2.5 Invertibility · Matrix Theory

To solve $𝑇 (𝐯) = 𝐰$ is to undo $𝑇$ . In $𝐹$ , multiplication by $𝑎 \neq 0$ is undone by multiplication by $𝑎^{- 1}$ , and $𝑎^{- 1} 𝑎 = 𝑎 𝑎^{- 1} = 1$ , so it does not matter on which side $𝑎^{- 1}$ stands. For linear maps the two sides come apart, and they do different jobs. Let $𝑇 \in L (𝑉, 𝑊)$ and $𝑆 \in L (𝑊, 𝑉)$ . If $𝑆 𝑇 = {id}_{𝑉}$ , then every solution of $𝑇 (𝐯) = 𝐰$ satisfies

𝐯 = 𝑆 (𝑇 (𝐯)) = 𝑆 (𝐰),

so there is at most one solution. If $𝑇 𝑆 = {id}_{𝑊}$ , then $𝑇 (𝑆 (𝐰)) = 𝐰$ , so $𝑆 (𝐰)$ is a solution. A map that undoes $𝑇$ on one side therefore answers one of the two questions about $𝑇 (𝐯) = 𝐰$ , uniqueness or existence, and the inverses of this section are the maps that answer both.

Throughout, $𝑈$ , $𝑉$ , and $𝑊$ are vector spaces over $𝐹$ , and as in Notation 2.1.18 we write $𝑇 𝑆$ for $𝑇 \circ 𝑆$ .

2.5.1Left and right inverses

Definition 2.5.1 (Left inverse). Let $𝑇 \in L (𝑉, 𝑊)$ . A linear map $𝑆 \in L (𝑊, 𝑉)$ is a left inverse of $𝑇$ if $𝑆 𝑇 = {id}_{𝑉}$ , that is, $𝑆 (𝑇 (𝐯)) = 𝐯$ for every $𝐯 \in 𝑉$ . If $𝑇$ has a left inverse, $𝑇$ is left-invertible.

Likewise, for $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ , a matrix $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ is a left inverse of $𝐴$ if $𝐵 𝐴 = 𝐼_{𝑛}$ , and $𝐴$ is left-invertible if it has a left inverse.

Definition 2.5.2 (Right inverse). Let $𝑇 \in L (𝑉, 𝑊)$ . A linear map $𝑆 \in L (𝑊, 𝑉)$ is a right inverse of $𝑇$ if $𝑇 𝑆 = {id}_{𝑊}$ , that is, $𝑇 (𝑆 (𝐰)) = 𝐰$ for every $𝐰 \in 𝑊$ . If $𝑇$ has a right inverse, $𝑇$ is right-invertible.

Likewise, for $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ , a matrix $𝐶 \in 𝐹^{𝑛 \times 𝑚}$ is a right inverse of $𝐴$ if $𝐴 𝐶 = 𝐼_{𝑚}$ , and $𝐴$ is right-invertible if it has a right inverse.

“Left” and “right” refer to the side on which $𝑆$ stands in the product that equals the identity. The two notions are mirror images: $𝑆$ is a left inverse of $𝑇$ exactly when $𝑇$ is a right inverse of $𝑆$ . Note the sizes. A one-sided inverse of an $𝑚 \times 𝑛$ matrix is $𝑛 \times 𝑚$ , so that $𝐵 𝐴$ and $𝐴 𝐶$ are both defined and square, but they are identity matrices of different sizes, $𝐼_{𝑛}$ and $𝐼_{𝑚}$ . Also, we insist that one-sided inverses be linear; Remark 2.5.8 explains what changes without this.

The definitions for matrices are the definitions for maps, applied to $𝑇_{𝐴}$ . Indeed $𝑇_{𝐵} 𝑇_{𝐴} = 𝑇_{𝐵 𝐴}$ by Theorem 2.4.2, and $𝑇_{𝐼_{𝑛}} = {id}_{𝐹^{𝑛}}$ by Proposition 2.3.13. A matrix is determined by its map (Proposition 2.3.17), so $𝐵 𝐴 = 𝐼_{𝑛}$ if and only if $𝑇_{𝐵} 𝑇_{𝐴} = {id}_{𝐹^{𝑛}}$ . Thus $𝐵$ is a left inverse of $𝐴$ exactly when $𝑇_{𝐵}$ is a left inverse of $𝑇_{𝐴}$ . Since every linear map $𝐹^{𝑚} \to 𝐹^{𝑛}$ is $𝑇_{𝐵}$ for some $𝐵$ , it follows that $𝐴$ is left-invertible exactly when $𝑇_{𝐴}$ is. The same holds on the right, and Proposition 2.5.20 extends all of this to matrices of maps in arbitrary bases.

Example 2.5.3 (The $𝑥$ -axis in the plane). Let $𝑇 : 𝐹 \to 𝐹^{2}$ , $𝑇 (𝑥) = (𝑥, 0)$ , whose standard matrix is the column $(\begin{matrix} 1 \\ 0 \end{matrix})$ . For $𝑐 \in 𝐹$ , let

𝑆_{𝑐} : 𝐹^{2} \to 𝐹, 𝑆_{𝑐} (𝑥, 𝑦) = 𝑥 + 𝑐 𝑦 .

