a zero vector of the vector space

𝑊

, so it is the zero vector of

𝑊

(Proposition 1.2.27, applied in

𝑊

). That is,

𝟎_{𝑊} = 𝟎_{𝑉}

. Similarly, the additive inverse of

𝐰

𝑊

is the vector

- 𝐰

𝑉

, by Proposition 1.2.28. So the vectors whose existence (A3) and (A4) promise are also inherited from

𝑉

, and from now on we write

𝟎

for the zero vector of every subspace of

𝑉

Remark 1.4.5 (Why nonemptiness matters). The empty set is closed under addition and scalar multiplication, because there are no elements to test: “for all $𝐰, 𝐰^{'} \in \emptyset$ ” is vacuously true. Likewise, it is stable under linear combinations of lists of length $𝑘 \geq 1$ . But $\emptyset$ is not a vector space, since (A3) requires an element. So the conditions $𝑊 \neq \emptyset$ in (II) and $𝟎 \in 𝑊$ in (III) cannot be dropped, and the proof shows where they enter: nonemptiness was used exactly once, to produce the zero vector of $𝑊$ as $0 𝐰_{0}$ .

In practice, check $𝟎 \in 𝑊$ . It proves nonemptiness, and it is also the fastest test in the other direction: a set that does not contain $𝟎$ is not a subspace, whatever else it does. If (II) also allowed $𝑘 = 0$ , the empty linear combination, which represents $𝟎$ (Definition 1.3.4), would force $𝟎 \in 𝑊$ , and the condition $𝑊 \neq \emptyset$ would be automatic. Exercise 1.4.15 states the criterion in this form.

Exercise 1.4.6.routine Show that

𝑊 \subseteq 𝑉

is a subspace of

𝑉

if and only if

𝟎 \in 𝑊

and

𝑎 𝐰 + 𝐰^{'} \in 𝑊

for all

𝐰, 𝐰^{'} \in 𝑊

and all

𝑎 \in 𝐹

Solution.

Uses. Theorem 1.4.3 (Subspace criterion)

Hint 1. Compare with (II) and (III) of Theorem 1.4.3. The vector

𝑎 𝐰 + 𝐰^{'}

is a linear combination of length

2

Hint 2. For the converse, special values of

𝑎

and

𝐰^{'}

recover the two closure conditions of (III).

Sketch. If

𝑊

is a subspace, then

𝟎 \in 𝑊

by (III), and

𝑎 𝐰 + 1 𝐰^{'} \in 𝑊

by (II). Conversely,

𝑎 = 1

gives

𝐰 + 𝐰^{'} \in 𝑊

, and

𝐰^{'} = 𝟎

gives

𝑎 𝐰 = 𝑎 𝐰 + 𝟎 \in 𝑊

, so (III) holds.

Suppose $𝑊$ is a subspace. Then $𝟎 \in 𝑊$ by (III) of Theorem 1.4.3, and for $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ , the vector $𝑎 𝐰 +$ $𝐰^{'} =$ $𝑎 𝐰 +$ $1 𝐰^{'}$ is a linear combination of elements of $𝑊$ , so it lies in $𝑊$ by (II).

Conversely, suppose the condition holds, and let $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ . Taking the scalar $1$ gives $𝐰 + 𝐰^{'} = 1 𝐰 + 𝐰^{'} \in 𝑊$ , by (A5). Taking $𝟎$ as the second vector, which is allowed since $𝟎 \in 𝑊$ , gives $𝑎 𝐰 = 𝑎 𝐰 + 𝟎 \in 𝑊$ , by (A3). So $𝑊$ satisfies (III), and it is a subspace by Theorem 1.4.3.

This one-line form is convenient when $𝑊$ is defined by a condition that is visibly compatible with linear combinations, as for kernels in Section 2.6.

1.4.2Examples

To show that a subset is a subspace, we check (III). To show that it is not, one failure suffices: $𝟎 \notin 𝑊$ , or particular vectors and scalars whose sum or scalar multiple leaves $𝑊$ .

Example 1.4.7 (Trivial subspaces).

{𝟎}

and

𝑉

are subspaces of

𝑉

. For

𝑉

itself there is nothing to check. The set

{𝟎}

contains

𝟎

, and

𝟎 + 𝟎 = 𝟎

and

𝑎 𝟎 = 𝟎

by (A3) and Proposition 1.2.33. With the operations of

𝑉

{𝟎}

is the zero space (Example 1.2.3). By (III), every subspace of

𝑉

contains it.

Example 1.4.8 (One equation in $𝐹^{𝑛}$ ). Let $𝑎_{1}, \dots, 𝑎_{𝑛} \in 𝐹$ , and let

𝐻 = {𝐱 \in 𝐹^{𝑛} | 𝑎_{1} 𝑥_{1} + \dots + 𝑎_{𝑛} 𝑥_{𝑛} = 0} .

Then $𝐻$ is a subspace of $𝐹^{𝑛}$ . It contains $𝟎$ , since $𝑎_{1} 0 + \dots + 𝑎_{𝑛} 0 = 0$ . If $𝐱, 𝐲 \in 𝐻$ and $𝑐 \in 𝐹$ , then by the coordinatewise operations, distributivity, and regrouping,

\begin{aligned} \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} (𝑥_{𝑖} + 𝑦_{𝑖}) & = \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝑥_{𝑖} + \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝑦_{𝑖} = 0 + 0 = 0, \\ \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} (𝑐 𝑥_{𝑖}) & = 𝑐 \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝑥_{𝑖} = 𝑐 \cdot 0 = 0, \end{aligned}

so $𝐱 + 𝐲 \in 𝐻$ and $𝑐 𝐱 \in 𝐻$ . For $𝑛 = 2$ and the equation $𝑥_{1} - 𝑥_{2} = 0$ over $ℝ$ , $𝐻$ is the line ${(𝑟, 𝑟) | 𝑟 \in ℝ}$ . In $ℝ^{3}$ , the equation $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ gives a plane through $𝟎$ .

With a nonzero right-hand side, the solution set of $𝑎_{1} 𝑥_{1} +$ $\dots +$ $𝑎_{𝑛} 𝑥_{𝑛} =$ $𝑏$ is never a subspace, since it does not contain $𝟎$ : the left side at $𝟎$ is $0 \neq 𝑏$ . The solution set of a homogeneous system of $𝑚$ equations is the set of vectors that lie in $𝑚$ subspaces of the form $𝐻$ at once, one for each equation; it is a subspace by Proposition 1.4.16 below. Corollary 3.1.16 reaches the same conclusion by recognizing it as a kernel.

Example 1.4.9 (Polynomials). For $𝑛 \geq 0$ , ${𝐹 [𝑥]}_{\leq 𝑛}$ is a subspace of $𝐹 [𝑥]$ . It contains $𝟎$ , whose degree is $- \infty$ , and by Proposition 1.2.18(a) and (b), sums and nonzero multiples of polynomials of degree at most $𝑛$ have degree at most $𝑛$ ; the multiple $0 𝑝 = 𝟎$ does too.

Fix $𝑐 \in 𝐹$ . The polynomials that vanish at $𝑐$ form a subspace ${𝑝 \in 𝐹 [𝑥] | 𝑝 (𝑐) = 0}$ of $𝐹 [𝑥]$ . Indeed $𝟎 (𝑐) = 0$ , and if $𝑝 (𝑐) = 𝑞 (𝑐) = 0$ and $𝑎 \in 𝐹$ , then $(𝑝 + 𝑞) (𝑐) =$ $𝑝 (𝑐) +$ $𝑞 (𝑐) =$ $0$ and $(𝑎 𝑝) (𝑐) =$ $𝑎 𝑝 (𝑐) =$ $0$ by Proposition 1.2.18(c).

The polynomials of degree exactly $𝑛$ do not form a subspace, since $𝟎$ has degree $- \infty$ . Neither do the polynomials with $𝑝 (𝑐) = 1$ , since $𝟎 (𝑐) = 0$ .

Example 1.4.10 (Functions). Let $𝑋$ be a set and $𝑡_{0} \in 𝑋$ . The functions vanishing at $𝑡_{0}$ form a subspace of $𝐹^{𝑋}$ : the zero function vanishes at $𝑡_{0}$ , and if $𝑓 (𝑡_{0}) = 𝑔 (𝑡_{0}) = 0$ , then $(𝑓 + 𝑔) (𝑡_{0}) =$ $0 +$ $0 =$ $0$ and $(𝑎 𝑓) (𝑡_{0}) =$ $𝑎 \cdot$ $0 =$ $0$ by the pointwise operations. The functions with $𝑓 (𝑡_{0}) = 1$ do not, since the zero function is not among them.

In $ℝ^{ℝ}$ , the even functions $𝐸 =$ ${𝑓 | 𝑓 (- 𝑡) = 𝑓 (𝑡) for all 𝑡}$ and the odd functions $𝑂 =$ ${𝑓 | 𝑓 (- 𝑡) = - 𝑓 (𝑡) for all 𝑡}$ form subspaces. For $𝑂$ : the zero function is odd, as $0 = - 0$ . If $𝑓, 𝑔 \in 𝑂$ and $𝑎 \in ℝ$ , then for every $𝑡$ ,

(𝑓 + 𝑔) (- 𝑡) = 𝑓 (- 𝑡) + 𝑔 (- 𝑡) = - 𝑓 (𝑡) - 𝑔 (𝑡) = - (𝑓 + 𝑔) (𝑡), (𝑎 𝑓) (- 𝑡) = 𝑎 𝑓 (- 𝑡) = - (𝑎 𝑓) (𝑡) .

The argument for $𝐸$ is the same without the signs. Finally, $𝐶 [0, 1]$ is a subspace of $ℝ^{[0, 1]}$ (Example 1.2.12), and so are the continuous functions with integral $0$ .

Example 1.4.11 (Matrices). In $𝐹^{𝑛 \times 𝑛}$ , the symmetric matrices, those with $𝑎_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ for all $𝑖, 𝑗$ , form a subspace $𝑆$ , and so do the trace-zero matrices, those with $𝑎_{11} + \dots + 𝑎_{𝑛 𝑛} = 0$ . For $𝑛 = 2$ ,

𝑆 = {(\begin{matrix} 𝑎 & 𝑏 \\ 𝑏 & 𝑑 \end{matrix}) | 𝑎, 𝑏, 𝑑 \in 𝐹}, 𝑇 = {(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & - 𝑎 \end{matrix}) | 𝑎, 𝑏, 𝑐 \in 𝐹} .

Both are cut out by equations among the entries that the zero matrix satisfies, and that survive entrywise sums and scalar multiples: if $𝑎_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ and $𝑏_{𝑖 𝑗} = 𝑏_{𝑗 𝑖}$ , then $𝑎_{𝑖 𝑗} +$ $𝑏_{𝑖 𝑗} =$ $𝑎_{𝑗 𝑖} +$ $𝑏_{𝑗 𝑖}$ and $𝑐 𝑎_{𝑖 𝑗} = 𝑐 𝑎_{𝑗 𝑖}$ , and the trace-zero condition is the equation of Example 1.4.8 in the $𝑛^{2}$ entries. In the language of Section 2.4, $𝑆$ consists of the $𝐴$ with $𝐴^{𝖳} = 𝐴$ , and $𝑇$ of the $𝐴$ with $tr 𝐴 = 0$ .

Example 1.4.12 (The subspaces of $𝔽_{2}^{2}$ ). Over $𝔽_{2}$ , the only scalars are $0$ and $1$ , and $0 𝐰 = 𝟎$ , $1 𝐰 = 𝐰$ . So a subset of an $𝔽_{2}$ -vector space that contains $𝟎$ is automatically closed under scalar multiplication, and the subspaces are exactly the subsets that contain $𝟎$ and are closed under addition.

The space $𝔽_{2}^{2}$ has four elements, $𝟎$ , $𝐞_{1} = (1, 0)$ , $𝐞_{2} = (0, 1)$ , and $𝐞_{1} + 𝐞_{2} = (1, 1)$ . Each vector satisfies $𝐯 + 𝐯 = 𝟎$ , and the sum of any two different nonzero vectors is the third one. So a subset containing $𝟎$ and at most one nonzero vector $𝐯$ is closed under addition, since $𝐯 + 𝐯 = 𝟎$ and $𝐯 + 𝟎 = 𝐯$ . A subset containing two different nonzero vectors contains their sum, hence all three nonzero vectors, and it is all of $𝔽_{2}^{2}$ . Hence $𝔽_{2}^{2}$ has exactly five subspaces:

{𝟎}, {𝟎, (1, 0)}, {𝟎, (0, 1)}, {𝟎, (1, 1)}, 𝔽_{2}^{2} .

The middle three are the spans of the nonzero vectors (Example 1.3.11), the analogues of lines through the origin. Section 1.4.6 shows that every $𝐹^{2}$ looks like this.

Figure 1.4.1. Three subsets of

ℝ^{2}

that fail to be subspaces, each in a different way. Left: the line

𝑐_{1} + 𝑐_{2} = 1

misses

𝟎

; it also contains

(1, 0)

and

(0, 1)

but not their sum. Middle: the union of the axes contains

𝟎

and every multiple of its vectors, but not

(1, 0) + (0, 1)

. Right: the closed first quadrant contains

𝟎

and all sums of its vectors, but not

(- 1) 𝐯

Non-example 1.4.13 (Three ways to fail). Each of the following subsets of $ℝ^{2}$ violates a different condition of (III) (Figure 1.4.1).

(a)The line $𝐿 =$ ${(𝑐_{1}, 𝑐_{2}) | 𝑐_{1} + 𝑐_{2} = 1}$ does not contain $𝟎$ . (It fails the other two conditions as well.)
(b)The union of the axes, $𝐴 =$ ${(𝑥, 𝑦) | 𝑥 𝑦 = 0}$ , contains $𝟎$ and is closed under scalar multiplication: if $𝑥 𝑦 = 0$ , then $(𝑎 𝑥) (𝑎 𝑦) =$ $𝑎^{2} 𝑥 𝑦 =$ $0$ . But $(1, 0)$ and $(0, 1)$ lie in $𝐴$ , while their sum $(1, 1)$ does not.
(c)The first quadrant $𝑄 =$ ${(𝑥, 𝑦) | 𝑥 \geq 0, 𝑦 \geq 0}$ contains $𝟎$ and is closed under addition, since sums of nonnegative numbers are nonnegative. But $(1, 0) \in 𝑄$ , while $(- 1) (1, 0) =$ $(- 1, 0) \notin$ $𝑄$ .

The empty set satisfies both closure conditions and fails only $𝟎 \in 𝑊$ (Remark 1.4.5). Together with (b) and (c), this shows that none of the three conditions in (III) follows from the other two.

The set $𝐴$ in (b) is the union of two subspaces, the two axes. We return to it in Section 1.4.4.

1.4.3The span is the smallest subspace

Intuition. Let

𝑈

be a subspace of

𝑉

that contains the entries of a list

ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}

. By (II) of Theorem 1.4.3,

𝑈

contains every linear combination of

𝐯_{1}, \dots, 𝐯_{𝑘}

, so

span (ℒ︀) \subseteq 𝑈

. If

span (ℒ︀)

is itself a subspace, it is therefore the smallest subspace containing the entries of

ℒ︀

: it lies inside every other one. And it is a subspace, because sums and scalar multiples of linear combinations of

ℒ︀

are linear combinations of

ℒ︀

. So the span can be described from the inside, as the set of all linear combinations, or from the outside, as the least subspace that the entries of

ℒ︀

force on us.

Proposition 1.4.14 (The span is the smallest subspace containing a list). Let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ be a list in $𝑉$ .

(a) $span (ℒ︀)$ is a subspace of $𝑉$ , and $ℒ︀ \subseteq span (ℒ︀)$ .
(b)If $𝑈$ is a subspace of $𝑉$ and $ℒ︀ \subseteq 𝑈$ , then $span (ℒ︀) \subseteq 𝑈$ .

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion), Proposition 1.3.17 (A list lies in its span), Lemma 1.3.9 (Arithmetic of linear combinations), Definition 1.3.10 (Span)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. For (a), check (III) of Theorem 1.4.3. What is the sum of two linear combinations of

ℒ︀

Hint 2. For (b), an element of

span (ℒ︀)

is a linear combination of vectors that lie in

𝑈

. Which condition of Theorem 1.4.3 speaks about linear combinations? Treat the empty list separately.

Sketch. (a)

𝟎

and the entries of

ℒ︀

lie in

span (ℒ︀)

by Proposition 1.3.17, and by Lemma 1.3.9,

\sum 𝑎_{𝑖} 𝐯_{𝑖} +

\sum 𝑏_{𝑖} 𝐯_{𝑖} =

\sum (𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖}

and

𝑐 \sum 𝑎_{𝑖} 𝐯_{𝑖} =

\sum (𝑐 𝑎_{𝑖}) 𝐯_{𝑖}

. (b) The subspace

𝑈

satisfies (II) of Theorem 1.4.3 and contains

𝐯_{1}, \dots, 𝐯_{𝑘}

, so it contains every linear combination of them; if

𝑘 = 0

span (ℒ︀) = {𝟎} \subseteq 𝑈

(a) We check (III) of Theorem 1.4.3. By Proposition 1.3.17, $𝟎 \in span (ℒ︀)$ , and also $ℒ︀ \subseteq span (ℒ︀)$ . Let $𝐮, 𝐰 \in span (ℒ︀)$ and $𝑐 \in 𝐹$ , say $𝐮 =$ $\sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} 𝐯_{𝑖}$ and $𝐰 =$ $\sum_{𝑖 = 1}^{𝑘} 𝑏_{𝑖} 𝐯_{𝑖}$ . By Lemma 1.3.9,

𝐮 + 𝐰 = \sum_{𝑖 = 1}^{𝑘} (𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖} and 𝑐 𝐮 = \sum_{𝑖 = 1}^{𝑘} (𝑐 𝑎_{𝑖}) 𝐯_{𝑖},

which are linear combinations of $ℒ︀$ . So $𝐮 + 𝐰$ and $𝑐 𝐮$ lie in $span (ℒ︀)$ , and $span (ℒ︀)$ is a subspace.

(b) Let $𝑈$ be a subspace with $ℒ︀ \subseteq 𝑈$ , and let $𝐰 \in span (ℒ︀)$ . If $𝑘 = 0$ , then $𝐰 = 𝟎$ , which lies in $𝑈$ by (III) of Theorem 1.4.3. If $𝑘 \geq 1$ , then $𝐰 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘}$ for some scalars $𝑎_{𝑖}$ , where each $𝐯_{𝑖}$ lies in $𝑈$ . By (II) of Theorem 1.4.3, $𝐰 \in 𝑈$ . Hence $span (ℒ︀) \subseteq 𝑈$ . □

In particular, $span (ℒ︀)$ is a vector space in its own right, and $ℒ︀$ spans it. Part (b) gives the second proof of Proposition 1.3.18 promised in Section 1.3.4. If $ℳ︀ \subseteq span (ℒ︀)$ , then $span (ℒ︀)$ is a subspace, by (a), that contains the entries of $ℳ︀$ , so it contains $span (ℳ︀)$ , by (b). The first proof expanded every linear combination of $ℳ︀$ into one of $ℒ︀$ and collected coefficients. The second needs only that $span (ℒ︀)$ is closed under sums and scalar multiples; Theorem 1.4.3 upgrades this to closure under all linear combinations, by an induction done once and for all.

Part (b) also changes how we prove that a span lies in a given subspace: it suffices to check the finitely many entries of the list. For instance, the plane $𝑃 =$ ${𝐱 \in ℝ^{3} | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0}$ is a subspace (Example 1.4.8) that contains $(1, 0, - 1)$ and $(0, 1, - 1)$ , so it contains their span. In Exercise 1.3.16 the same containment took a computation with an arbitrary linear combination.

Exercise 1.4.15.routine Show that a subset

𝑊 \subseteq 𝑉

is a subspace of

𝑉

if and only if

span (ℒ︀) \subseteq 𝑊

for every list

ℒ︀

𝑊

, the empty list included.

Solution.

Uses. Proposition 1.4.14 (The span is the smallest subspace containing a list), Theorem 1.4.3 (Subspace criterion), Example 1.3.11 (Small spans)

Hint 1. One direction is Proposition 1.4.14(b).

Hint 2. For the other, apply the condition to three kinds of list: the empty list, a list

𝐰, 𝐰^{'}

of length

2

, and a list

𝐰

of length

1

Sketch. (

\Rightarrow

) is Proposition 1.4.14(b). (

\Leftarrow

)

span (()) = {𝟎}

gives

𝟎 \in 𝑊

;

𝐰 +

𝐰^{'} \in

span (𝐰, 𝐰^{'})

and

𝑎 𝐰 \in span (𝐰)

give the closure conditions, so (III) of Theorem 1.4.3 holds.

If $𝑊$ is a subspace and $ℒ︀$ is a list in $𝑊$ , then $span (ℒ︀) \subseteq 𝑊$ by Proposition 1.4.14(b).

Conversely, suppose $span (ℒ︀) \subseteq 𝑊$ for every list $ℒ︀$ in $𝑊$ . For the empty list, $span (()) = {𝟎}$ (Definition 1.3.10), so $𝟎 \in 𝑊$ . Let $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ . Then $𝐰 +$ $𝐰^{'} =$ $1 𝐰 +$ $1 𝐰^{'} \in$ $span (𝐰, 𝐰^{'}) \subseteq$ $𝑊$ and $𝑎 𝐰 \in span (𝐰) \subseteq 𝑊$ . So $𝑊$ satisfies (III) of Theorem 1.4.3, and it is a subspace.

In this form the criterion needs no separate nonemptiness condition: the empty list supplies $𝟎$ (Remark 1.4.5).

1.4.4Intersections and unions

Given two subspaces, we would like to build new ones from them. Intersections always work; unions almost never do.

Proposition 1.4.16 (Intersections of subspaces). Let ${(𝑈_{𝑖})}_{𝑖 \in 𝐼}$ be a family of subspaces of $𝑉$ , indexed by any set $𝐼$ . Then

⋂_{𝑖 \in 𝐼} 𝑈_{𝑖} = {𝐯 \in 𝑉 | 𝐯 \in 𝑈_{𝑖} for every 𝑖 \in 𝐼}

is a subspace of $𝑉$ .

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. Check the three conditions of (III) in Theorem 1.4.3 for the intersection.

Hint 2. A vector lies in the intersection exactly when it lies in every

𝑈_{𝑖}

, and each

𝑈_{𝑖}

satisfies (III) on its own.

Sketch.

𝟎

lies in every

𝑈_{𝑖}

. If

𝐮, 𝐰

lie in every

𝑈_{𝑖}

, then so do

𝐮 + 𝐰

and

𝑎 𝐮

, since each

𝑈_{𝑖}

is closed under the operations.

Write

𝑋 = ⋂_{𝑖 \in 𝐼} 𝑈_{𝑖}

. We check (III) of Theorem 1.4.3. Each

𝑈_{𝑖}

is a subspace, so

𝟎 \in 𝑈_{𝑖}

for every

𝑖

, and

𝟎 \in 𝑋

. Let

𝐮, 𝐰 \in 𝑋

and

𝑎 \in 𝐹

, and fix

𝑖 \in 𝐼

. Then

𝐮, 𝐰 \in 𝑈_{𝑖}

, and since

𝑈_{𝑖}

is closed under addition and scalar multiplication,

𝐮 + 𝐰 \in 𝑈_{𝑖}

and

𝑎 𝐮 \in 𝑈_{𝑖}

. As

𝑖

was arbitrary,

𝐮 + 𝐰 \in 𝑋

and

𝑎 𝐮 \in 𝑋

. So

𝑋

is a subspace. □

The family may be infinite, or even empty; for $𝐼 = \emptyset$ the condition “for every $𝑖 \in 𝐼$ ” is vacuous, the intersection is $𝑉$ , and the proof still applies. The polynomials of degree at most $𝑛$ that vanish at $𝑐$ form a subspace, as the intersection of two subspaces from Example 1.4.9, and the solution set of a homogeneous system is the intersection of the subspaces $𝐻$ of Example 1.4.8, one for each equation.

Intersections give a second description of the span, from the outside.

Corollary 1.4.17 (The span as an intersection). Let

ℒ︀

be a list in

𝑉

. Then

span (ℒ︀)

is the intersection of all subspaces of

𝑉

that contain the entries of

ℒ︀

Proof.routine

Uses. Proposition 1.4.14 (The span is the smallest subspace containing a list), Proposition 1.4.16 (Intersections of subspaces)

Techniques. Double containment (Technique 1.3.14)

Hint 1. Let

𝑋

be the intersection. Is

span (ℒ︀)

one of the subspaces being intersected?

Hint 2. Each subspace in the family contains

span (ℒ︀)

, by Proposition 1.4.14(b).

Sketch. Since

span (ℒ︀)

belongs to the family,

𝑋 \subseteq span (ℒ︀)

. Since every member contains

span (ℒ︀)

span (ℒ︀) \subseteq 𝑋

Let

𝑋

be the intersection of all subspaces

𝑈

𝑉

with

ℒ︀ \subseteq 𝑈

. We proceed by double containment. By Proposition 1.4.14(a),

span (ℒ︀)

is a member of the family, so

𝑋 \subseteq span (ℒ︀)

. By Proposition 1.4.14(b), every member

𝑈

satisfies

span (ℒ︀) \subseteq 𝑈

, so

span (ℒ︀) \subseteq 𝑋

. □

The description in Definition 1.3.10 builds the span by listing its elements, and it is the one we compute with. The description in Corollary 1.4.17 uses no linear combinations at all, and so it makes sense for any subset $𝑆 \subseteq 𝑉$ , finite or not: by Proposition 1.4.16, the intersection of all subspaces containing $𝑆$ is a subspace, and it is the smallest one containing $𝑆$ .

Unions

Now take two subspaces $𝑈$ and $𝑊$ . The union $𝑈 \cup 𝑊$ contains $𝟎$ , and it is closed under scalar multiplication, since a multiple of a vector of $𝑈$ lies in $𝑈$ and a multiple of a vector of $𝑊$ lies in $𝑊$ . Addition is the problem. The union of the two axes of $ℝ^{2}$ (Figure 1.4.1, middle) is not closed under addition: a vector on one axis plus a vector on the other lies on neither. Reading off exactly why this example fails is a move worth naming.

Technique 1.4.18 (Finding counterexamples and minimal examples). To disprove a claim about all objects of some kind, a single object for which it fails is enough. Look for one by shrinking everything: the smallest space, the fewest vectors, the simplest field (Technique 1.2.38). Then read the counterexample. Ask which features made it work; often they point to the exact hypothesis under which the claim does hold, and the counterexample, run on general objects, turns into a proof.

Reach for it whenever a claim seems plausible but no proof suggests itself, and in every “prove or disprove”. The canonical example is the claim that the union of two subspaces is a subspace. It fails for the two axes of $ℝ^{2}$ , since $(1, 0) + (0, 1)$ lies on neither (Non-example 1.4.13). The failure used only one vector of the first subspace outside the second and one vector of the second outside the first. So nested subspaces escape the counterexample, and Proposition 1.4.20 shows that only they do.

Used in 88 places and 24 practice exercises

Proposition 1.4.20, Exercise 1.4.26, Exercise 1.4.27, Exercise 1.4.37, Exercise 1.4.40, Exercise 1.5.16, Exercise 1.5.34, Exercise 1.6.35, Exercise 1.7.12, Exercise 1.7.15, Problem 1.10, Exercise 2.4.44, Exercise 2.5.34, Exercise 2.5.35, Exercise 2.6.36, Exercise 2.6.37, Exercise 2.6.38, Exercise 2.6.40, Exercise 2.6.50, Exercise 2.6.52, Exercise 2.7.21, Exercise 2.7.22, Exercise 2.7.27, Exercise 2.8.14, Exercise 2.8.15, Problem 2.10, Problem 2.14, Problem 2.20, Problem 3.7, Problem 3.8, Exercise 4.1.41, Exercise 4.1.48, Exercise 4.2.38, Exercise 4.2.45, Exercise 4.3.29, Exercise 4.3.32, Exercise 4.3.34, Exercise 4.3.35, Exercise 4.3.36, Exercise 4.4.34, Exercise 4.5.39, Exercise 4.5.46, Exercise 4.6.6, Exercise 4.6.51, Exercise 4.7.51, Exercise 4.8.5, Exercise 4.8.6, Exercise 4.8.11, Exercise 5.5.38, Exercise 5.5.49, Problem 5.4, Exercise 6.2.34, Exercise 6.3.46, Exercise 6.3.47, Exercise 6.3.55, Exercise 6.4.48, Exercise 6.4.50, Exercise 6.4.63, Exercise 6.5.35, Exercise 6.5.44, Exercise 6.6.36, Exercise 6.7.15, Problem 6.6, Problem 6.7, Exercise 7.1.53, Exercise 7.2.43, Exercise 7.2.53, Exercise 7.2.54, Exercise 7.3.38, Exercise 7.4.38, Exercise 7.5.33, Exercise 7.5.37, Exercise 7.6.43, Exercise 7.7.4, Exercise 7.7.5, Exercise 7.7.6, Exercise 7.7.8, Problem 7.5, Exercise 8.2.34, Exercise 8.3.29, Exercise 8.3.38, Exercise A.1.15, Exercise A.1.25, Exercise A.2.13, Exercise A.3.14, Exercise A.3.17, Exercise A.3.18, Exercise B.1.38; 24 practice exercises.

The key step in the counterexample is that a vector on one axis plus a vector off that axis lands off that axis. This holds for every subspace.

Lemma 1.4.19 (Adding a vector outside a subspace). Let

𝑈

be a subspace of

𝑉

, let

𝐮 \in 𝑈

, and let

𝐯 \in 𝑉

with

𝐯 \notin 𝑈

. Then

𝐮 + 𝐯 \notin 𝑈

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion), Proposition 1.2.34 (Negatives and the scalar $- 1$ )

Hint 1. Suppose

𝐮 + 𝐯 \in 𝑈

. Can you recover

𝐯

from two vectors of

𝑈

Hint 2.

𝐯 = (𝐮 + 𝐯) + (- 1) 𝐮

, a linear combination of two vectors of

𝑈

Sketch. If

𝐮 + 𝐯 \in 𝑈

, then

𝐯 = (𝐮 + 𝐯) + (- 1) 𝐮

lies in

𝑈

by closure, a contradiction.

Suppose, to the contrary, that $𝐮 + 𝐯 \in 𝑈$ . By Proposition 1.2.34 and the axioms,

(𝐮 + 𝐯) + (- 1) 𝐮 = (𝐯 + 𝐮) + (- 𝐮) = 𝐯 + (𝐮 + (- 𝐮)) = 𝐯 .

The left side is a linear combination of $𝐮 + 𝐯$ and $𝐮$ , which lie in $𝑈$ , so $𝐯 \in 𝑈$ by (II) of Theorem 1.4.3. This contradicts $𝐯 \notin 𝑈$ . □

Proposition 1.4.20 (Unions of two subspaces). Let

𝑈

and

𝑊

be subspaces of

𝑉

. Then

𝑈 \cup 𝑊

is a subspace of

𝑉

if and only if

𝑈 \subseteq 𝑊

𝑊 \subseteq 𝑈

Intuition. The backward direction is immediate, since

𝑈 \cup 𝑊

is then

𝑊

𝑈

. The forward direction concludes “

𝑈 \subseteq 𝑊

𝑊 \subseteq 𝑈

”, and there are two natural ways to prove a disjunction. One is to assume that the first alternative fails and deduce the second: a vector

𝐮 \in 𝑈

outside

𝑊

is a lever, and adding it to each vector of

𝑊

forces that vector into

𝑈

. The other is the contrapositive: assume both alternatives fail, and build the counterexample of the two axes from one witness on each side.

Proof (assume one alternative fails).standard

Uses. Theorem 1.4.3 (Subspace criterion), Lemma 1.4.19 (Adding a vector outside a subspace)

Hint 1. The backward direction is immediate: if

𝑈 \subseteq 𝑊

, then

𝑈 \cup 𝑊 = 𝑊

Hint 2. For the forward direction, assume

𝑈 ⊈ 𝑊

and prove

𝑊 \subseteq 𝑈

. Fix

𝐮 \in 𝑈

with

𝐮 \notin 𝑊

Hint 3. For

𝐰 \in 𝑊

, the sum

𝐮 + 𝐰

lies in

𝑈 \cup 𝑊

. Which of the two can it lie in? Use Lemma 1.4.19.