Each $𝑆_{𝑐}$ is linear, and $𝑆_{𝑐} (𝑇 (𝑥)) =$ $𝑆_{𝑐} (𝑥, 0) =$ $𝑥$ , so every $𝑆_{𝑐}$ is a left inverse of $𝑇$ . There are no others. By Corollary 2.2.17, a linear map $𝑆 : 𝐹^{2} \to 𝐹$ is $𝑆 (𝑥, 𝑦) =$ $𝑎 𝑥 +$ $𝑏 𝑦$ with $𝑎 = 𝑆 (𝐞_{1})$ and $𝑏 = 𝑆 (𝐞_{2})$ , and $𝑆 (𝑇 (1)) =$ $𝑆 (𝐞_{1}) =$ $𝑎$ , so $𝑆 𝑇 = {id}_{𝐹}$ forces $𝑎 = 1$ . In matrix terms, the left inverses of $(\begin{matrix} 1 \\ 0 \end{matrix})$ are the rows $(\begin{matrix} 1 & 𝑐 \end{matrix})$ with $𝑐 \in 𝐹$ .

On the other hand, $𝑇$ has no right inverse. If $𝑇 𝑆 = {id}_{𝐹^{2}}$ for some $𝑆 \in L (𝐹^{2}, 𝐹)$ , then

(0, 1) = 𝑇 (𝑆 (0, 1)) = (𝑆 (0, 1), 0),

and comparing second coordinates gives $1 = 0$ , which is false. The obstruction is that $(0, 1)$ is not a value of $𝑇$ at all.

Figure 2.5.1. Two left inverses of

𝑇 (𝑥) = (𝑥, 0)

over

ℝ

. The values of

𝑇

form the

𝑥

-axis (blue).

𝑆_{𝑐}

is constant along each line

𝑥 + 𝑐 𝑦 = 𝑡

(grey) and sends the whole line to its intersection

(𝑡, 0)

with the axis, identified with

𝑡

. Both maps fix the axis pointwise, as a left inverse must, but they treat the point

𝐩

differently.

Intuition (Choices off the image). A left inverse

𝑆

𝑇

must send

𝑇 (𝐯)

back to

𝐯

, so its values on the set

𝑇 (𝑉)

of values of

𝑇

are forced. Off that set,

𝑆

is constrained only by linearity, and in Example 2.5.3 this leaves a free choice:

𝑆_{𝑐}

collapses the plane onto the

𝑥

-axis along the direction

(- 𝑐, 1)

, and every direction other than that of the axis itself works (Figure 2.5.1). A right inverse has the opposite problem. It must produce, for every

𝐰 \in 𝑊

, a vector

𝑆 (𝐰)

with

𝑇 (𝑆 (𝐰)) = 𝐰

, and this is impossible as soon as some

𝐰

is not a value of

𝑇

. So left inverses are about

𝑇

losing no information, and right inverses about

𝑇

reaching everything.

To make this precise, recall that a function $𝑓 : 𝑋 \to 𝑌$ is injective if $𝑓 (𝑥) = 𝑓 (𝑥^{'})$ implies $𝑥 = 𝑥^{'}$ , and surjective if every $𝑦 \in 𝑌$ equals $𝑓 (𝑥)$ for some $𝑥 \in 𝑋$ (Section A.1). These properties of linear maps are the subject of Section 2.6.

Proposition 2.5.4 (Left inverses give uniqueness). Suppose $𝑇 \in L (𝑉, 𝑊)$ has a left inverse $𝑆$ . If $𝑇 (𝐯) = 𝐰$ , then $𝐯 = 𝑆 (𝐰)$ . In particular, $𝑇$ is injective.

Likewise, if $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ has a left inverse $𝐵$ and $𝐴 𝐱 = 𝐲$ , then $𝐱 = 𝐵 𝐲$ , and $𝑇_{𝐴}$ is injective.

Proof.routine

Uses. Definition 2.5.1 (Left inverse)

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

Hint 1. Apply

𝑆

to both sides of

𝑇 (𝐯) = 𝐰

Hint 2. For injectivity, suppose

𝑇 (𝐯) = 𝑇 (𝐯^{'})

. Both

𝐯

and

𝐯^{'}

solve

𝑇 (𝐱) = 𝐰

for one and the same

𝐰

Sketch. Applying

𝑆

𝑇 (𝐯) = 𝐰

gives

𝐯 =

𝑆 (𝑇 (𝐯)) =

𝑆 (𝐰)

. If

𝑇 (𝐯) = 𝑇 (𝐯^{'}) = 𝐰

, then

𝐯

and

𝐯^{'}

both equal

𝑆 (𝐰)

. The matrix statement is the same argument with

𝐵 𝐴 = 𝐼_{𝑛}

Suppose $𝑇 (𝐯) = 𝐰$ . Applying $𝑆$ to both sides and using $𝑆 𝑇 = {id}_{𝑉}$ ,

𝐯 = {id}_{𝑉} (𝐯) = 𝑆 (𝑇 (𝐯)) = 𝑆 (𝐰) .

Now let $𝐯, 𝐯^{'} \in 𝑉$ with $𝑇 (𝐯) = 𝑇 (𝐯^{'})$ , and call this vector $𝐰$ . By what we just showed, $𝐯 = 𝑆 (𝐰) = 𝐯^{'}$ . So $𝑇$ is injective.

For the matrix statement, if $𝐴 𝐱 = 𝐲$ , then $𝐱 =$ $𝐼_{𝑛} 𝐱 =$ $(𝐵 𝐴) 𝐱 =$ $𝐵 (𝐴 𝐱) =$ $𝐵 𝐲$ by Theorem 2.4.2, and injectivity of $𝑇_{𝐴}$ follows as before. □

Proposition 2.5.5 (Right inverses give existence). Suppose $𝑇 \in L (𝑉, 𝑊)$ has a right inverse $𝑆$ . Then for every $𝐰 \in 𝑊$ , the vector $𝐯 = 𝑆 (𝐰)$ satisfies $𝑇 (𝐯) = 𝐰$ . In particular, $𝑇$ is surjective.

Likewise, if $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ has a right inverse $𝐶$ , then $𝐱 = 𝐶 𝐲$ satisfies $𝐴 𝐱 = 𝐲$ for every $𝐲 \in 𝐹^{𝑚}$ , and $𝑇_{𝐴}$ is surjective.

Proof.routine

Uses. Definition 2.5.2 (Right inverse)

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

Hint 1. Fix

𝐰 \in 𝑊

. Which vector could possibly map to

𝐰

, if all you know about

𝑇

𝑇 𝑆 = {id}_{𝑊}

Hint 2. Compute

𝑇 (𝑆 (𝐰))

Sketch.

𝑇 (𝑆 (𝐰)) =

(𝑇 𝑆) (𝐰) =

{id}_{𝑊} (𝐰) =

𝐰

, so every

𝐰

is a value of

𝑇

. For matrices,

𝐴 (𝐶 𝐲) =

(𝐴 𝐶) 𝐲 =

𝐼_{𝑚} 𝐲 =

𝐲

Let $𝐰 \in 𝑊$ and $𝐯 = 𝑆 (𝐰)$ . Then

𝑇 (𝐯) = 𝑇 (𝑆 (𝐰)) = (𝑇 𝑆) (𝐰) = {id}_{𝑊} (𝐰) = 𝐰 .

As $𝐰$ was arbitrary, every vector of $𝑊$ is a value of $𝑇$ , so $𝑇$ is surjective. For matrices, $𝐴 (𝐶 𝐲) =$ $(𝐴 𝐶) 𝐲 =$ $𝐼_{𝑚} 𝐲 =$ $𝐲$ by Theorem 2.4.2 and Proposition 2.3.13. □

Neither proposition says more than it claims. A left inverse produces the only possible solution, but that candidate need not be a solution. With $𝐴 = (\begin{matrix} 1 \\ 0 \end{matrix})$ , its left inverse $𝐵 = (\begin{matrix} 1 & 0 \end{matrix})$ from Example 2.5.3, and $𝐲 = (0, 1)$ , the candidate is $𝐵 𝐲 = 0$ , but $𝐴 𝑥 = (𝑥, 0) \neq 𝐲$ for every $𝑥$ : the system has no solution at all. A right inverse produces a solution, but not necessarily the only one. The row $𝐴 = (\begin{matrix} 1 & 0 \end{matrix})$ has the right inverses $(\begin{matrix} 1 \\ 0 \end{matrix})$ and $(\begin{matrix} 1 \\ 1 \end{matrix})$ , and applied to $𝑦 = 1$ they give the two different solutions $(1, 0)$ and $(1, 1)$ of $𝐴 𝐱 = 1$ . So “at most one solution” and “at least one solution” really are separate properties, detected by inverses on separate sides.

The uniqueness half has a useful reformulation for products.

Corollary 2.5.6 (One-sided cancellation). Let

𝑇 \in L (𝑉, 𝑊)

(a)If $𝑇$ is left-invertible, then $𝑇 𝑅_{1} = 𝑇 𝑅_{2}$ implies $𝑅_{1} = 𝑅_{2}$ , for all $𝑅_{1}, 𝑅_{2} \in L (𝑈, 𝑉)$ .
(b)If $𝑇$ is right-invertible, then $𝑅_{1} 𝑇 = 𝑅_{2} 𝑇$ implies $𝑅_{1} = 𝑅_{2}$ , for all $𝑅_{1}, 𝑅_{2} \in L (𝑊, 𝑈)$ .

Likewise, if

𝐴 \in 𝐹^{𝑚 \times 𝑛}

is left-invertible, then

𝐴 𝑋 = 𝐴 𝑌

implies

𝑋 = 𝑌

for

𝑋, 𝑌 \in 𝐹^{𝑛 \times 𝑝}

, and if

𝐴

is right-invertible, then

𝑋 𝐴 = 𝑌 𝐴

implies

𝑋 = 𝑌

for

𝑋, 𝑌 \in 𝐹^{𝑝 \times 𝑚}

Proof.routine

Uses. Definition 2.5.1 (Left inverse), Definition 2.5.2 (Right inverse), Proposition 2.1.24 (Algebra of composition), Proposition 2.4.14 (Algebra of matrices)

Hint 1. Multiply both sides of

𝑇 𝑅_{1} = 𝑇 𝑅_{2}

by a left inverse

𝑆

𝑇

, on the left.