Sketch. (

\Leftarrow

)

𝑈 \cup 𝑊

𝑊

𝑈

. (

\Rightarrow

) Suppose

𝑈 \cup 𝑊

is a subspace and

𝑈 ⊈ 𝑊

; fix

𝐮 \in 𝑈 ∖ 𝑊

. For

𝐰 \in 𝑊

𝐮 + 𝐰 \in 𝑈 \cup 𝑊

by closure, and

𝐮 + 𝐰 \notin 𝑊

by Lemma 1.4.19 (applied to

𝑊

, with

𝐰 \in 𝑊

and

𝐮 \notin 𝑊

). So

𝐮 + 𝐰 \in 𝑈

, and

𝐰 = (𝐮 + 𝐰) - 𝐮 \in 𝑈

. Hence

𝑊 \subseteq 𝑈

( $\Leftarrow$ ) If $𝑈 \subseteq 𝑊$ , then $𝑈 \cup 𝑊 = 𝑊$ , and if $𝑊 \subseteq 𝑈$ , then $𝑈 \cup 𝑊 = 𝑈$ . Either way, $𝑈 \cup 𝑊$ is a subspace.

( $\Rightarrow$ ) Suppose $𝑈 \cup 𝑊$ is a subspace. If $𝑈 \subseteq 𝑊$ , we are done. Otherwise, fix a vector $𝐮 \in 𝑈$ with $𝐮 \notin 𝑊$ ; we show $𝑊 \subseteq 𝑈$ . Let $𝐰 \in 𝑊$ . Both $𝐮$ and $𝐰$ lie in the subspace $𝑈 \cup 𝑊$ , so $𝐮 + 𝐰 \in 𝑈 \cup 𝑊$ . By Lemma 1.4.19, applied to the subspace $𝑊$ , the vector $𝐰 \in 𝑊$ , and the vector $𝐮 \notin 𝑊$ , we have $𝐰 + 𝐮 \notin 𝑊$ . Hence $𝐮 + 𝐰 \in 𝑈$ . Since $𝐮 \in 𝑈$ as well, $𝐰 =$ $(𝐮 + 𝐰) +$ $(- 1) 𝐮 \in$ $𝑈$ by (II) of Theorem 1.4.3. As $𝐰$ was arbitrary, $𝑊 \subseteq 𝑈$ . □

Proof (generalize the counterexample).standard

Uses. Theorem 1.4.3 (Subspace criterion), Lemma 1.4.19 (Adding a vector outside a subspace)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. The backward direction is immediate. For the forward direction, prove the contrapositive: if neither subspace contains the other, then

𝑈 \cup 𝑊

is not closed under addition.

Hint 2. Imitate the two axes: take

𝐮 \in 𝑈

outside

𝑊

and

𝐰 \in 𝑊

outside

𝑈

, and locate

𝐮 + 𝐰

Sketch. If

𝑈 ⊈ 𝑊

and

𝑊 ⊈ 𝑈

, pick

𝐮 \in 𝑈 ∖ 𝑊

and

𝐰 \in 𝑊 ∖ 𝑈

. By Lemma 1.4.19, applied once to

𝑈

and once to

𝑊

𝐮 + 𝐰

lies in neither

𝑈

nor

𝑊

. So

𝑈 \cup 𝑊

is not closed under addition.

The backward direction is as in the first proof. For the forward direction, we prove the contrapositive. Suppose

𝑈 ⊈ 𝑊

and

𝑊 ⊈ 𝑈

, and pick

𝐮 \in 𝑈

with

𝐮 \notin 𝑊

and

𝐰 \in 𝑊

with

𝐰 \notin 𝑈

. By Lemma 1.4.19 applied to

𝑈

, with

𝐮 \in 𝑈

and

𝐰 \notin 𝑈

, we get

𝐮 + 𝐰 \notin 𝑈

. By Lemma 1.4.19 applied to

𝑊

, with

𝐰 \in 𝑊

and

𝐮 \notin 𝑊

, we get

𝐰 + 𝐮 \notin 𝑊

. So

𝐮

and

𝐰

lie in

𝑈 \cup 𝑊

, but

𝐮 + 𝐰

does not. Hence

𝑈 \cup 𝑊

is not closed under addition, and it is not a subspace by Theorem 1.4.3. □

The second proof is the counterexample of the two axes, with $𝑈$ and $𝑊$ in place of the axes and the two witnesses in place of $(1, 0)$ and $(0, 1)$ . The first proof shows a different habit: to prove “ $𝑃$ or $𝑄$ ”, assume $𝑃$ fails and use the failure. Here the failure of $𝑈 \subseteq 𝑊$ supplied a single vector $𝐮$ , and that one vector moved every vector of $𝑊$ into $𝑈$ .

Exercise 1.4.21.routine Let

𝑈

and

𝑊

be subspaces of

𝑉

with

𝑈 \cup 𝑊 = 𝑉

. Show that

𝑈 = 𝑉

𝑊 = 𝑉

Solution.

Uses. Proposition 1.4.20 (Unions of two subspaces), Example 1.4.7 (Trivial subspaces)

Hint 1.

𝑉

is a subspace of itself, so

𝑈 \cup 𝑊

is a subspace.

Hint 2. Apply Proposition 1.4.20. If

𝑈 \subseteq 𝑊

, what is

𝑈 \cup 𝑊

Sketch.

𝑈 \cup 𝑊 = 𝑉

is a subspace, so

𝑈 \subseteq 𝑊

𝑊 \subseteq 𝑈

by Proposition 1.4.20. In the first case

𝑉 = 𝑈 \cup 𝑊 = 𝑊

, in the second

𝑉 = 𝑈

Since $𝑈 \cup 𝑊 = 𝑉$ is a subspace of $𝑉$ , Proposition 1.4.20 gives $𝑈 \subseteq 𝑊$ or $𝑊 \subseteq 𝑈$ . If $𝑈 \subseteq 𝑊$ , then $𝑉 = 𝑈 \cup 𝑊 = 𝑊$ . If $𝑊 \subseteq 𝑈$ , then $𝑉 = 𝑈 \cup 𝑊 = 𝑈$ .

So a vector space is never the union of two proper subspaces. Three can suffice: $𝔽_{2}^{2}$ is the union of its three lines ${𝟎, (1, 0)}$ , ${𝟎, (0, 1)}$ , ${𝟎, (1, 1)}$ (Example 1.4.12). Exercise 1.4.42 and Exercise 1.4.43 show how this depends on the size of the field.

Unions of chains

A union of subspaces can still be a subspace when the subspaces are nested. Closure under addition involves only two vectors at a time, so only two members of the family at a time, and for nested subspaces Proposition 1.4.20 applies to each pair.

Proposition 1.4.22 (Unions of chains). Let

{(𝑈_{𝑖})}_{𝑖 \in 𝐼}

be a family of subspaces of

𝑉

, indexed by a nonempty set

𝐼

, such that for all

𝑖, 𝑗 \in 𝐼

𝑈_{𝑖} \subseteq 𝑈_{𝑗}

𝑈_{𝑗} \subseteq 𝑈_{𝑖}

. Then

⋃_{𝑖 \in 𝐼} 𝑈_{𝑖}

is a subspace of

𝑉

. In particular, if

𝑈_{1} \subseteq 𝑈_{2} \subseteq 𝑈_{3} \subseteq \dots

are subspaces of

𝑉

, then

⋃_{𝑛 \geq 1} 𝑈_{𝑛}

is a subspace of

𝑉

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. Check (III) of Theorem 1.4.3. Two vectors of the union lie in two members of the family.

Hint 2. If

𝐮 \in 𝑈_{𝑖}

and

𝐰 \in 𝑈_{𝑗}

, one of

𝑈_{𝑖}

𝑈_{𝑗}

contains both vectors.

Sketch.

𝟎

lies in any member. If

𝐮 \in 𝑈_{𝑖}

and

𝐰 \in 𝑈_{𝑗}

, say with

𝑈_{𝑖} \subseteq 𝑈_{𝑗}

, then

𝐮 + 𝐰 \in 𝑈_{𝑗}

. Multiples of

𝐮

stay in

𝑈_{𝑖}

. For an increasing sequence,

𝑈_{𝑚} \subseteq 𝑈_{𝑛}

whenever

𝑚 \leq 𝑛

, so any two members are nested.

Write $𝑋 = ⋃_{𝑖 \in 𝐼} 𝑈_{𝑖}$ . We check (III) of Theorem 1.4.3. Since $𝐼$ is nonempty, there is some $𝑖_{0} \in 𝐼$ , and $𝟎 \in 𝑈_{𝑖_{0}} \subseteq 𝑋$ . Let $𝐮, 𝐰 \in 𝑋$ and $𝑎 \in 𝐹$ , say $𝐮 \in 𝑈_{𝑖}$ and $𝐰 \in 𝑈_{𝑗}$ . Then $𝑎 𝐮 \in 𝑈_{𝑖} \subseteq 𝑋$ . By hypothesis, $𝑈_{𝑖} \subseteq 𝑈_{𝑗}$ or $𝑈_{𝑗} \subseteq 𝑈_{𝑖}$ ; exchanging the names of $𝐮$ and $𝐰$ if necessary, we may assume $𝑈_{𝑖} \subseteq 𝑈_{𝑗}$ . Then $𝐮$ and $𝐰$ both lie in the subspace $𝑈_{𝑗}$ , so $𝐮 + 𝐰 \in 𝑈_{𝑗} \subseteq 𝑋$ . So $𝑋$ is a subspace.

For an increasing sequence $𝑈_{1} \subseteq 𝑈_{2} \subseteq \dots$ , we have $𝑈_{𝑚} \subseteq 𝑈_{𝑛}$ whenever $𝑚 \leq 𝑛$ , by induction on $𝑛 - 𝑚$ , so any two members are nested and the first part applies. □

Example 1.4.23 (

𝐹 [𝑥]

as a union of a chain). The subspaces

{𝐹 [𝑥]}_{\leq 0} \subseteq

{𝐹 [𝑥]}_{\leq 1} \subseteq

{𝐹 [𝑥]}_{\leq 2} \subseteq

\dots

of Example 1.4.9 form an increasing sequence, and their union is all of

𝐹 [𝑥]

, since every polynomial has a degree, which is

- \infty

or some

𝑛 \geq 0

. Each

{𝐹 [𝑥]}_{\leq 𝑛}

is a proper subspace, as

𝑥^{𝑛 + 1} \notin {𝐹 [𝑥]}_{\leq 𝑛}

. So

𝐹 [𝑥]

is the union of infinitely many proper subspaces. By Exercise 1.4.21, two proper subspaces never suffice, and over an infinite field no finite number does (Exercise 1.4.43).

1.4.5Sums of subspaces

Since $𝑈 \cup 𝑊$ is usually not a subspace, we ask instead for the smallest subspace that contains both $𝑈$ and $𝑊$ . Any subspace containing $𝑈$ and $𝑊$ contains every sum $𝐮 + 𝐰$ with $𝐮 \in 𝑈$ and $𝐰 \in 𝑊$ , and these sums already form a subspace.

Proposition 1.4.24 (The sum of two subspaces). Let $𝑈$ and $𝑊$ be subspaces of $𝑉$ , and let

𝑈 + 𝑊 = {𝐮 + 𝐰 | 𝐮 \in 𝑈, 𝐰 \in 𝑊} .

(a) $𝑈 + 𝑊$ is a subspace of $𝑉$ that contains $𝑈$ and $𝑊$ .
(b)If $𝑋$ is a subspace of $𝑉$ with $𝑈 \subseteq 𝑋$ and $𝑊 \subseteq 𝑋$ , then $𝑈 + 𝑊 \subseteq 𝑋$ .

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion), Definition 1.2.1 (Vector space)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. For (a), check (III) of Theorem 1.4.3. Regroup

(𝐮 + 𝐰) + (𝐮^{'} + 𝐰^{'})

so that the vectors of

𝑈

are together.

Hint 2. To see

𝑈 \subseteq 𝑈 + 𝑊

, write

𝐮 = 𝐮 + 𝟎

. For (b), use that

𝑋

is closed under addition.

Sketch. (a)

𝟎 = 𝟎 + 𝟎

;

(𝐮 + 𝐰) +

(𝐮^{'} + 𝐰^{'}) =

(𝐮 + 𝐮^{'}) +

(𝐰 + 𝐰^{'})

and

𝑎 (𝐮 + 𝐰) =

𝑎 𝐮 +

𝑎 𝐰

; and

𝐮 = 𝐮 + 𝟎

𝐰 = 𝟎 + 𝐰

. (b) If

𝐮, 𝐰 \in 𝑋

, then

𝐮 + 𝐰 \in 𝑋

(a) We check (III) of Theorem 1.4.3. Since $𝟎 \in 𝑈$ and $𝟎 \in 𝑊$ , $𝟎 = 𝟎 + 𝟎 \in 𝑈 + 𝑊$ . Let $𝐮 + 𝐰$ and $𝐮^{'} + 𝐰^{'}$ be elements of $𝑈 + 𝑊$ , with $𝐮, 𝐮^{'} \in 𝑈$ and $𝐰, 𝐰^{'} \in 𝑊$ , and let $𝑎 \in 𝐹$ . By (A1) and (A2),

(𝐮 + 𝐰) + (𝐮^{'} + 𝐰^{'}) = (𝐮 + 𝐮^{'}) + (𝐰 + 𝐰^{'}),

with $𝐮 + 𝐮^{'} \in 𝑈$ and $𝐰 + 𝐰^{'} \in 𝑊$ , and by (A7), $𝑎 (𝐮 + 𝐰) =$ $𝑎 𝐮 +$ $𝑎 𝐰$ , with $𝑎 𝐮 \in 𝑈$ and $𝑎 𝐰 \in 𝑊$ . So both lie in $𝑈 + 𝑊$ , and $𝑈 + 𝑊$ is a subspace. Finally, for $𝐮 \in 𝑈$ we have $𝐮 = 𝐮 + 𝟎 \in 𝑈 + 𝑊$ , and for $𝐰 \in 𝑊$ we have $𝐰 = 𝟎 + 𝐰 \in 𝑈 + 𝑊$ , by (A3) and (A1).

(b) Let $𝐮 + 𝐰 \in 𝑈 + 𝑊$ , with $𝐮 \in 𝑈 \subseteq 𝑋$ and $𝐰 \in 𝑊 \subseteq 𝑋$ . Since $𝑋$ is closed under addition, $𝐮 + 𝐰 \in 𝑋$ . □

So $𝑈 + 𝑊$ is the smallest subspace containing $𝑈 \cup 𝑊$ , and $𝑈 \cup 𝑊 \subseteq 𝑈 + 𝑊$ with equality exactly when one of $𝑈$ , $𝑊$ contains the other (Proposition 1.4.20). Sums of spans are spans: $span (ℒ︀) +$ $span (ℳ︀) =$ $span (ℒ︀, ℳ︀)$ by Exercise 1.3.36. For example, in $𝐹^{3}$ the sum of the $𝑥_{1}$ -axis $span (𝐞_{1})$ and the $𝑥_{2}$ -axis $span (𝐞_{2})$ is $span (𝐞_{1}, 𝐞_{2})$ , the plane $𝑥_{3} = 0$ , while their union is just the two axes. Sums of more than two subspaces, the case where every vector of $𝑈 + 𝑊$ has only one decomposition $𝐮 + 𝐰$ , and the dimension of $𝑈 + 𝑊$ are the subject of Section 4.4.

1.4.6Subspaces of the plane

In $ℝ^{2}$ we have met three kinds of subspace: ${𝟎}$ , lines through the origin, and $ℝ^{2}$ itself. There are no others, over any field.

Intuition. A subspace that contains one nonzero vector contains the whole line through it, and a subspace that contains two vectors pointing in different directions contains the whole plane. There is no room in between. The proof makes this precise with Proposition 1.4.14(b), which turns “contains these vectors” into “contains their span”, and with Exercise 1.3.37, which says when two vectors span

𝐹^{2}

Theorem 1.4.25 (The subspaces of

𝐹^{2}

). The subspaces of

𝐹^{2}

are

{𝟎}

, the lines

span (𝐯)

with

𝐯 \neq 𝟎

, and

𝐹^{2}

. Two such lines

span (𝐯)

and

span (𝐯^{'})

are equal if and only if

𝐯^{'} = 𝑐 𝐯

for some

𝑐 \neq 0

Proof.standard

Uses. Proposition 1.4.14 (The span is the smallest subspace containing a list), Exercise 1.3.37, Exercise 1.3.28, Example 1.4.7 (Trivial subspaces)

Hint 1. Each of the listed sets is a subspace. For the converse, let

𝑈

be a subspace and split into cases according to how much of

𝐹^{2}

it contains.