Hint 2. Regroup with associativity:

𝑆 (𝑇 𝑅_{𝑖}) =

(𝑆 𝑇) 𝑅_{𝑖} =

𝑅_{𝑖}

Sketch. With

𝑆 𝑇 = {id}_{𝑉}

𝑅_{1} =

(𝑆 𝑇) 𝑅_{1} =

𝑆 (𝑇 𝑅_{1}) =

𝑆 (𝑇 𝑅_{2}) =

(𝑆 𝑇) 𝑅_{2} =

𝑅_{2}

. With

𝑇 𝑆 = {id}_{𝑊}

𝑅_{1} =

𝑅_{1} (𝑇 𝑆) =

(𝑅_{1} 𝑇) 𝑆 =

(𝑅_{2} 𝑇) 𝑆 =

𝑅_{2}

. Matrices are the same with Proposition 2.4.14.

(a) Let $𝑆$ be a left inverse of $𝑇$ . By Proposition 2.1.24,

\begin{aligned} 𝑅_{1} & = {id}_{𝑉} 𝑅_{1} = (𝑆 𝑇) 𝑅_{1} = 𝑆 (𝑇 𝑅_{1}) \\ = 𝑆 (𝑇 𝑅_{2}) = (𝑆 𝑇) 𝑅_{2} = {id}_{𝑉} 𝑅_{2} = 𝑅_{2} . \end{aligned}

(b) Let $𝑆$ be a right inverse of $𝑇$ . Then

\begin{aligned} 𝑅_{1} & = 𝑅_{1} {id}_{𝑊} = 𝑅_{1} (𝑇 𝑆) = (𝑅_{1} 𝑇) 𝑆 \\ = (𝑅_{2} 𝑇) 𝑆 = 𝑅_{2} (𝑇 𝑆) = 𝑅_{2} . \end{aligned}

The matrix statements are the same computations, with $𝐵 𝐴 = 𝐼_{𝑛}$ or $𝐴 𝐶 = 𝐼_{𝑚}$ and associativity and identities from Proposition 2.4.14. □

This answers the question left open in Example 2.4.19: a left-invertible matrix can be cancelled on the left, and in particular $𝐴 𝑋 = 0$ forces $𝑋 = 0$ . The side matters. $𝐴 = (\begin{matrix} 1 \\ 0 \end{matrix})$ is left-invertible, yet $(\begin{matrix} 0 & 1 \end{matrix}) 𝐴 = 0 = (\begin{matrix} 0 & 0 \end{matrix}) 𝐴$ ; cancelling $𝐴$ on the right would need a right inverse.

Remark 2.5.7 (The converses). Every injective linear map between finite-dimensional spaces has a left inverse, and every surjective one has a right inverse. For

𝑇_{𝐴}

these statements are among the Fundamental Theorems (Theorem 2.7.2, Theorem 2.7.4), which are stated in Section 2.7, where one-sided inverses are also built directly from a basis, and completed in Section 3.5, where they are constructed by row reduction. The converses are harder than the two propositions because a left inverse must be defined, linearly, on all of

𝑊

, while injectivity of

𝑇

only says what it has to do on

𝑇 (𝑉)

Remark 2.5.8 (Why one-sided inverses must be linear). Every injective function

𝑓 : 𝑋 \to 𝑌

with

𝑋

nonempty has a left inverse function, and every surjective function has a right inverse function (Section A.1). For a linear map we ask for more, a linear one-sided inverse. Without linearity there are many more left inverses: for

𝑇 (𝑥) = (𝑥, 0)

from

ℝ

ℝ^{2}

, the function

𝑔 (𝑥, 𝑦) = 𝑥 + 𝑦^{2}

satisfies

𝑔 (𝑇 (𝑥)) = 𝑥

for all

𝑥

, but

𝑔

is not linear, since

𝑔 (0, 2) =

4 \neq

2 𝑔 (0, 1)

. For two-sided inverses the distinction disappears: the inverse function of a bijective linear map is automatically linear (Proposition 2.7.7).

Exercise 2.5.9.routine Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ .

(a)Show that if some column of $𝐴$ is zero, then $𝐴$ has no left inverse.
(b)Show that if some row of $𝐴$ is zero, then $𝐴$ has no right inverse.

Solution.

Uses. Proposition 2.5.4 (Left inverses give uniqueness), Proposition 2.5.5 (Right inverses give existence), Proposition 2.4.8 (Four views of the product), Proposition 2.3.13 (Linearity of the product)

Hint 1. If column

𝑗

is zero, then

𝐴 𝐞_{𝑗} = 𝟎 = 𝐴 𝟎

. What does that say about

𝑇_{𝐴}

Hint 2. If row

𝑖

is zero, then row

𝑖

𝐴 𝐶

is zero for every

𝐶

, by the row view. Compare with row

𝑖

𝐼_{𝑚}

Sketch. (a)

𝐴 𝐞_{𝑗} = 𝟎 = 𝐴 𝟎

with

𝐞_{𝑗} \neq 𝟎

, so

𝑇_{𝐴}

is not injective, and Proposition 2.5.4 rules out a left inverse. (b) Row

𝑖

𝐴 𝐶

is row

𝑖

𝐴

times

𝐶

, which is zero, while row

𝑖

𝐼_{𝑚}

𝐞_{𝑖}^{𝖳} \neq 0

(a) Suppose column $𝑗$ of $𝐴$ is zero. By Proposition 2.3.13, $𝐴 𝐞_{𝑗}$ is column $𝑗$ of $𝐴$ , so $𝐴 𝐞_{𝑗} = 𝟎 = 𝐴 𝟎$ although $𝐞_{𝑗} \neq 𝟎$ . Hence $𝑇_{𝐴}$ is not injective, and by Proposition 2.5.4, $𝐴$ has no left inverse. (Directly: for every $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ , $(𝐵 𝐴) 𝐞_{𝑗} =$ $𝐵 (𝐴 𝐞_{𝑗}) =$ $𝟎$ , whereas $𝐼_{𝑛} 𝐞_{𝑗} = 𝐞_{𝑗}$ .)

(b) Suppose row $𝑖$ of $𝐴$ is zero, and let $𝐶 \in 𝐹^{𝑛 \times 𝑚}$ . By the row view (Proposition 2.4.8(c)), row $𝑖$ of $𝐴 𝐶$ is a combination of the rows of $𝐶$ with the entries of row $𝑖$ of $𝐴$ as coefficients, so it is zero. Row $𝑖$ of $𝐼_{𝑚}$ has a $1$ in position $𝑖$ , so $𝐴 𝐶 \neq 𝐼_{𝑚}$ . (In the language of Proposition 2.5.5: every vector $𝐴 𝐱$ has $𝑖$ -th entry $0$ , so $𝐞_{𝑖}$ is not a value of $𝑇_{𝐴}$ .)

2.5.2Invertible maps and matrices

In Example 2.5.3 the map $𝑇$ had many left inverses and no right inverse. That is no accident. A left inverse $𝑆_{𝑙}$ and a right inverse $𝑆_{𝑟}$ of $𝑇$ are both maps $𝑊 \to 𝑉$ , and to compare them we want a single expression that involves both. The composite $𝑆_{𝑙} 𝑇 𝑆_{𝑟}$ is one: grouping it as $𝑆_{𝑙} (𝑇 𝑆_{𝑟})$ uses that $𝑆_{𝑟}$ is a right inverse, and grouping it as $(𝑆_{𝑙} 𝑇) 𝑆_{𝑟}$ uses that $𝑆_{𝑙}$ is a left inverse.

Theorem 2.5.10 (Left and right inverses agree). Let $𝑇 \in L (𝑉, 𝑊)$ . If $𝑆_{𝑙} \in L (𝑊, 𝑉)$ is a left inverse of $𝑇$ and $𝑆_{𝑟} \in L (𝑊, 𝑉)$ is a right inverse of $𝑇$ , then $𝑆_{𝑙} = 𝑆_{𝑟}$ .

Consequently, if $𝑇$ has both a left inverse and a right inverse, then it has exactly one left inverse and exactly one right inverse, these are equal, and this map is both a left and a right inverse of $𝑇$ .

The same holds for matrices: if $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵, 𝐶 \in 𝐹^{𝑛 \times 𝑚}$ satisfy $𝐵 𝐴 = 𝐼_{𝑛}$ and $𝐴 𝐶 = 𝐼_{𝑚}$ , then $𝐵 = 𝐶$ .

Proof.routine

Uses. Definition 2.5.1 (Left inverse), Definition 2.5.2 (Right inverse), Proposition 2.1.24 (Algebra of composition), Proposition 2.4.14 (Algebra of matrices)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Find one composite that involves

𝑆_{𝑙}

𝑇

, and

𝑆_{𝑟}

, and evaluate it in two ways.

Hint 2. Group

𝑆_{𝑙} 𝑇 𝑆_{𝑟}

𝑆_{𝑙} (𝑇 𝑆_{𝑟})

, and then as

(𝑆_{𝑙} 𝑇) 𝑆_{𝑟}

Hint 3. For the second paragraph: if

𝑅

is another left inverse, apply the first paragraph to

𝑅

and

𝑆_{𝑟}

Sketch. By associativity,

𝑆_{𝑙} =

𝑆_{𝑙} {id}_{𝑊} =

𝑆_{𝑙} (𝑇 𝑆_{𝑟}) =

(𝑆_{𝑙} 𝑇) 𝑆_{𝑟} =

{id}_{𝑉} 𝑆_{𝑟} =

𝑆_{𝑟}

. Every left inverse therefore equals the fixed right inverse

𝑆_{𝑟}

, so there is only one left inverse; symmetrically there is only one right inverse, and the two coincide. Matrices:

𝐵 =

𝐵 (𝐴 𝐶) =

(𝐵 𝐴) 𝐶 =

𝐶

By Proposition 2.1.24,

\begin{aligned} 𝑆_{𝑙} & = 𝑆_{𝑙} {id}_{𝑊} & (identity) \\ = 𝑆_{𝑙} (𝑇 𝑆_{𝑟}) & (as 𝑆_{𝑟} is a right inverse) \\ = (𝑆_{𝑙} 𝑇) 𝑆_{𝑟} & (associativity) \\ = {id}_{𝑉} 𝑆_{𝑟} & (as 𝑆_{𝑙} is a left inverse) \\ = 𝑆_{𝑟} & (identity) . \end{aligned}