Hint 2. If

𝑈 \neq {𝟎}

, pick

𝐯 \in 𝑈

with

𝐯 \neq 𝟎

. Then

span (𝐯) \subseteq 𝑈

. If this is not all of

𝑈

, pick

𝐰 \in 𝑈

outside

span (𝐯)

Hint 3. By Exercise 1.3.37(b),

𝐯, 𝐰

spans

𝐹^{2}

. Now apply Proposition 1.4.14(b) again.

Sketch. The listed sets are subspaces by Example 1.4.7 and Proposition 1.4.14(a). Let

𝑈

be a subspace other than

{𝟎}

, and pick

𝐯 \neq 𝟎

𝑈

, so

span (𝐯) \subseteq 𝑈

. Either

𝑈 = span (𝐯)

, or there is

𝐰 \in 𝑈 ∖ span (𝐯)

, and then

𝐹^{2} =

span (𝐯, 𝐰) \subseteq

𝑈

by Exercise 1.3.37(b). The statement about equal lines is Exercise 1.3.28(b).

The sets ${𝟎}$ and $𝐹^{2}$ are subspaces (Example 1.4.7), and so is every span (Proposition 1.4.14(a)). Conversely, let $𝑈$ be a subspace of $𝐹^{2}$ . If $𝑈 = {𝟎}$ , we are done. Otherwise, choose $𝐯 \in 𝑈$ with $𝐯 \neq 𝟎$ . By Proposition 1.4.14(b), $span (𝐯) \subseteq 𝑈$ . If $𝑈 = span (𝐯)$ , then $𝑈$ is a line. Otherwise, choose $𝐰 \in 𝑈$ with $𝐰 \notin span (𝐯)$ . Since $𝐯 \neq 𝟎$ and $𝐰 \notin span (𝐯)$ , the list $𝐯, 𝐰$ spans $𝐹^{2}$ by Exercise 1.3.37(b). Both entries lie in $𝑈$ , so by Proposition 1.4.14(b),

𝐹^{2} = span (𝐯, 𝐰) \subseteq 𝑈,

and $𝑈 = 𝐹^{2}$ . The criterion for two lines to be equal is Exercise 1.3.28(b). □

Both vectors lie on the line

span ((2, 1))

, so their span is that line. In the interactive version, drag the second vector off the line: the span jumps from the line to all of

ℝ^{2}

, with nothing in between.

Over $ℝ$ , the lines $span (𝐯)$ are the lines through the origin. Over $𝔽_{2}$ each line has two elements, and the theorem recovers the five subspaces of Example 1.4.12. Over a field with $𝑞$ elements, $𝐹^{2}$ has $𝑞 + 1$ lines (Exercise 1.4.33).

The proof was a count in disguise: $𝑈$ contains no direction, one direction, or two. Section 4.1 makes this count precise. Every subspace $𝑈$ of $𝐹^{𝑛}$ has a dimension between $0$ and $𝑛$ , with $𝑈 = 𝐹^{𝑛}$ only for dimension $𝑛$ (Proposition 4.1.21). In particular, the subspaces of $ℝ^{3}$ are ${𝟎}$ , the lines and the planes through $𝟎$ , and $ℝ^{3}$ .

Exercises

Exercise 1.4.26.routine Decide which of the following are subspaces, and justify each answer.

(a) ${𝐱 \in ℝ^{3} | 𝑥_{1} = 2 𝑥_{3}}$ .
(b) ${𝐱 \in ℝ^{3} | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 1}$ .
(c) ${𝐱 \in ℝ^{3} | 𝑥_{1} 𝑥_{2} = 0}$ .
(d) ${𝐱 \in ℝ^{3} | 𝑥_{1} \leq 𝑥_{2}}$ .
(e) ${𝐱 \in ℝ^{3} | 𝑥_{1}^{2} + 𝑥_{2}^{2} = 0}$ , and the subset of $ℂ^{3}$ defined by the same equation.

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Example 1.4.8 (One equation in $𝐹^{𝑛}$ ), Proposition 1.4.16 (Intersections of subspaces)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Check whether

𝟎

lies in the set first. When it does, test closure on simple vectors, such as standard vectors and their negatives.

Hint 2. For (e), over

ℝ

a sum of two squares vanishes only when both squares do. Over

ℂ

, look for a nonzero solution of

𝑧^{2} + 𝑤^{2} = 0

Sketch. (a) Yes, an equation as in Example 1.4.8. (b) No,

𝟎

is missing. (c) No:

𝐞_{1} + 𝐞_{2}

. (d) No:

(- 1) 𝐞_{2}

. (e) Over

ℝ

the set is

{𝑥_{1} = 𝑥_{2} = 0}

, a subspace. Over

ℂ

it contains

(1, 𝑖, 0)

and

(1, - 𝑖, 0)

but not their sum

(2, 0, 0)

(a) Yes. It is the solution set of $𝑥_{1} + 0 𝑥_{2} - 2 𝑥_{3} = 0$ , a subspace by Example 1.4.8.

(b) No. The coordinates of $𝟎$ add up to $0 \neq 1$ .

(c) No. The vectors $(1, 0, 0)$ and $(0, 1, 0)$ lie in the set, but their sum $(1, 1, 0)$ does not, since $1 \cdot 1 \neq 0$ .

(d) No. The vector $(0, 1, 0)$ lies in the set, since $0 \leq 1$ , but $(- 1) (0, 1, 0) =$ $(0, - 1, 0)$ does not, since $0 > - 1$ .

(e) For real numbers, $𝑥_{1}^{2} \geq 0$ and $𝑥_{2}^{2} \geq 0$ , so $𝑥_{1}^{2} + 𝑥_{2}^{2} = 0$ holds exactly when $𝑥_{1} = 𝑥_{2} = 0$ . So the subset of $ℝ^{3}$ is ${𝐱 | 𝑥_{1} = 0} \cap$ ${𝐱 | 𝑥_{2} = 0}$ , an intersection of two subspaces of the form in Example 1.4.8, hence a subspace by Proposition 1.4.16. In $ℂ^{3}$ , the vectors $(1, 𝑖, 0)$ and $(1, - 𝑖, 0)$ satisfy the equation, since $1 + 𝑖^{2} = 1 + {(- 𝑖)}^{2} = 0$ , but their sum $(2, 0, 0)$ gives $4 + 0 \neq 0$ . So the subset of $ℂ^{3}$ is not a subspace. The same equation defines a subspace over one field and not over the other.

Exercise 1.4.27.routine Decide which of the following are subspaces, and justify each answer. Write $𝑝 = \sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ for polynomials.

(a) ${𝑓 \in ℝ^{ℝ} | 𝑓 (0) = 𝑓 (1)}$ .
(b) ${𝑓 \in ℝ^{ℝ} | 𝑓 (1) = {𝑓 (0)}^{2}}$ .
(c)The bounded functions in $ℝ^{ℝ}$ , those $𝑓$ for which there is $𝑀$ with $| 𝑓 (𝑡) | \leq 𝑀$ for all $𝑡$ .
(d) ${𝑝 \in ℝ [𝑥] | 𝑝 (1) 𝑝 (2) = 0}$ .
(e) ${𝑝 \in ℝ [𝑥] | 𝑎_{𝑘} = 0 for every odd 𝑘}$ .
(f) ${𝐴 \in 𝐹^{2 \times 2} | 𝑎_{11} 𝑎_{22} = 𝑎_{12} 𝑎_{21}}$ .

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Definition 1.2.9 (Function spaces), Definition 1.2.13 (Polynomials), Definition 1.2.17 (Evaluation), Proposition 1.2.18 (Degree and evaluation versus the operations), Proposition 1.4.20 (Unions of two subspaces)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. The sets defined by linear conditions, such as an equation between values or coefficients, are subspaces. For the others, look for constants, simple polynomials, or diagonal matrices that break closure.

Hint 2. For (b), try the constant function

1

and its double. For (d), try

𝑥 - 1

and

𝑥 - 2

. For (f), try the two diagonal matrices with a single

1

Sketch. (a) Yes: the condition survives sums and multiples. (b) No: the constant

1

is in it, the constant

2

is not. (c) Yes:

| 𝑓 + 𝑔 | \leq 𝑀 + 𝑁

and

| 𝑎 𝑓 | \leq | 𝑎 | 𝑀

. (d) No:

(𝑥 - 1) +

(𝑥 - 2) =

2 𝑥 -

3

is nonzero at

1

and

2

. (e) Yes: coefficientwise operations. (f) No:

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

(a) Yes. The zero function satisfies $0 = 0$ . If $𝑓 (0) = 𝑓 (1)$ and $𝑔 (0) = 𝑔 (1)$ , then $(𝑓 + 𝑔) (0) =$ $𝑓 (1) +$ $𝑔 (1) =$ $(𝑓 + 𝑔) (1)$ and $(𝑎 𝑓) (0) =$ $𝑎 𝑓 (1) =$ $(𝑎 𝑓) (1)$ .

(b) No. The constant function $1$ lies in the set, since $1 = 1^{2}$ , but $2 \cdot 1$ is the constant function $2$ , and $2 \neq 2^{2}$ .

(c) Yes. The zero function is bounded by $𝑀 = 0$ . If $| 𝑓 (𝑡) | \leq 𝑀$ and $| 𝑔 (𝑡) | \leq 𝑁$ for all $𝑡$ , then $| 𝑓 (𝑡) +$ $𝑔 (𝑡) | \leq$ $𝑀 +$ $𝑁$ and $| 𝑎 𝑓 (𝑡) | \leq | 𝑎 | 𝑀$ for all $𝑡$ , by the triangle inequality for real numbers.

(d) No. The polynomials $𝑥 - 1$ and $𝑥 - 2$ lie in the set, but their sum $𝑝 = 2 𝑥 - 3$ has $𝑝 (1) 𝑝 (2) =$ $(- 1) \cdot$ $1 =$ $- 1$ . In fact, since $ℝ$ has no zero divisors, the set is the union of the subspaces ${𝑝 (1) = 0}$ and ${𝑝 (2) = 0}$ of Example 1.4.9, and neither contains the other ( $𝑥 - 1$ is in the first only, $𝑥 - 2$ in the second only), so Proposition 1.4.20 also shows that it is not a subspace.

(e) Yes. The operations of $ℝ [𝑥]$ are coefficientwise, so if $𝑎_{𝑘} = 𝑏_{𝑘} = 0$ for every odd $𝑘$ , then $𝑎_{𝑘} + 𝑏_{𝑘} = 0$ and $𝑐 𝑎_{𝑘} = 0$ for every odd $𝑘$ ; and $𝟎$ has all coefficients $0$ .

(f) No, over every field. The matrices $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $(\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ satisfy the equation, since $1 \cdot 0 = 0 \cdot 0$ , but their sum $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ gives $1 \cdot 1 \neq 0 \cdot 0$ .

Exercise 1.4.28.standard By Proposition 1.2.19, $ℂ^{2}$ is a vector space over $ℂ$ and also over $ℝ$ , with the same addition. For $𝑧 = 𝑎 + 𝑏 𝑖$ with $𝑎, 𝑏 \in ℝ$ , write $𝑧 = 𝑎 - 𝑏 𝑖$ .

(a)Show that every subspace of $ℂ^{2}$ over $ℂ$ is also a subspace of $ℂ^{2}$ over $ℝ$ .
(b)For each of the sets $ℝ^{2} =$ ${(𝑎, 𝑏) | 𝑎, 𝑏 \in ℝ}$ and $𝐷 = {(𝑧, 𝑧) | 𝑧 \in ℂ}$ , decide whether it is a subspace of $ℂ^{2}$ over $ℝ$ , and whether it is a subspace over $ℂ$ .

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Proposition 1.2.19 (Restriction of scalars)

Hint 1. The conditions

𝟎 \in 𝑊

and closure under addition do not depend on the field. Only closure under scalar multiplication does, and over

ℝ

there are fewer scalars to test.

Hint 2. For (b), test multiplication by

𝑖

. For

𝐷

over

ℝ

, check that

𝑧 + 𝑤 = 𝑧 + 𝑤

and

𝑟 𝑧 = 𝑟 𝑧

for real

𝑟

Sketch. (a) A set closed under multiplication by all complex scalars is closed under multiplication by the real ones. (b)

ℝ^{2}

is a real subspace but not a complex one:

𝑖 (1, 0) = (𝑖, 0)

𝐷

is a real subspace, since conjugation is additive and commutes with real scalars, but not a complex one:

(1, 1) \in 𝐷

and

𝑖 (1, 1) =

(𝑖, 𝑖) \notin

𝐷

(a) Let $𝑊$ be a subspace of $ℂ^{2}$ over $ℂ$ . Then $𝟎 \in 𝑊$ , $𝑊$ is closed under addition, and $𝑐 𝐰 \in 𝑊$ for all $𝑐 \in ℂ$ and $𝐰 \in 𝑊$ , in particular for all real $𝑐$ . The real vector space $ℂ^{2}$ has the same addition, and its scalar multiplication is the complex one restricted to real scalars. So $𝑊$ satisfies (III) of Theorem 1.4.3 over $ℝ$ .

(b) $ℝ^{2}$ . It contains $𝟎 = (0, 0)$ , sums of pairs of real numbers are pairs of real numbers, and so are real multiples. So $ℝ^{2}$ is a subspace of $ℂ^{2}$ over $ℝ$ . It is not a subspace over $ℂ$ : $(1, 0) \in ℝ^{2}$ but $𝑖 (1, 0) = (𝑖, 0) \notin ℝ^{2}$ .

$𝐷$ . For $𝑧 = 𝑎 + 𝑏 𝑖$ and $𝑤 = 𝑐 + 𝑑 𝑖$ ,

𝑧 + 𝑤 = (𝑎 + 𝑐) - (𝑏 + 𝑑) 𝑖 = 𝑧 + 𝑤,

and for real $𝑟$ , $𝑟 𝑧 =$ $𝑟 𝑎 -$ $𝑟 𝑏 𝑖 =$ $𝑟 𝑧$ . So $𝐷$ contains $(0, 0) = 𝟎$ , and

(𝑧, 𝑧) + (𝑤, 𝑤) = (𝑧 + 𝑤, 𝑧 + 𝑤), 𝑟 (𝑧, 𝑧) = (𝑟 𝑧, 𝑟 𝑧)

lie in $𝐷$ . Hence $𝐷$ is a subspace over $ℝ$ . It is not a subspace over $ℂ$ : $(1, 1) = (1, 1) \in 𝐷$ , but $𝑖 (1, 1) = (𝑖, 𝑖)$ , and $𝑖 = - 𝑖 \neq 𝑖$ .

So the converse of (a) fails.

Exercise 1.4.29.routine Find the error in the following argument.

Claim. The set $𝑃 = {𝑡 \in ℝ | 𝑡 > 0}$ is a subspace of the real vector space $ℝ$ .

Proof. For $𝑠, 𝑡 \in 𝑃$ and $𝑐 \in ℝ$ , define $𝑠 \oplus 𝑡 = 𝑠 𝑡$ and $𝑐 ⊙ 𝑡 = 𝑡^{𝑐}$ . With these operations, $𝑃$ is a real vector space, with zero vector $1$ . Hence $𝑃$ is a subspace of $ℝ$ .

(You may take for granted that $𝑃$ is a real vector space under $\oplus$ and $⊙$ .)

Solution.

Uses. Definition 1.4.1 (Subspace), Theorem 1.4.3 (Subspace criterion), Remark 1.4.4 (The zero vector of a subspace)

Hint 1. Reread Definition 1.4.1. Under which operations must a subspace be a vector space?

Hint 2. Test

𝑃

against (III) of Theorem 1.4.3, with the usual operations of

ℝ

Sketch. A subspace must be a vector space under the operations of

ℝ

itself, not under new ones. Under the usual operations,

0 \notin 𝑃

and

(- 1) \cdot 1 = - 1 \notin 𝑃

. The claimed zero vector

1

is also a giveaway: a subspace’s zero is the zero of

ℝ

(Remark 1.4.4).

The argument shows that $𝑃$ is a vector space, but under operations that are not those of $ℝ$ . Definition 1.4.1 requires $𝑃$ to be a vector space under the addition and scalar multiplication of $ℝ$ , restricted to $𝑃$ . Under these, $𝑃$ fails (III) of Theorem 1.4.3: $0 \notin 𝑃$ , and $1 \in 𝑃$ while $(- 1) \cdot 1 = - 1 \notin 𝑃$ . So $𝑃$ is not a subspace of $ℝ$ .

The zero vector $1$ is a symptom of the error: by Remark 1.4.4, the zero vector of a subspace of $ℝ$ must be the real number $0$ .

Exercise 1.4.30.routine Find the error in the following argument.