Now suppose $𝑇$ has a left inverse $𝑆_{𝑙}$ and a right inverse $𝑆_{𝑟}$ . If $𝑅$ is any left inverse, the first part applied to $𝑅$ and $𝑆_{𝑟}$ gives $𝑅 = 𝑆_{𝑟}$ . So every left inverse equals $𝑆_{𝑟}$ , and $𝑇$ has exactly one left inverse, namely $𝑆_{𝑟}$ . In the same way every right inverse equals $𝑆_{𝑙}$ , so $𝑆_{𝑙}$ is the only right inverse. Finally, the common map $𝑆_{𝑙} = 𝑆_{𝑟}$ is a left inverse (as $𝑆_{𝑙}$ ) and a right inverse (as $𝑆_{𝑟}$ ).

For matrices, by Proposition 2.4.14,

𝐵 = 𝐵 𝐼_{𝑚} = 𝐵 (𝐴 𝐶) = (𝐵 𝐴) 𝐶 = 𝐼_{𝑛} 𝐶 = 𝐶,

and the rest follows as for maps. □

This is Technique 1.1.8 once more. The expression $𝑆_{𝑙} 𝑇 𝑆_{𝑟}$ plays the role that $0 + 0^{'}$ played for the zero of a field (Theorem 1.1.9): it is evaluated twice, once using the property of each candidate. Read contrapositively, the theorem explains Example 2.5.3.

Corollary 2.5.11 (Many one-sided inverses). If

𝑇 \in L (𝑉, 𝑊)

has two different left inverses, then

𝑇

has no right inverse. If

𝑇

has two different right inverses, then

𝑇

has no left inverse. The same holds for matrices.

Proof.routine

Uses. Theorem 2.5.10 (Left and right inverses agree)

Techniques. Contrapositive (Technique 1.5.4)

Hint 1. Prove the contrapositive.

Hint 2. If

𝑇

has a right inverse and a left inverse, how many left inverses can it have?

Sketch. If

𝑇

had a right inverse

𝑆_{𝑟}

, then by Theorem 2.5.10 every left inverse would equal

𝑆_{𝑟}

, so any two left inverses would be equal. The second statement is symmetric.

Suppose

𝑆

and

𝑆^{'}

are left inverses of

𝑇

and

𝑇

has a right inverse

𝑆_{𝑟}

. By Theorem 2.5.10,

𝑆 = 𝑆_{𝑟}

and

𝑆^{'} = 𝑆_{𝑟}

, so

𝑆 = 𝑆^{'}

. Hence if

𝑇

has two different left inverses, it has no right inverse. The statement about right inverses is proved the same way, and the matrix statements follow from the matrix part of Theorem 2.5.10. □

In Example 2.5.3, $𝑆_{0} (𝑥, 𝑦) = 𝑥$ and $𝑆_{1} (𝑥, 𝑦) = 𝑥 + 𝑦$ are different left inverses of $𝑇$ , since $𝑆_{0} (0, 1) =$ $0 \neq$ $1 =$ $𝑆_{1} (0, 1)$ . So Corollary 2.5.11 gives a second proof that $𝑇$ has no right inverse, one that never needed to find a vector outside $𝑇 (𝐹)$ .

Definition 2.5.12 (Invertible). A linear map $𝑇 \in L (𝑉, 𝑊)$ is invertible if there is a linear map $𝑆 \in L (𝑊, 𝑉)$ that is both a left and a right inverse of $𝑇$ :

𝑆 𝑇 = {id}_{𝑉} and 𝑇 𝑆 = {id}_{𝑊} .

Such an $𝑆$ is a two-sided inverse of $𝑇$ . By Theorem 2.5.10, $𝑇$ is invertible if and only if it is both left-invertible and right-invertible, and in that case it has exactly one two-sided inverse, called the inverse of $𝑇$ and written $𝑇^{- 1}$ .

Likewise, $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ is invertible if there is $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ with $𝐵 𝐴 = 𝐼_{𝑛}$ and $𝐴 𝐵 = 𝐼_{𝑚}$ . Again $𝐵$ is unique; it is the inverse of $𝐴$ , written $𝐴^{- 1}$ , and $𝐴$ is invertible if and only if it is both left-invertible and right-invertible.

So an invertible map satisfies $𝑇^{- 1} 𝑇 = {id}_{𝑉}$ and $𝑇 𝑇^{- 1} = {id}_{𝑊}$ . By Proposition 2.5.4 and Proposition 2.5.5 it is injective and surjective, so for every $𝐰 \in 𝑊$ the equation $𝑇 (𝐯) = 𝐰$ has exactly one solution, $𝐯 = 𝑇^{- 1} (𝐰)$ . In particular $𝑇^{- 1}$ is the inverse function of the bijection $𝑇$ . Conversely, every bijective linear map is invertible, because its inverse function turns out to be linear (Proposition 2.7.7). By the discussion after Definition 2.5.2, a matrix $𝐴$ is invertible exactly when $𝑇_{𝐴}$ is, and then ${(𝑇_{𝐴})}^{- 1} = 𝑇_{𝐴^{- 1}}$ ; this is also part (c) of Proposition 2.5.20 below.