Claim. If $𝑈$ and $𝑊$ are subspaces of $𝑉$ , then $𝑈 \cup 𝑊$ is a subspace of $𝑉$ .

Proof. Since $𝟎 \in 𝑈$ , we have $𝟎 \in 𝑈 \cup 𝑊$ . Let $𝐯 \in 𝑈 \cup 𝑊$ and $𝑎 \in 𝐹$ . Then $𝐯$ lies in $𝑈$ or in $𝑊$ , and in either case $𝑎 𝐯$ lies in the same subspace, hence in $𝑈 \cup 𝑊$ . Finally, let $𝐯, 𝐰 \in 𝑈 \cup 𝑊$ . Each of them lies in $𝑈$ or in $𝑊$ , and $𝑈$ and $𝑊$ are closed under addition, so $𝐯 + 𝐰 \in 𝑈$ or $𝐯 + 𝐰 \in 𝑊$ . Hence $𝐯 + 𝐰 \in 𝑈 \cup 𝑊$ .

Solution.

Uses. Proposition 1.4.20 (Unions of two subspaces), Lemma 1.4.19 (Adding a vector outside a subspace), Non-example 1.4.13 (Three ways to fail)

Hint 1. Two of the three steps are correct. Which one quietly assumes something about where

𝐯

and

𝐰

lie?

Hint 2. Closure of

𝑈

under addition says something about two vectors of

𝑈

. What if

𝐯 \in 𝑈

and

𝐰 \in 𝑊

Sketch. The last step applies closure of

𝑈

(or of

𝑊

) to two vectors that need not both lie in

𝑈

(or both in

𝑊

). In the mixed case

𝐯 \in 𝑈 ∖ 𝑊

𝐰 \in 𝑊 ∖ 𝑈

, the sum lies in neither (Lemma 1.4.19); the axes of

ℝ^{2}

are a counterexample to the claim.

The first two steps are correct:

𝑈 \cup 𝑊

always contains

𝟎

and is closed under scalar multiplication. The error is in the last step. Closure of

𝑈

under addition applies when

𝐯

and

𝐰

both lie in

𝑈

, and closure of

𝑊

when both lie in

𝑊

. The argument ignores the mixed case,

𝐯 \in 𝑈

and

𝐰 \in 𝑊

with

𝐯 \notin 𝑊

and

𝐰 \notin 𝑈

, and in that case

𝐯 + 𝐰

lies in neither subspace, by Lemma 1.4.19. The claim is false: for the two axes of

ℝ^{2}

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

lies on neither axis (Non-example 1.4.13). By Proposition 1.4.20, it holds exactly when one subspace contains the other, which is exactly when the mixed case cannot occur.

Exercise 1.4.31.standard

(a)Let $𝑝$ be a prime, and let $𝑉$ be a vector space over $𝔽_{𝑝}$ . Show that a nonempty subset $𝑊 \subseteq 𝑉$ that is closed under addition is a subspace.
(b)Show that (a) fails over $ℚ$ : find a nonempty subset of $ℚ^{2}$ that is closed under addition and under $𝐯 \mapsto - 𝐯$ , but is not a subspace of the $ℚ$ -vector space $ℚ^{2}$ .

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Definition 1.1.19 (Integers modulo $𝑛$ ), Definition 1.2.1 (Vector space)

Hint 1. In

𝔽_{𝑝}

, every scalar is a sum of

1

′s:

𝑐 = 1 + \dots + 1

(

𝑐

terms) for

1 \leq 𝑐 \leq 𝑝 - 1

, and

0

is the sum of

𝑝

ones.

Hint 2. By (A8) and (A5),

(1 + \dots + 1) 𝐰 = 𝐰 + \dots + 𝐰

. For (b), think of the integers inside

ℚ

Sketch. (a) For

𝐰 \in 𝑊

and

𝑚 \geq 1

𝑚 * 𝐰 \in 𝑊

by closure, and

(𝑚 * 1) 𝐰 = 𝑚 * 𝐰

by (A8) and (A5). Every

𝑐 \in 𝔽_{𝑝}

equals

𝑚 * 1

for some

1 \leq 𝑚 \leq 𝑝

, so

𝑐 𝐰 \in 𝑊

; with

𝑐 = 0

this gives

𝟎 \in 𝑊

. (b)

ℤ^{2}

: closed under sums and negatives, but

\frac{1}{2} (1, 0) \notin ℤ^{2}

(a) For a positive integer $𝑚$ , write $𝑚 * 𝐰 = 𝐰 + \dots + 𝐰$ ( $𝑚$ terms) and $𝑚 * 1 = 1 + \dots + 1$ in $𝔽_{𝑝}$ . We claim that $(𝑚 * 1) 𝐰 = 𝑚 * 𝐰$ , by induction on $𝑚$ . For $𝑚 = 1$ this is (A5). For $𝑚 \geq 2$ , by (A8), the inductive hypothesis, and (A5),

(𝑚 * 1) 𝐰 = ((𝑚 - 1) * 1 + 1) 𝐰 = (𝑚 - 1) * 𝐰 + 1 𝐰 = 𝑚 * 𝐰 .

Now let $𝐰 \in 𝑊$ , which exists since $𝑊 \neq \emptyset$ . By closure under addition and induction on $𝑚$ , $𝑚 * 𝐰 \in 𝑊$ for every $𝑚 \geq 1$ . Every $𝑐 \in 𝔽_{𝑝}$ is of the form $𝑚 * 1$ with $1 \leq 𝑚 \leq 𝑝$ : for $𝑐 \in {1, \dots, 𝑝 - 1}$ , the sum of $𝑐$ ones in $𝔽_{𝑝}$ is $𝑐$ , and the sum of $𝑝$ ones is $𝑝 mod 𝑝 = 0$ . So $𝑐 𝐰 = 𝑚 * 𝐰 \in 𝑊$ for every $𝑐 \in 𝔽_{𝑝}$ . With $𝑐 = 0$ this gives $𝟎 = 0 𝐰 \in 𝑊$ . Hence $𝑊$ satisfies (III) of Theorem 1.4.3.

(b) Let $𝑊 =$ $ℤ^{2} =$ ${(𝑚, 𝑛) | 𝑚, 𝑛 \in ℤ}$ . It contains $(0, 0)$ , and sums and negatives of pairs of integers are pairs of integers. But $(1, 0) \in 𝑊$ and $\frac{1}{2} (1, 0) = (\frac{1}{2}, 0) \notin 𝑊$ , so $𝑊$ is not a subspace. The proof of (a) fails because $\frac{1}{2}$ is not a sum of $1$ ′s in $ℚ$ .

Exercise 1.4.32.standard For which primes

𝑝

𝑆_{𝑝} =

{(𝑥, 𝑦) \in 𝔽_{𝑝}^{2} | 𝑥^{2} = 𝑦^{2}}

a subspace of

𝔽_{𝑝}^{2}

Solution.

Uses. Exercise 1.1.40, Proposition 1.4.20 (Unions of two subspaces), Example 1.4.8 (One equation in $𝐹^{𝑛}$ )

Techniques. Try small cases and small fields (Technique 1.2.38)

Hint 1. Factor:

𝑥^{2} = 𝑦^{2}

if and only if

𝑥 = 𝑦

𝑥 = - 𝑦

(Exercise 1.1.40). So

𝑆_{𝑝}

is a union of two lines.

Hint 2. When are the two lines

𝑥 = 𝑦

and

𝑥 = - 𝑦

the same line? If they differ, apply Proposition 1.4.20.

Sketch. By Exercise 1.1.40,

𝑆_{𝑝} = 𝐷 \cup 𝐷^{'}

with

𝐷 = {𝑥 = 𝑦}

and

𝐷^{'} = {𝑥 = - 𝑦}

, both subspaces. For

𝑝 = 2

- 1 = 1

, so

𝑆_{2} = 𝐷

is a subspace. For odd

𝑝

1 \neq - 1

, so

(1, 1) \in 𝐷 ∖ 𝐷^{'}

and

(1, - 1) \in 𝐷^{'} ∖ 𝐷

, and

𝑆_{𝑝}

is not a subspace by Proposition 1.4.20; indeed

(1, 1) +

(1, - 1) =

(2, 0) \notin

𝑆_{𝑝}

. Answer:

𝑝 = 2

only.

By Exercise 1.1.40, $𝑥^{2} = 𝑦^{2}$ holds if and only if $𝑥 = 𝑦$ or $𝑥 = - 𝑦$ . So $𝑆_{𝑝} = 𝐷 \cup 𝐷^{'}$ , where

𝐷 = {(𝑥, 𝑦) | 𝑥 - 𝑦 = 0}, 𝐷^{'} = {(𝑥, 𝑦) | 𝑥 + 𝑦 = 0}

are subspaces of $𝔽_{𝑝}^{2}$ by Example 1.4.8.

If $𝑝 = 2$ , then $- 1 = 1$ in $𝔽_{2}$ , so $𝐷^{'} = 𝐷$ and $𝑆_{2} = 𝐷$ is a subspace. (It is ${𝟎, (1, 1)}$ .)

If $𝑝$ is odd, then $1 + 1 = 2 \neq 0$ in $𝔽_{𝑝}$ , so $1 \neq - 1$ . Then $(1, 1) \in 𝐷$ but $(1, 1) \notin 𝐷^{'}$ , since $1 + 1 \neq 0$ ; and $(1, - 1) \in 𝐷^{'}$ but $(1, - 1) \notin 𝐷$ , since $1 - (- 1) = 2 \neq 0$ . So neither of $𝐷$ , $𝐷^{'}$ contains the other, and $𝑆_{𝑝} = 𝐷 \cup 𝐷^{'}$ is not a subspace by Proposition 1.4.20. Concretely, $(1, 1) +$ $(1, - 1) =$ $(2, 0)$ , and $2^{2} = 4 \neq 0 = 0^{2}$ in $𝔽_{𝑝}$ .

So $𝑆_{𝑝}$ is a subspace exactly when $𝑝 = 2$ .

Exercise 1.4.33.standard

(a)List all subspaces of $𝔽_{3}^{2}$ , writing each line as a set of three vectors.
(b)Let $𝐹$ be a finite field with $𝑞$ elements. Show that $𝐹^{2}$ has exactly $𝑞 + 1$ lines $span (𝐯)$ , $𝐯 \neq 𝟎$ , and hence exactly $𝑞 + 3$ subspaces.

Solution.

Uses. Theorem 1.4.25 (The subspaces of $𝐹^{2}$ ), Exercise 1.3.28, Corollary 1.2.36 (Cancelling scalars and vectors)

Hint 1. By Theorem 1.4.25, it suffices to count lines. How many elements does a line

span (𝐯)

have, and how many of them are nonzero?

Hint 2. Every nonzero vector lies on exactly one line. So the lines split the

𝑞^{2} - 1

nonzero vectors into groups of equal size.

Sketch. (a)

{𝟎}

, four lines

{𝟎, (1, 0), (2, 0)}

{𝟎, (0, 1), (0, 2)}

{𝟎, (1, 1), (2, 2)}

{𝟎, (1, 2), (2, 1)}

, and

𝔽_{3}^{2}

. (b) A line has

𝑞

elements,

𝑞 - 1

of them nonzero, and each nonzero

𝐯

lies on exactly one line,

span (𝐯)

. So there are

(𝑞^{2} - 1) / (𝑞 - 1) =

𝑞 +

1

lines, plus

{𝟎}

and

𝐹^{2}

(b) By Theorem 1.4.25, the subspaces of $𝐹^{2}$ are ${𝟎}$ , the lines, and $𝐹^{2}$ , and these are different: ${𝟎}$ has $1$ element, a line has $𝑞$ (see below), and $𝐹^{2}$ has $𝑞^{2}$ , with $1 < 𝑞 < 𝑞^{2}$ as $𝑞 \geq 2$ . So it suffices to count the lines.

Let $𝐯 \neq 𝟎$ . By Corollary 1.2.36(b), the vectors $𝑎 𝐯$ , $𝑎 \in 𝐹$ , are distinct, so $span (𝐯)$ has exactly $𝑞$ elements, of which $𝑞 - 1$ are nonzero. Every nonzero vector $𝐰$ lies on exactly one line. It lies on $span (𝐰)$ . If it also lies on $span (𝐯)$ , then $𝐰 = 𝑐 𝐯$ with $𝑐 \neq 0$ (as $𝐰 \neq 𝟎$ ), and $span (𝐰) = span (𝐯)$ by Exercise 1.3.28(b). So the lines partition the $𝑞^{2} - 1$ nonzero vectors of $𝐹^{2}$ into blocks of $𝑞 - 1$ vectors, and there are

\frac{𝑞^{2} - 1}{𝑞 - 1} = 𝑞 + 1

lines. Together with ${𝟎}$ and $𝐹^{2}$ , this gives $𝑞 + 3$ subspaces.

(a) For $𝑞 = 3$ there are $4$ lines. Listing the multiples $0 𝐯, 1 𝐯, 2 𝐯$ of $𝐯 =$ $(1, 0),$ $(0, 1),$ $(1, 1),$ $(1, 2)$ in $𝔽_{3}^{2}$ , the six subspaces are

\begin{aligned} {𝟎}, {𝟎, (1, 0), (2, 0)}, {𝟎, (0, 1), (0, 2)}, \\ {𝟎, (1, 1), (2, 2)}, {𝟎, (1, 2), (2, 1)}, 𝔽_{3}^{2} . \end{aligned}

The four lines contain $4 \cdot 2 = 8$ nonzero vectors, which are all the nonzero vectors of $𝔽_{3}^{2}$ , as (b) predicts.

Exercise 1.4.34.standard Let

𝜎 : 𝐹^{2} \to 𝐹^{2}

swap the coordinates,

𝜎 (𝑥, 𝑦) = (𝑦, 𝑥)

. Find all subspaces

𝑈

𝐹^{2}

such that

𝜎 (𝐮) \in 𝑈

for every

𝐮 \in 𝑈

. The answer depends on whether

1 + 1 = 0

𝐹

Solution.

Uses. Theorem 1.4.25 (The subspaces of $𝐹^{2}$ ), Exercise 1.1.40, Exercise 1.3.28

Hint 1. By Theorem 1.4.25, only the lines

span (𝐯)

need thought. Show that such a line qualifies exactly when

𝜎 (𝐯) = 𝑐 𝐯

for some

𝑐 \in 𝐹

Hint 2. Write

𝐯 = (𝑎, 𝑏)

. From

𝑏 = 𝑐 𝑎

and

𝑎 = 𝑐 𝑏

, deduce

𝑐^{2} = 1

, so

𝑐 = \pm 1

Sketch.

{𝟎}

and

𝐹^{2}

qualify. A line

span (𝐯)

qualifies iff

𝜎 (𝐯) \in span (𝐯)

, since

𝜎 (𝑡 𝐯) = 𝑡 𝜎 (𝐯)

. With

𝐯 = (𝑎, 𝑏)

and

𝜎 (𝐯) = 𝑐 𝐯

𝑎 = 𝑐^{2} 𝑎

𝑏 = 𝑐^{2} 𝑏

, so

𝑐^{2} = 1

and

𝑐 = \pm 1

, giving the lines

span ((1, 1))

and

span ((1, - 1))

. These are different iff

1 + 1 \neq 0

. So there are four such subspaces if

1 + 1 \neq 0

, and three if

1 + 1 = 0

Call $𝑈$ swap-closed if $𝜎 (𝐮) \in 𝑈$ for all $𝐮 \in 𝑈$ . Clearly ${𝟎}$ and $𝐹^{2}$ are swap-closed. By Theorem 1.4.25, every other subspace is a line $span (𝐯)$ with $𝐯 = (𝑎, 𝑏) \neq 𝟎$ . For $𝑡 \in 𝐹$ , $𝜎 (𝑡 𝐯) =$ $(𝑡 𝑏, 𝑡 𝑎) =$ $𝑡 𝜎 (𝐯)$ . So if $𝜎 (𝐯) \in span (𝐯)$ , then $𝜎$ maps all of $span (𝐯)$ into $span (𝐯)$ ; and conversely a swap-closed line contains $𝜎 (𝐯)$ . Hence $span (𝐯)$ is swap-closed if and only if $𝜎 (𝐯) = 𝑐 𝐯$ for some $𝑐 \in 𝐹$ , that is, $𝑏 = 𝑐 𝑎$ and $𝑎 = 𝑐 𝑏$ .

Suppose $𝑏 = 𝑐 𝑎$ and $𝑎 = 𝑐 𝑏$ . Then $𝑎 = 𝑐^{2} 𝑎$ and $𝑏 = 𝑐^{2} 𝑏$ . Since $𝑎$ or $𝑏$ is nonzero, $𝑐^{2} = 1$ , and by Exercise 1.1.40, $𝑐 = 1$ or $𝑐 = - 1$ . If $𝑐 = 1$ , then $𝑎 = 𝑏$ and $𝐯 = 𝑎 (1, 1)$ . If $𝑐 = - 1$ , then $𝑏 = - 𝑎$ and $𝐯 = 𝑎 (1, - 1)$ . Either way $𝑎 \neq 0$ , so $span (𝐯)$ is $span ((1, 1))$ or $span ((1, - 1))$ by Exercise 1.3.28(b). Conversely, both lines are swap-closed, since $𝜎 (1, 1) = (1, 1)$ and $𝜎 (1, - 1) =$ $(- 1, 1) =$ $- (1, - 1)$ .

If $1 + 1 \neq 0$ , the two lines are different, because $(1, - 1) = 𝑐 (1, 1)$ would force $𝑐 = 1$ and $𝑐 = - 1$ . Then the swap-closed subspaces are the four subspaces ${𝟎}$ , $span ((1, 1))$ , $span ((1, - 1))$ , and $𝐹^{2}$ . If $1 + 1 = 0$ , then $- 1 = 1$ , the two lines coincide, and there are three: ${𝟎}$ , $span ((1, 1))$ , and $𝐹^{2}$ . Over $ℝ$ , the two lines are the diagonals, the mirror of $𝜎$ and the line it reverses. Vectors $𝐯$ with $𝜎 (𝐯) = 𝑐 𝐯$ are eigenvectors, the subject of Section 7.2.

Exercise 1.4.35.standard Let $𝐸$ and $𝑂$ be the subspaces of even and odd functions in $ℝ^{ℝ}$ (Example 1.4.10).

(a)Show that $𝐸 \cap 𝑂 = {𝟎}$ and $𝐸 + 𝑂 = ℝ^{ℝ}$ .
(b)Over $𝔽_{2}$ , call $𝑓 : 𝔽_{2} \to 𝔽_{2}$ even if $𝑓 (- 𝑡) = 𝑓 (𝑡)$ for all $𝑡$ , and odd if $𝑓 (- 𝑡) = - 𝑓 (𝑡)$ for all $𝑡$ . Show that every such $𝑓$ is both even and odd. Which steps of (a) fail over $𝔽_{2}$ ?

Solution.

Uses. Proposition 1.4.24 (The sum of two subspaces), Example 1.4.10 (Functions)

Techniques. Double containment (Technique 1.3.14)

Hint 1. For

𝐸 \cap 𝑂

: if

𝑓

is both even and odd, compare

𝑓 (- 𝑡)

computed both ways.

Hint 2. For

𝐸 + 𝑂

, guess the even part of

𝑓

by symmetrizing: average

𝑓 (𝑡)

and

𝑓 (- 𝑡)

. The odd part is what remains.

Sketch. (a) If

𝑓 \in 𝐸 \cap 𝑂

, then

𝑓 (𝑡) =

𝑓 (- 𝑡) =

- 𝑓 (𝑡)

, so

2 𝑓 (𝑡) = 0

. Every

𝑓

𝑓_{𝑒} + 𝑓_{𝑜}

with

𝑓_{𝑒 (𝑡)} =

(𝑓 (𝑡) + 𝑓 (- 𝑡)) / 2

even and

𝑓_{𝑜 (𝑡)} =

(𝑓 (𝑡) - 𝑓 (- 𝑡)) / 2

odd. (b) In

𝔽_{2}

- 𝑡 = 𝑡

and

- 𝑓 (𝑡) = 𝑓 (𝑡)

, so both conditions read

𝑓 (𝑡) = 𝑓 (𝑡)

. Then

𝐸 + 𝑂

is still everything, but

𝐸 \cap 𝑂 \neq {𝟎}

: the step

2 𝑓 (𝑡) =

0 \Rightarrow

𝑓 (𝑡) =

0

divides by

2 = 0

(a) If $𝑓 \in 𝐸 \cap 𝑂$ , then for every $𝑡$ , $𝑓 (𝑡) =$ $𝑓 (- 𝑡) =$ $- 𝑓 (𝑡)$ , so $2 𝑓 (𝑡) = 0$ and $𝑓 (𝑡) = 0$ . Hence $𝐸 \cap 𝑂 = {𝟎}$ .

For the sum, $𝐸 + 𝑂 \subseteq ℝ^{ℝ}$ holds trivially. Conversely, let $𝑓 \in ℝ^{ℝ}$ , and define

𝑓_{𝑒} (𝑡) = \frac{𝑓 (𝑡) + 𝑓 (- 𝑡)}{2}, 𝑓_{𝑜} (𝑡) = \frac{𝑓 (𝑡) - 𝑓 (- 𝑡)}{2} .

Then $𝑓_{𝑒} (- 𝑡) =$ $(𝑓 (- 𝑡) + 𝑓 (𝑡)) / 2 =$ $𝑓_{𝑒} (𝑡)$ , so $𝑓_{𝑒} \in 𝐸$ ; $𝑓_{𝑜} (- 𝑡) =$ $(𝑓 (- 𝑡) - 𝑓 (𝑡)) / 2 =$ $- 𝑓_{𝑜} (𝑡)$ , so $𝑓_{𝑜} \in 𝑂$ ; and $𝑓_{𝑒} (𝑡) +$ $𝑓_{𝑜} (𝑡) =$ $𝑓 (𝑡)$ for every $𝑡$ . So $𝑓 =$ $𝑓_{𝑒} +$ $𝑓_{𝑜} \in$ $𝐸 +$ $𝑂$ , and $𝐸 + 𝑂 = ℝ^{ℝ}$ .

(b) In $𝔽_{2}$ , $- 𝑎 = 𝑎$ for every $𝑎$ , since $𝑎 + 𝑎 = (1 + 1) 𝑎 = 0$ . So for every $𝑓 : 𝔽_{2} \to 𝔽_{2}$ and every $𝑡$ , $𝑓 (- 𝑡) = 𝑓 (𝑡)$ and $- 𝑓 (𝑡) = 𝑓 (𝑡)$ , and both conditions read $𝑓 (𝑡) = 𝑓 (𝑡)$ . Every function is even and odd: $𝐸 = 𝑂 = 𝔽_{2}^{𝔽_{2}}$ . So $𝐸 + 𝑂 = 𝔽_{2}^{𝔽_{2}}$ still holds, trivially, but $𝐸 \cap 𝑂 \neq {𝟎}$ , and a function no longer has just one decomposition into an even and an odd part. The first argument in (a) fails because it passes from $2 𝑓 (𝑡) = 0$ to $𝑓 (𝑡) = 0$ , and the formulas for $𝑓_{𝑒}$ and $𝑓_{𝑜}$ in the second involve $1 / 2$ . In $𝔽_{2}$ , $2 = 0$ has no inverse.

Exercise 1.4.36.standard Let $𝑛 \geq 2$ . In $𝐹^{𝑛 \times 𝑛}$ , let $𝑆$ be the symmetric matrices, $𝑁$ the strictly upper triangular matrices ( $𝑎_{𝑖 𝑗} = 0$ whenever $𝑖 \geq 𝑗$ ), and $𝐾$ the skew-symmetric matrices ( $𝑎_{𝑖 𝑗} = - 𝑎_{𝑗 𝑖}$ for all $𝑖, 𝑗$ ).

(a)Show that $𝑁$ and $𝐾$ are subspaces, that $𝑆 \cap 𝑁 = {𝟎}$ , and that $𝑆 + 𝑁 = 𝐹^{𝑛 \times 𝑛}$ . This holds over every field.
(b)Show that $𝑆 + 𝐾 = 𝐹^{𝑛 \times 𝑛}$ if and only if $1 + 1 \neq 0$ in $𝐹$ .

Solution.

Uses. Proposition 1.4.24 (The sum of two subspaces), Example 1.4.11 (Matrices), Theorem 1.4.3 (Subspace criterion)

Techniques. Double containment (Technique 1.3.14)

Hint 1.

𝑁

and

𝐾

are cut out by entrywise equations, as in Example 1.4.11. For

𝑆 + 𝑁 = 𝐹^{𝑛 \times 𝑛}

, build the symmetric part of

𝐴

from the entries of

𝐴

on and below the diagonal.

Hint 2. For (b) with

1 + 1 \neq 0

, use

(𝐴 + 𝐴^{𝖳}) / 2

and

(𝐴 - 𝐴^{𝖳}) / 2

, written entrywise. If

1 + 1 = 0

, compare the conditions defining

𝐾

and

𝑆

Sketch. (a) A matrix in

𝑆 \cap 𝑁

vanishes on and below the diagonal, and its entries above equal entries below. For

𝐴

, let

𝐵

copy

𝐴

on and below the diagonal and be symmetric; then

𝐴 - 𝐵 \in 𝑁

. (b) If

1 + 1 \neq 0

𝑎_{𝑖 𝑗} =

ℎ (𝑎_{𝑖 𝑗} + 𝑎_{𝑗 𝑖}) +

ℎ (𝑎_{𝑖 𝑗} - 𝑎_{𝑗 𝑖})

with

ℎ = {(1 + 1)}^{- 1}

. If

1 + 1 = 0

, then

- 𝑎 = 𝑎

, so

𝐾 = 𝑆

and

𝑆 + 𝐾 = 𝑆

, which misses the matrix with a single

1

in position

(1, 2)

(a) The zero matrix satisfies the conditions defining $𝑁$ and $𝐾$ . If $𝐴, 𝐵 \in 𝑁$ and $𝑐 \in 𝐹$ , then for $𝑖 \geq 𝑗$ , $𝑎_{𝑖 𝑗} + 𝑏_{𝑖 𝑗} = 0$ and $𝑐 𝑎_{𝑖 𝑗} = 0$ , so $𝐴 + 𝐵, 𝑐 𝐴 \in 𝑁$ . If $𝐴, 𝐵 \in 𝐾$ , then $𝑎_{𝑖 𝑗} +$ $𝑏_{𝑖 𝑗} =$ $- 𝑎_{𝑗 𝑖} -$ $𝑏_{𝑗 𝑖} =$ $- (𝑎_{𝑗 𝑖} + 𝑏_{𝑗 𝑖})$ and $𝑐 𝑎_{𝑖 𝑗} =$ $- (𝑐 𝑎_{𝑗 𝑖})$ , so $𝐴 + 𝐵, 𝑐 𝐴 \in 𝐾$ . So $𝑁$ and $𝐾$ are subspaces.

Let $𝐴 \in 𝑆 \cap 𝑁$ . For $𝑖 \geq 𝑗$ , $𝑎_{𝑖 𝑗} = 0$ because $𝐴 \in 𝑁$ . For $𝑖 < 𝑗$ , $𝑎_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ because $𝐴 \in 𝑆$ , and $𝑎_{𝑗 𝑖} = 0$ because $𝑗 > 𝑖$ . So $𝐴 = 𝟎$ .

Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , and define $𝐵$ by $𝑏_{𝑖 𝑗} = 𝑎_{𝑖 𝑗}$ if $𝑖 \geq 𝑗$ and $𝑏_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ if $𝑖 < 𝑗$ . Then $𝐵$ is symmetric: for $𝑖 > 𝑗$ , $𝑏_{𝑖 𝑗} = 𝑎_{𝑖 𝑗}$ , and $𝑏_{𝑗 𝑖} = 𝑎_{𝑖 𝑗}$ since $𝑗 < 𝑖$ . Let $𝐶 = 𝐴 - 𝐵$ . For $𝑖 \geq 𝑗$ , $𝑐_{𝑖 𝑗} =$ $𝑎_{𝑖 𝑗} -$ $𝑎_{𝑖 𝑗} =$ $0$ , so $𝐶 \in 𝑁$ . Hence $𝐴 = 𝐵 + 𝐶 \in 𝑆 + 𝑁$ , and $𝑆 + 𝑁 = 𝐹^{𝑛 \times 𝑛}$ .

(b) Suppose $1 + 1 \neq 0$ , and let $ℎ = {(1 + 1)}^{- 1}$ . For $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , define $𝐵$ and $𝐶$ by

𝑏_{𝑖 𝑗} = ℎ (𝑎_{𝑖 𝑗} + 𝑎_{𝑗 𝑖}), 𝑐_{𝑖 𝑗} = ℎ (𝑎_{𝑖 𝑗} - 𝑎_{𝑗 𝑖}) .

Then $𝑏_{𝑗 𝑖} = 𝑏_{𝑖 𝑗}$ and $𝑐_{𝑗 𝑖} = - 𝑐_{𝑖 𝑗}$ , so $𝐵 \in 𝑆$ and $𝐶 \in 𝐾$ , and $𝑏_{𝑖 𝑗} +$ $𝑐_{𝑖 𝑗} =$ $ℎ (1 + 1) 𝑎_{𝑖 𝑗} =$ $𝑎_{𝑖 𝑗}$ . So $𝐴 = 𝐵 + 𝐶$ , and $𝑆 + 𝐾 = 𝐹^{𝑛 \times 𝑛}$ .

Suppose $1 + 1 = 0$ . Then $- 𝑎 = 𝑎$ for all $𝑎 \in 𝐹$ , so the condition $𝑎_{𝑖 𝑗} = - 𝑎_{𝑗 𝑖}$ is the condition $𝑎_{𝑖 𝑗} = 𝑎_{𝑗 𝑖}$ , and $𝐾 = 𝑆$ . Hence $𝑆 + 𝐾 = 𝑆$ by Proposition 1.4.24. Since $𝑛 \geq 2$ , the matrix $𝐸$ with $1$ in position $(1, 2)$ and $0$ elsewhere is not symmetric ( $𝑒_{12} = 1 \neq 0 = 𝑒_{21}$ ), so $𝑆 + 𝐾 \neq 𝐹^{𝑛 \times 𝑛}$ .

Exercise 1.4.37.standard Let $𝑈$ , $𝑊$ , and $𝑋$ be subspaces of $𝑉$ . Prove or disprove each statement.

(a)If $𝑈 + 𝑊 = 𝑈 + 𝑋$ , then $𝑊 = 𝑋$ .
(b) $𝑈 \cap$ $(𝑊 + 𝑋) =$ $(𝑈 \cap 𝑊) +$ $(𝑈 \cap 𝑋)$ .
(c)If $𝑊 \subseteq 𝑈$ , then $𝑈 \cap$ $(𝑊 + 𝑋) =$ $𝑊 +$ $(𝑈 \cap 𝑋)$ .

Solution.

Uses. Proposition 1.4.24 (The sum of two subspaces), Proposition 1.4.16 (Intersections of subspaces)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Double containment (Technique 1.3.14)

Hint 1. Look for counterexamples among

{𝟎}

ℝ^{2}

, and lines in

ℝ^{2}

. For (b), use three different lines.

Hint 2. For (c), one containment follows from Proposition 1.4.24(b). For the other, write

𝐯 \in 𝑈 \cap (𝑊 + 𝑋)

𝐰 + 𝐱

and show that

𝐱 \in 𝑈

Sketch. (a) False:

𝑈 = ℝ^{2}

𝑊 = {𝟎}

𝑋 = ℝ^{2}

. (b) False: with the lines

𝑈 = span ((1, 1))

𝑊 = span (𝐞_{1})

𝑋 = span (𝐞_{2})

, the left side is

𝑈

and the right side is

{𝟎}

. (c) True.

𝑊

and

𝑈 \cap 𝑋

lie in

𝑈 \cap (𝑊 + 𝑋)

, which gives

\supseteq

. If

𝐯 = 𝐰 + 𝐱 \in 𝑈

, then

𝐱 = 𝐯 - 𝐰 \in 𝑈

because

𝐰 \in 𝑊 \subseteq 𝑈

, so

𝐯 \in 𝑊 + (𝑈 \cap 𝑋)

(a) False. In $𝑉 = ℝ^{2}$ , take $𝑈 = ℝ^{2}$ , $𝑊 = {𝟎}$ , and $𝑋 = ℝ^{2}$ . Then $𝑈 + 𝑊 = ℝ^{2} = 𝑈 + 𝑋$ , but $𝑊 \neq 𝑋$ .