Remark 2.5.13 (Invertible matrices are square). The definition does not ask

𝐴

to be square, and nothing so far rules out an invertible

𝑚 \times 𝑛

matrix with

𝑚 \neq 𝑛

: Theorem 2.5.10 shows that a left inverse and a right inverse coincide, not that

𝑚 = 𝑛

. In fact every invertible matrix is square (Corollary 3.5.15). The proof uses rank (Section 3.5), although over

ℚ

ℝ

, and

ℂ

the trace gives a short argument (Exercise 2.5.44). Nothing below depends on it.

Example 2.5.14 (Invertible maps and matrices).

(a) ${id}_{𝑉}$ is its own inverse, and $𝐼_{𝑛}^{- 1} = 𝐼_{𝑛}$ . For $𝑐 \in 𝐹$ with $𝑐 \neq 0$ , $𝑐 {id}_{𝑉}$ is invertible with inverse $𝑐^{- 1} {id}_{𝑉}$ , since $(𝑐^{- 1} {id}_{𝑉}) (𝑐 {id}_{𝑉}) =$ $(𝑐^{- 1} 𝑐) {id}_{𝑉} =$ ${id}_{𝑉}$ in both orders (Proposition 2.1.24(d)).
(b)Rotating by $𝜃$ and then by $- 𝜃$ in either order is rotating by $0$ , so $𝑅_{𝜃} 𝑅_{- 𝜃} =$ $𝑅_{- 𝜃} 𝑅_{𝜃} =$ ${id}_{ℝ^{2}}$ (Exercise 2.1.43), and $𝑅_{𝜃}^{- 1} = 𝑅_{- 𝜃}$ . In matrices,
${(\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix})}^{- 1} = (\begin{matrix} cos 𝜃 & sin 𝜃 \\ - sin 𝜃 & cos 𝜃 \end{matrix}) .$
A reflection satisfies $𝑆_{ℓ}^{2} = {id}_{ℝ^{2}}$ (Exercise 2.1.42), so it is its own inverse: reflecting twice returns every point.
(c)For a basis $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ of $𝑉$ , the coordinate map $𝑉 \to 𝐹^{𝑛}$ , $𝐯 \mapsto {[𝐯]}_{ℬ︀}$ , is invertible, and its inverse is $(𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto$ $𝑎_{1} 𝐛_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐛_{𝑛}$ (Example 2.1.8).
(d)If $𝑑_{1}, \dots, 𝑑_{𝑛} \in 𝐹$ are all nonzero, then $diag (𝑑_{1}, \dots, 𝑑_{𝑛})$ is invertible with inverse $diag (𝑑_{1}^{- 1}, \dots, 𝑑_{𝑛}^{- 1})$ , because diagonal matrices multiply entry by entry (Example 2.4.10). If some $𝑑_{𝑖} = 0$ , then column $𝑖$ is zero, and Exercise 2.5.9 shows that the matrix has no left inverse.
(e)Suppose $1 + 1 \neq 0$ in $𝐹$ , and write $2 = 1 + 1$ . Let $𝑇 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be $\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘} \mapsto$ $\sum_{𝑘} 2^{𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ , the substitution $𝑝 (𝑥) \mapsto 𝑝 (2 𝑥)$ . The map $𝑆 : \sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘} \mapsto$ $\sum_{𝑘} 2^{- 𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ is linear for the same reason as $𝑇$ , and both $𝑆 𝑇$ and $𝑇 𝑆$ multiply the coefficient of each $𝑥^{𝑘}$ by $2^{𝑘} 2^{- 𝑘} = 1$ . So $𝑇^{- 1} = 𝑆$ , that is, $𝑇^{- 1} (𝑝) = 𝑝 (𝑥 / 2)$ .
(f)The map $𝑇 (𝑥, 𝑦) =$ $(𝑦, 𝑥 - 𝑦)$ of Exercise 2.1.25 satisfies $𝑇 (𝑇 + id) =$ $(𝑇 + id) 𝑇 =$ $id$ , so $𝑇^{- 1} = 𝑇 + id$ , that is, $𝑇^{- 1} (𝑥, 𝑦) =$ $(𝑥 + 𝑦, 𝑥)$ .

Non-example 2.5.15 (Maps that are not invertible). The projection

{Pr}_{ℓ}

onto a line

ℓ

through

𝟎

ℝ^{2}

(Example 2.1.11) is neither left- nor right-invertible: it sends every vector perpendicular to

ℓ

𝟎

, so it is not injective, and its values all lie on

ℓ

, so it is not surjective. The zero map

𝑉 \to 𝑊

is not left-invertible unless

𝑉 = {𝟎}

, and not right-invertible unless

𝑊 = {𝟎}

. Differentiation

𝐷

𝐹 [𝑥]

is not left-invertible, since

𝐷 (1) = 0 = 𝐷 (0)

; whether it is right-invertible is taken up in Example 2.5.24.

2.5.3The algebra of inverses

To show that a map is invertible, we never have to solve for its inverse. It is enough to name a candidate and check it.