(b) False. In $ℝ^{2}$ , take $𝑈 = span ((1, 1))$ , $𝑊 = span (𝐞_{1})$ , and $𝑋 = span (𝐞_{2})$ . Then $𝑊 +$ $𝑋 =$ $span (𝐞_{1}, 𝐞_{2}) =$ $ℝ^{2}$ (Exercise 1.3.36), so the left side is $𝑈$ . On the other hand, $𝑈 \cap 𝑊 = {𝟎}$ : if $(𝑡, 𝑡) = (𝑠, 0)$ then $𝑡 = 0$ . Likewise $𝑈 \cap 𝑋 = {𝟎}$ . So the right side is ${𝟎} + {𝟎} = {𝟎} \neq 𝑈$ . (The containment $\supseteq$ always holds: $𝑈 \cap 𝑊$ lies in $𝑈$ and in $𝑊 \subseteq 𝑊 + 𝑋$ , likewise $𝑈 \cap 𝑋$ , and Proposition 1.4.24(b) applies.)

(c) True. We proceed by double containment.

$\supseteq$ . $𝑊 \subseteq 𝑈$ by hypothesis and $𝑊 \subseteq 𝑊 + 𝑋$ by Proposition 1.4.24(a), so $𝑊 \subseteq 𝑈 \cap (𝑊 + 𝑋)$ . Also $𝑈 \cap 𝑋 \subseteq 𝑈$ , and $𝑈 \cap 𝑋 \subseteq 𝑋 \subseteq 𝑊 + 𝑋$ , so $𝑈 \cap 𝑋 \subseteq 𝑈 \cap (𝑊 + 𝑋)$ . The set $𝑈 \cap (𝑊 + 𝑋)$ is a subspace by Proposition 1.4.24(a) and Proposition 1.4.16, so it contains $𝑊 + (𝑈 \cap 𝑋)$ by Proposition 1.4.24(b).

$\subseteq$ . Let $𝐯 \in 𝑈 \cap (𝑊 + 𝑋)$ , and write $𝐯 = 𝐰 + 𝐱$ with $𝐰 \in 𝑊$ and $𝐱 \in 𝑋$ . Since $𝐰 \in 𝑊 \subseteq 𝑈$ and $𝐯 \in 𝑈$ , we get $𝐱 = 𝐯 - 𝐰 \in 𝑈$ . So $𝐱 \in 𝑈 \cap 𝑋$ , and $𝐯 =$ $𝐰 +$ $𝐱 \in$ $𝑊 +$ $(𝑈 \cap 𝑋)$ .

Statement (c) is called the modular law. The counterexample in (b) shows that the hypothesis $𝑊 \subseteq 𝑈$ cannot simply be dropped.

Exercise 1.4.38.standard Let

𝑈

be a proper subspace of

𝑉

. Show that the only subspace of

𝑉

that contains every vector outside

𝑈

𝑉

itself.

Solution.

Uses. Lemma 1.4.19 (Adding a vector outside a subspace), Theorem 1.4.3 (Subspace criterion)

Hint 1. Let

𝑋

be a subspace containing

𝑉 ∖ 𝑈

. You need to show that

𝑈 \subseteq 𝑋

too. Fix a vector

𝐯

outside

𝑈

Hint 2. For

𝐮 \in 𝑈

, where does

𝐮 + 𝐯

lie? Then write

𝐮

in terms of

𝐮 + 𝐯

and

𝐯

Sketch. Let

𝑋

contain

𝑉 ∖ 𝑈

, and fix

𝐯 \notin 𝑈

. For

𝐮 \in 𝑈

𝐮 + 𝐯 \notin 𝑈

by Lemma 1.4.19, so

𝐮 + 𝐯 \in 𝑋

; also

𝐯 \in 𝑋

, so

𝐮 = (𝐮 + 𝐯) - 𝐯 \in 𝑋

. Hence

𝑋 \supseteq 𝑈 \cup (𝑉 ∖ 𝑈) = 𝑉

Let $𝑋$ be a subspace of $𝑉$ that contains every vector of $𝑉 ∖ 𝑈$ . Since $𝑈 \neq 𝑉$ , we may fix $𝐯 \in 𝑉$ with $𝐯 \notin 𝑈$ ; then $𝐯 \in 𝑋$ . Let $𝐮 \in 𝑈$ . By Lemma 1.4.19, $𝐮 + 𝐯 \notin 𝑈$ , so $𝐮 + 𝐯 \in 𝑋$ . Hence

𝐮 = (𝐮 + 𝐯) + (- 1) 𝐯 \in 𝑋

by (II) of Theorem 1.4.3. So $𝑈 \subseteq 𝑋$ , and with $𝑉 ∖ 𝑈 \subseteq 𝑋$ this gives $𝑋 = 𝑉$ .

In other words, the vectors outside a proper subspace generate everything: every $𝐮 \in 𝑈$ is the difference of two vectors outside $𝑈$ .

Exercise 1.4.39.standard

(a)Let $𝑈$ be a subspace of $𝐹 [𝑥]$ that contains a polynomial of degree $𝑘$ for every $𝑘 \geq 0$ . Show that $𝑈 = 𝐹 [𝑥]$ .
(b)Find a proper subspace of $𝐹 [𝑥]$ that contains polynomials of infinitely many different degrees.

Solution.

Uses. Exercise 1.3.33, Proposition 1.4.14 (The span is the smallest subspace containing a list), Example 1.4.23 ( $𝐹 [𝑥]$ as a union of a chain), Exercise 1.4.27

Hint 1. Choose

𝑝_{𝑘} \in 𝑈

of degree

𝑘

for each

𝑘

. What is

span (𝑝_{0}, \dots, 𝑝_{𝑛})

? See Exercise 1.3.33.

Hint 2. Every polynomial lies in some

{𝐹 [𝑥]}_{\leq 𝑛}

. For (b), consider the even polynomials.

Sketch. (a) With

deg 𝑝_{𝑘} = 𝑘

{𝐹 [𝑥]}_{\leq 𝑛} =

span (𝑝_{0}, \dots, 𝑝_{𝑛}) \subseteq

𝑈

for every

𝑛

, and every polynomial lies in some

{𝐹 [𝑥]}_{\leq 𝑛}

. (b) The polynomials with

𝑎_{𝑘} = 0

for all odd

𝑘

: a subspace containing

𝑥^{2 𝑘}

for every

𝑘

, but not

𝑥

(a) For each $𝑘 \geq 0$ , choose $𝑝_{𝑘} \in 𝑈$ with $deg 𝑝_{𝑘} = 𝑘$ . Let $𝑛 \geq 0$ . By Exercise 1.3.33(a), $span (𝑝_{0}, \dots, 𝑝_{𝑛}) =$ ${𝐹 [𝑥]}_{\leq 𝑛}$ , and since $𝑝_{0}, \dots, 𝑝_{𝑛} \in 𝑈$ , Proposition 1.4.14(b) gives ${𝐹 [𝑥]}_{\leq 𝑛} \subseteq 𝑈$ . Every polynomial lies in ${𝐹 [𝑥]}_{\leq 𝑛}$ for some $𝑛$ : take $𝑛 = deg 𝑝$ if $𝑝 \neq 𝟎$ and $𝑛 = 0$ if $𝑝 = 𝟎$ . Hence $𝐹 [𝑥] =$ $⋃_{𝑛 \geq 0} {𝐹 [𝑥]}_{\leq 𝑛} \subseteq$ $𝑈$ .

(b) Let $𝑈$ be the set of polynomials whose coefficients $𝑎_{𝑘}$ vanish for every odd $𝑘$ . It is a subspace (the argument of Exercise 1.4.27(e) works over every field), it contains $𝑥^{2 𝑘}$ for every $𝑘 \geq 0$ , and it does not contain $𝑥$ . So it is proper. Part (a) shows that it must miss some degree, and it misses all the odd ones.

Exercise 1.4.40.hard A function $𝑓 : ℝ \to ℝ$ has period $𝑝 > 0$ if $𝑓 (𝑡 + 𝑝) = 𝑓 (𝑡)$ for all $𝑡 \in ℝ$ , and $𝑓$ is periodic if it has some period.

(a)Show that the functions with period $1$ form a subspace of $ℝ^{ℝ}$ .
(b)Is the set of all periodic functions a subspace of $ℝ^{ℝ}$ ?

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Definition 1.2.9 (Function spaces)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. For (b), the trouble is addition: two functions with periods that have no common multiple. You need a sum whose lack of a period you can actually prove.

Hint 2. Let

𝑔

1

ℚ

and

0

elsewhere, and let

ℎ

1

\sqrt{2} ℚ = {\sqrt{2} 𝑟 | 𝑟 \in ℚ}

and

0

elsewhere. Find periods of

𝑔

and

ℎ

Hint 3. At which points does

𝑔 + ℎ

take the value

2

? A period would move one such point to another.

Sketch. (a) The condition

𝑓 (𝑡 + 1) = 𝑓 (𝑡)

survives sums and multiples. (b) No.

𝑔 = 1

ℚ

has period

1

, and

ℎ = 1

\sqrt{2} ℚ

has period

\sqrt{2}

. Since

ℚ \cap \sqrt{2} ℚ = {0}

, the function

𝑔 + ℎ

equals

2

0

and at most

1

elsewhere. If

𝑝 > 0

were a period, then

(𝑔 + ℎ) (𝑝) =

(𝑔 + ℎ) (0) =

2

, forcing

𝑝 = 0

(a) The zero function has period $1$ . If $𝑓 (𝑡 + 1) = 𝑓 (𝑡)$ and $𝑔 (𝑡 + 1) = 𝑔 (𝑡)$ for all $𝑡$ , then $(𝑓 + 𝑔) (𝑡 + 1) =$ $𝑓 (𝑡) +$ $𝑔 (𝑡) =$ $(𝑓 + 𝑔) (𝑡)$ and $(𝑎 𝑓) (𝑡 + 1) =$ $𝑎 𝑓 (𝑡) =$ $(𝑎 𝑓) (𝑡)$ .

(b) No: the periodic functions are not closed under addition. Let $𝑔 (𝑡) = 1$ if $𝑡 \in ℚ$ and $𝑔 (𝑡) = 0$ otherwise, and let $ℎ (𝑡) = 1$ if $𝑡 \in \sqrt{2} ℚ$ and $ℎ (𝑡) = 0$ otherwise, where $\sqrt{2} ℚ = {\sqrt{2} 𝑟 | 𝑟 \in ℚ}$ .

The function $𝑔$ has period $1$ , since $𝑡 + 1 \in ℚ$ if and only if $𝑡 \in ℚ$ . The function $ℎ$ has period $\sqrt{2}$ , since $𝑡 + \sqrt{2} = \sqrt{2} 𝑟$ with $𝑟 \in ℚ$ if and only if $𝑡 = \sqrt{2} (𝑟 - 1)$ . So $𝑔$ and $ℎ$ are periodic.

Next, $ℚ \cap \sqrt{2} ℚ = {0}$ . Indeed, if $𝑟 = \sqrt{2} 𝑠$ with $𝑟, 𝑠 \in ℚ$ and $𝑠 \neq 0$ , then $\sqrt{2} = 𝑟 / 𝑠$ would be rational; so $𝑠 = 0$ and $𝑟 = 0$ . Hence $(𝑔 + ℎ) (0) = 2$ , while $(𝑔 + ℎ) (𝑡) \leq 1$ for every $𝑡 \neq 0$ , because $𝑡$ lies in at most one of $ℚ$ and $\sqrt{2} ℚ$ .

Suppose $𝑔 + ℎ$ had a period $𝑝 > 0$ . Then $(𝑔 + ℎ) (𝑝) =$ $(𝑔 + ℎ) (0 + 𝑝) =$ $(𝑔 + ℎ) (0) =$ $2$ , so $𝑝 = 0$ , a contradiction. So $𝑔 + ℎ$ is not periodic, although $𝑔$ and $ℎ$ are.

The same happens for many nicer functions, such as $sin 𝑡 + sin (\sqrt{2} 𝑡)$ , but proving that this one has no period takes more work. The indicator functions are chosen so that the proof is short.

Exercise 1.4.41.hard Let $𝑛 \geq 2$ . Call a subspace $𝑈$ of $𝐹^{𝑛}$ symmetric if permuting the coordinates of any vector of $𝑈$ gives a vector of $𝑈$ . Let $𝐬 = (1, 1, \dots, 1)$ and $𝑍 =$ ${𝐱 \in 𝐹^{𝑛} | 𝑥_{1} + \dots + 𝑥_{𝑛} = 0}$ .

(a)Show that the symmetric subspaces of $𝐹^{𝑛}$ are exactly ${𝟎}$ , $span (𝐬)$ , $𝑍$ , and $𝐹^{𝑛}$ .
(b)When is $span (𝐬) \subseteq 𝑍$ ?

Solution.

Uses. Proposition 1.4.14 (The span is the smallest subspace containing a list), Example 1.4.8 (One equation in $𝐹^{𝑛}$ ), Exercise 1.2.45

Hint 1. The four subspaces are symmetric. For the converse, split into two cases: every vector of

𝑈

has all coordinates equal, or some vector

𝐱 \in 𝑈

has

𝑥_{𝑖} \neq 𝑥_{𝑗}

Hint 2. In the second case, subtract from

𝐱

the vector obtained by swapping coordinates

𝑖

and

𝑗

. Then move the result around with other permutations to get every

𝐞_{𝑘} - 𝐞_{𝑙}

into

𝑈

Hint 3. The vectors

𝐞_{𝑘} - 𝐞_{𝑛}

span

𝑍

. If

𝑈

is bigger than

𝑍

, pick

𝐮 \in 𝑈

with nonzero coordinate sum and write any

𝐲

as an element of

𝑍

plus a multiple of

𝐮

Sketch. If all vectors of

𝑈

are constant,

𝑈 \subseteq span (𝐬)

, so

𝑈 = {𝟎}

span (𝐬)

. Otherwise some

𝐱 \in 𝑈

has

𝑥_{𝑖} \neq 𝑥_{𝑗}

; subtracting its swap gives

(𝑥_{𝑖} - 𝑥_{𝑗}) (𝐞_{𝑖} - 𝐞_{𝑗}) \in

𝑈

, and permuting gives every

𝐞_{𝑘} - 𝐞_{𝑙} \in 𝑈

, so

𝑍 =

span (𝐞_{1} - 𝐞_{𝑛}, \dots, 𝐞_{𝑛 - 1} - 𝐞_{𝑛}) \subseteq

𝑈

. If some

𝐮 \in 𝑈

has coordinate sum

𝜎 \neq 0

, then

𝐲 =

(𝐲 - (𝜏 / 𝜎) 𝐮) +

(𝜏 / 𝜎) 𝐮 \in

𝑈

, where

𝜏

is the coordinate sum of

𝐲

. (b) Exactly when

𝑛 * 1 = 0

𝐹

For a bijection $𝜋$ of ${1, \dots, 𝑛}$ and $𝐱 \in 𝐹^{𝑛}$ , write $𝜋 𝐱 =$ $(𝑥_{𝜋 (1)}, \dots, 𝑥_{𝜋 (𝑛)})$ ; so $𝑈$ is symmetric if $𝜋 𝐱 \in 𝑈$ for all $𝐱 \in 𝑈$ and all $𝜋$ . The sum of the coordinates of $𝐱$ is denoted $Σ (𝐱)$ ; it does not change under permutations.

(a) The four subspaces are symmetric: for ${𝟎}$ and $𝐹^{𝑛}$ there is nothing to check, $𝜋 (𝑐 𝐬) = 𝑐 𝐬$ , and $Σ (𝜋 𝐱) = Σ (𝐱)$ . Note that $𝑍$ is a subspace by Example 1.4.8. Now let $𝑈$ be a symmetric subspace.

Case 1: every $𝐱 \in 𝑈$ has all coordinates equal. Then $𝑈 \subseteq span (𝐬)$ . If $𝑈 \neq {𝟎}$ , it contains some $𝑐 𝐬$ with $𝑐 \neq 0$ , hence $𝐬 = 𝑐^{- 1} (𝑐 𝐬)$ , hence $span (𝐬)$ by Proposition 1.4.14(b). So $𝑈 = {𝟎}$ or $𝑈 = span (𝐬)$ .

Case 2: some $𝐱 \in 𝑈$ has $𝑥_{𝑖} \neq 𝑥_{𝑗}$ for some $𝑖 \neq 𝑗$ . Let $𝐲$ be $𝐱$ with coordinates $𝑖$ and $𝑗$ swapped. Then $𝐲 \in 𝑈$ , and $𝐱 - 𝐲$ has $𝑖$ -th coordinate $𝑥_{𝑖} - 𝑥_{𝑗}$ , $𝑗$ -th coordinate $𝑥_{𝑗} - 𝑥_{𝑖}$ , and all others $0$ :

𝐱 - 𝐲 = (𝑥_{𝑖} - 𝑥_{𝑗}) (𝐞_{𝑖} - 𝐞_{𝑗}) \in 𝑈 .

Multiplying by ${(𝑥_{𝑖} - 𝑥_{𝑗})}^{- 1}$ gives $𝐞_{𝑖} - 𝐞_{𝑗} \in 𝑈$ . Now let $𝑘 \neq 𝑙$ , and choose a bijection $𝜋$ of ${1, \dots, 𝑛}$ with $𝜋 (𝑘) = 𝑖$ and $𝜋 (𝑙) = 𝑗$ . Then $𝜋 (𝐞_{𝑖} - 𝐞_{𝑗})$ has $𝑘$ -th coordinate $1$ , $𝑙$ -th coordinate $- 1$ , and all others $0$ ; that is, $𝐞_{𝑘} - 𝐞_{𝑙} \in 𝑈$ .

Every $𝐳 \in 𝑍$ satisfies $𝑧_{𝑛} =$ $- (𝑧_{1} + \dots + 𝑧_{𝑛 - 1})$ , so

𝐳 = 𝑧_{1} (𝐞_{1} - 𝐞_{𝑛}) + \dots + 𝑧_{𝑛 - 1} (𝐞_{𝑛 - 1} - 𝐞_{𝑛}),

as both sides have the same first $𝑛 - 1$ coordinates and the same $𝑛$ -th coordinate. So $𝑍 \subseteq$ $span (𝐞_{1} - 𝐞_{𝑛}, \dots, 𝐞_{𝑛 - 1} - 𝐞_{𝑛}) \subseteq$ $𝑈$ by Proposition 1.4.14(b). If $𝑈 = 𝑍$ , we are done. Otherwise there is $𝐮 \in 𝑈$ with $𝜎 = Σ (𝐮) \neq 0$ . For any $𝐲 \in 𝐹^{𝑛}$ , let $𝜏 = Σ (𝐲)$ . Then $Σ (𝐲 - (𝜏 / 𝜎) 𝐮) =$ $𝜏 -$ $(𝜏 / 𝜎) 𝜎 =$ $0$ , so

𝐲 = (𝐲 - (𝜏 / 𝜎) 𝐮) + (𝜏 / 𝜎) 𝐮 \in 𝑍 + 𝑈 \subseteq 𝑈 .

Hence $𝑈 = 𝐹^{𝑛}$ .

(b) $span (𝐬) \subseteq 𝑍$ if and only if $𝐬 \in 𝑍$ , that is, $Σ (𝐬) = 𝑛 * 1 = 0$ in $𝐹$ (in the notation of Exercise 1.2.45). This happens exactly when $char 𝐹$ divides $𝑛$ ; over $𝔽_{2}$ with $𝑛$ even, for example. In that case the four subspaces form a chain ${𝟎} \subseteq$ $span (𝐬) \subseteq$ $𝑍 \subseteq$ $𝐹^{𝑛}$ . For $𝐹 = 𝔽_{2}$ and $𝑛 = 2$ , $span (𝐬) =$ $𝑍 =$ ${𝟎, (1, 1)}$ , so there are only three symmetric subspaces.

Exercise 1.4.42.hard

(a)Show that $𝔽_{2}^{2}$ is the union of three subspaces, none of which contains the other two.
(b)Suppose $𝐹$ has at least three elements, and let $𝑈_{1}$ , $𝑈_{2}$ , $𝑈_{3}$ be subspaces of $𝑉$ whose union is a subspace. Show that one of the $𝑈_{𝑖}$ contains the other two.

Solution.

Uses. Proposition 1.4.20 (Unions of two subspaces), Lemma 1.4.19 (Adding a vector outside a subspace), Example 1.4.12 (The subspaces of $𝔽_{2}^{2}$ ), Corollary 1.2.36 (Cancelling scalars and vectors)

Techniques. Try small cases and small fields (Technique 1.2.38)

Hint 1. For (a), use the lines of Example 1.4.12. For (b), let

𝑈 = 𝑈_{1} \cup 𝑈_{2} \cup 𝑈_{3}

, and suppose no

𝑈_{𝑖}

equals

𝑈

. First show that

𝑈_{1}

is not contained in

𝑈_{2} \cup 𝑈_{3}

Hint 2. Pick

𝐮 \in 𝑈_{1}

outside

𝑈_{2} \cup 𝑈_{3}

and

𝐯 \in 𝑈

outside

𝑈_{1}

. Consider the vectors

𝐯 + 𝜆 𝐮

for

𝜆 \in 𝐹

. Which of the

𝑈_{𝑖}

can contain them?

Hint 3. None lies in

𝑈_{1}

, and

𝑈_{2}

contains at most one of them: if it contains two, it contains their difference, a nonzero multiple of

𝐮

Sketch. (a) The three lines of

𝔽_{2}^{2}

. (b) Suppose no

𝑈_{𝑖}

is all of

𝑈 = 𝑈_{1} \cup 𝑈_{2} \cup 𝑈_{3}

. If

𝑈_{1} \subseteq 𝑈_{2} \cup 𝑈_{3}

, then

𝑈 = 𝑈_{2} \cup 𝑈_{3}

, and Proposition 1.4.20 gives a contradiction; so pick

𝐮 \in 𝑈_{1}

outside

𝑈_{2} \cup 𝑈_{3}

, and

𝐯 \in 𝑈 ∖ 𝑈_{1}

. The vectors

𝐯 + 𝜆 𝐮

𝜆 \in 𝐹

, are at least three distinct vectors of

𝑈

, none in

𝑈_{1}

(Lemma 1.4.19), and at most one in each of

𝑈_{2}

𝑈_{3}

. Contradiction.

(a) By Example 1.4.12, the lines ${𝟎, (1, 0)}$ , ${𝟎, (0, 1)}$ , ${𝟎, (1, 1)}$ are subspaces of $𝔽_{2}^{2}$ , and their union is $𝔽_{2}^{2}$ . Each contains exactly one nonzero vector, and these are different, so no line contains another.

(b) Let $𝑈 = 𝑈_{1} \cup 𝑈_{2} \cup 𝑈_{3}$ , a subspace by hypothesis. One of the $𝑈_{𝑖}$ contains the other two if and only if it equals $𝑈$ , so suppose, to the contrary, that $𝑈_{𝑖} \neq 𝑈$ for each $𝑖$ .

First, $𝑈_{1} ⊈ 𝑈_{2} \cup 𝑈_{3}$ . Otherwise $𝑈 = 𝑈_{2} \cup 𝑈_{3}$ is a subspace, so by Proposition 1.4.20 one of $𝑈_{2}$ , $𝑈_{3}$ contains the other and equals $𝑈$ , contrary to our assumption. So we may pick $𝐮 \in 𝑈_{1}$ with $𝐮 \notin 𝑈_{2}$ and $𝐮 \notin 𝑈_{3}$ . In particular $𝐮 \neq 𝟎$ . Since $𝑈_{1} \neq 𝑈$ , we may also pick $𝐯 \in 𝑈$ with $𝐯 \notin 𝑈_{1}$ .

For $𝜆 \in 𝐹$ , the vector $𝐯 + 𝜆 𝐮$ lies in the subspace $𝑈$ . It does not lie in $𝑈_{1}$ , by Lemma 1.4.19, since $𝜆 𝐮 \in 𝑈_{1}$ and $𝐯 \notin 𝑈_{1}$ . Suppose $𝑈_{2}$ contained $𝐯 + 𝜆 𝐮$ and $𝐯 + 𝜇 𝐮$ with $𝜆 \neq 𝜇$ . Then it would contain their difference $(𝜆 - 𝜇) 𝐮$ , and hence $𝐮 =$ ${(𝜆 - 𝜇)}^{- 1} (𝜆 - 𝜇) 𝐮$ , which is false. So $𝑈_{2}$ contains at most one of the vectors $𝐯 + 𝜆 𝐮$ , and likewise $𝑈_{3}$ . These vectors are distinct for distinct $𝜆$ , by Corollary 1.2.36(b), since $𝐮 \neq 𝟎$ , and $𝐹$ has at least three elements. So at least three distinct vectors of $𝑈$ lie in $𝑈_{2} \cup 𝑈_{3}$ , which contains at most two of them, and none of them lies in $𝑈_{1}$ . This contradicts $𝑈 = 𝑈_{1} \cup 𝑈_{2} \cup 𝑈_{3}$ .

Part (a) shows that the hypothesis on $𝐹$ is needed. The proof needs three values of $𝜆$ to outnumber $𝑈_{2}$ and $𝑈_{3}$ ; over $𝔽_{2}$ there are only two.

Exercise 1.4.43.challengeBeyond 21-242 Let $𝑉$ be a vector space over $𝐹$ , and let $𝑈_{1}, \dots, 𝑈_{𝑛}$ be proper subspaces of $𝑉$ , where $𝑛 \geq 1$ .

(a)Show that if $𝐹$ has at least $𝑛$ elements, then $𝑈_{1} \cup \dots \cup 𝑈_{𝑛} \neq 𝑉$ .
(b)Deduce that over an infinite field, no vector space is a union of finitely many proper subspaces.
(c)Show that (a) is sharp: if $𝐹$ has $𝑞$ elements, then $𝐹^{2}$ is a union of $𝑞 + 1$ proper subspaces.

Solution.

Uses. Lemma 1.4.19 (Adding a vector outside a subspace), Exercise 1.4.33, Corollary 1.2.36 (Cancelling scalars and vectors)

Hint 1. Suppose

𝑉 = 𝑈_{1} \cup \dots \cup 𝑈_{𝑛}

, and choose such a covering with

𝑛

as small as possible. Why is

𝑈_{1}

then not contained in

𝑈_{2} \cup \dots \cup 𝑈_{𝑛}

Hint 2. Pick

𝐮 \in 𝑈_{1}

outside the other

𝑈_{𝑗}

, and

𝐯 \notin 𝑈_{1}

. Count how many of the vectors

𝐯 + 𝜆 𝐮

𝜆 \in 𝐹

, each

𝑈_{𝑗}

can contain.

Sketch. (a) Take a counterexample with

𝑛

minimal. Then

𝑛 \geq 2

and

𝑈_{1} ⊈ 𝑈_{2} \cup \dots \cup 𝑈_{𝑛}

; pick

𝐮

𝑈_{1}

only, and

𝐯 \notin 𝑈_{1}

. The

| 𝐹 | \geq 𝑛

distinct vectors

𝐯 + 𝜆 𝐮

avoid

𝑈_{1}

, and each

𝑈_{𝑗}

with

𝑗 \geq 2

contains at most one of them, so at most

𝑛 - 1

are covered. (b) An infinite field has at least

𝑛

elements for every

𝑛

. (c)

𝐹^{2}

is the union of its

𝑞 + 1

lines.

(a) Suppose the statement fails, and among all coverings $𝑉 = 𝑈_{1} \cup \dots \cup 𝑈_{𝑛}$ of $𝑉$ by proper subspaces with $𝑛 \leq | 𝐹 |$ , choose one with $𝑛$ as small as possible. Then $𝑛 \geq 2$ , since $𝑈_{1} \neq 𝑉$ . If $𝑈_{1} \subseteq 𝑈_{2} \cup \dots \cup 𝑈_{𝑛}$ , then $𝑉 = 𝑈_{2} \cup \dots \cup 𝑈_{𝑛}$ is a covering by $𝑛 - 1 \leq | 𝐹 |$ proper subspaces, contradicting minimality. So we may pick $𝐮 \in 𝑈_{1}$ with $𝐮 \notin 𝑈_{𝑗}$ for all $𝑗 \geq 2$ ; in particular $𝐮 \neq 𝟎$ . Since $𝑈_{1} \neq 𝑉$ , we may also pick $𝐯 \in 𝑉$ with $𝐯 \notin 𝑈_{1}$ .

Consider the vectors $𝐯 + 𝜆 𝐮$ for $𝜆 \in 𝐹$ . They are distinct, by Corollary 1.2.36(b), since $𝐮 \neq 𝟎$ , so there are $| 𝐹 | \geq 𝑛$ of them. None lies in $𝑈_{1}$ , by Lemma 1.4.19, since $𝜆 𝐮 \in 𝑈_{1}$ and $𝐯 \notin 𝑈_{1}$ . For $𝑗 \geq 2$ , $𝑈_{𝑗}$ contains at most one of them: if $𝐯 + 𝜆 𝐮$ and $𝐯 + 𝜇 𝐮$ lie in $𝑈_{𝑗}$ with $𝜆 \neq 𝜇$ , then so does ${(𝜆 - 𝜇)}^{- 1} ((𝐯 + 𝜆 𝐮) - (𝐯 + 𝜇 𝐮)) =$ $𝐮$ , which is false. So at most $𝑛 - 1$ of these at least $𝑛$ vectors lie in $𝑈_{1} \cup \dots \cup 𝑈_{𝑛} = 𝑉$ , a contradiction.

(b) If $𝐹$ is infinite, it has at least $𝑛$ elements for every $𝑛$ , so (a) applies to every finite collection of proper subspaces.

(c) By Exercise 1.4.33, $𝐹^{2}$ has $𝑞 + 1$ lines, each a proper subspace with $𝑞 < 𝑞^{2}$ elements. Every nonzero $𝐯 \in 𝐹^{2}$ lies on the line $span (𝐯)$ , and $𝟎$ lies on every line, so $𝐹^{2}$ is the union of the $𝑞 + 1$ lines. Since $𝑞 + 1 > 𝑞$ , this does not contradict (a), and it shows that the bound $𝑛 \leq | 𝐹 |$ in (a) cannot be relaxed to $𝑛 \leq | 𝐹 | + 1$ .

Exercise 1.4.44.routine

(a)Find the smallest subspace of $ℝ^{3}$ containing $(1, 1, 0)$ and $(1, - 1, 0)$ , and describe it by one equation.
(b)Find the smallest subspace of $ℝ [𝑥]$ containing $1 + 𝑥^{2}$ and $1 - 𝑥^{2}$ .

Solution.

Uses. Proposition 1.4.14 (The span is the smallest subspace containing a list), Example 1.4.8 (One equation in $𝐹^{𝑛}$ ), Corollary 1.3.19 (Equal spans)

Techniques. Double containment (Technique 1.3.14)

Hint 1. By Proposition 1.4.14, the smallest subspace containing a list is its span.

Hint 2. For (a), both vectors satisfy

𝑥_{3} = 0

. For the reverse containment, solve

𝑎 + 𝑏 = 𝑥_{1}

and

𝑎 - 𝑏 = 𝑥_{2}

. For (b), add and subtract the two polynomials.

Sketch. (a)

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

{𝑥_{3} = 0}

: the plane is a subspace containing both vectors, and

(𝑥_{1}, 𝑥_{2}, 0) =

\frac{𝑥_{1} + 𝑥_{2}}{2} (1, 1, 0) +

\frac{𝑥_{1} - 𝑥_{2}}{2} (1, - 1, 0)

. (b)

span (1 + 𝑥^{2}, 1 - 𝑥^{2}) =

span (1, 𝑥^{2}) =

{𝑎 + 𝑏 𝑥^{2} | 𝑎, 𝑏 \in ℝ}

By Proposition 1.4.14, the smallest subspace containing the entries of a list is the span of the list.

(a) Let $𝑃 = {𝐱 \in ℝ^{3} | 𝑥_{3} = 0}$ , a subspace by Example 1.4.8. Both vectors lie in $𝑃$ , so $span ((1, 1, 0), (1, - 1, 0)) \subseteq$ $𝑃$ by Proposition 1.4.14(b). Conversely, for $(𝑥_{1}, 𝑥_{2}, 0) \in 𝑃$ ,

\frac{𝑥_{1} + 𝑥_{2}}{2} (1, 1, 0) + \frac{𝑥_{1} - 𝑥_{2}}{2} (1, - 1, 0) = (𝑥_{1}, 𝑥_{2}, 0) .

So the smallest subspace is the plane $𝑥_{3} = 0$ .

(b) Since $1 =$ $\frac{1}{2} ((1 + 𝑥^{2}) + (1 - 𝑥^{2}))$ and $𝑥^{2} =$ $\frac{1}{2} ((1 + 𝑥^{2}) - (1 - 𝑥^{2}))$ , and conversely $1 \pm 𝑥^{2}$ are combinations of $1$ and $𝑥^{2}$ , Corollary 1.3.19 gives

span (1 + 𝑥^{2}, 1 - 𝑥^{2}) = span (1, 𝑥^{2}) = {𝑎 + 𝑏 𝑥^{2} | 𝑎, 𝑏 \in ℝ} .

More practice

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