Technique 2.5.16 (Guess the inverse, then check both products). To show that $𝑇$ is invertible and to identify $𝑇^{- 1}$ , produce a candidate $𝑆$ and verify $𝑆 𝑇 = id$ and $𝑇 𝑆 = id$ . Then $𝑇$ is invertible and $𝑆 = 𝑇^{- 1}$ , by the uniqueness in Theorem 2.5.10. How the candidate was found plays no role in the proof.

Candidates come from undoing: to undo a rotation by $𝜃$ , rotate by $- 𝜃$ , and to undo a composite, undo its steps in reverse order. A polynomial identity also produces one: if $𝑇^{2} + 𝑇 = id$ , then $𝑇 (𝑇 + id) =$ $id =$ $(𝑇 + id) 𝑇$ , and if $𝑁^{𝑘} = 0$ , then $(id - 𝑁) (id + 𝑁 + \dots + 𝑁^{𝑘 - 1}) =$ $id -$ $𝑁^{𝑘} =$ $id$ , which is the geometric series $1 / (1 - 𝑥) =$ $1 +$ $𝑥 +$ $𝑥^{2} +$ $\dots$ cut off (Exercise 2.5.39). Heuristics of this kind may be as informal as you like, because the verification is what proves the result.

Check both products, unless you already know that $𝑇$ has an inverse on the other side. One product shows only that $𝑆$ is a one-sided inverse, and one-sided inverses need not be two-sided (Example 2.5.24). The canonical example is Proposition 2.5.17.

Used in 19 places and 1 practice exercise

Proposition 2.5.17, Exercise 2.5.19, Exercise 2.5.22, Proposition 2.5.26, Proposition 2.5.28, Exercise 2.5.35, Exercise 2.5.38, Exercise 2.5.39, Exercise 2.5.40, Exercise 2.5.47, Exercise 2.7.31, Exercise 2.8.9, Exercise 2.8.10, Exercise 2.8.14, Problem 2.5, Problem 2.10, Exercise 4.3.37, Proposition 6.6.18, Exercise 8.1.35; 1 practice exercise.

Intuition (Socks and shoes). To undo “first put on socks, then shoes”, take off the shoes first, then the socks. Likewise, to undo

𝑇 𝑆

, which is “first

𝑆

, then

𝑇

”, first undo

𝑇

and then undo

𝑆

. So the candidate for

{(𝑇 𝑆)}^{- 1}

𝑆^{- 1} 𝑇^{- 1}

, with the order reversed, as it was for transposes in Proposition 2.4.23.

Proposition 2.5.17 (Algebra of inverses).

(a)If $𝑇 \in L (𝑉, 𝑊)$ is invertible, then so is $𝑇^{- 1}$ , and ${(𝑇^{- 1})}^{- 1} = 𝑇$ .
(b)If $𝑆 \in L (𝑈, 𝑉)$ and $𝑇 \in L (𝑉, 𝑊)$ are invertible, then so is $𝑇 𝑆 \in L (𝑈, 𝑊)$ , and
${(𝑇 𝑆)}^{- 1} = 𝑆^{- 1} 𝑇^{- 1} .$
(c)If $𝑇 \in L (𝑉, 𝑊)$ is invertible and $𝑎 \in 𝐹$ is nonzero, then $𝑎 𝑇$ is invertible, and ${(𝑎 𝑇)}^{- 1} = 𝑎^{- 1} 𝑇^{- 1}$ .
(d)Statements (a)–(c) hold for matrices of compatible sizes. Moreover, if $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ is invertible, then so is $𝐴^{𝖳}$ , and ${(𝐴^{𝖳})}^{- 1} = {(𝐴^{- 1})}^{𝖳}$ .

Proof.routine

Uses. Definition 2.5.12 (Invertible), Theorem 2.5.10 (Left and right inverses agree), Proposition 2.1.24 (Algebra of composition), Proposition 2.4.14 (Algebra of matrices), Proposition 2.4.23 (Transpose of a product), Proposition 2.4.22 (Transpose is a linear involution)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. In each part the formula names the candidate. Multiply it by the given map on both sides and simplify.

Hint 2. For (b), regroup

(𝑇 𝑆) (𝑆^{- 1} 𝑇^{- 1})

𝑇 (𝑆 𝑆^{- 1}) 𝑇^{- 1}

Hint 3. For (d), transpose the equations

𝐴^{- 1} 𝐴 = 𝐼_{𝑛}

and

𝐴 𝐴^{- 1} = 𝐼_{𝑚}

with Proposition 2.4.23.

Sketch. (a) The equations

𝑇^{- 1} 𝑇 = {id}_{𝑉}

and

𝑇 𝑇^{- 1} = {id}_{𝑊}

say that

𝑇

is a two-sided inverse of

𝑇^{- 1}

. (b)

(𝑇 𝑆) (𝑆^{- 1} 𝑇^{- 1}) =

𝑇 (𝑆 𝑆^{- 1}) 𝑇^{- 1} =

𝑇 𝑇^{- 1} =

{id}_{𝑊}

, and similarly in the other order. (c)

(𝑎^{- 1} 𝑇^{- 1}) (𝑎 𝑇) =

(𝑎^{- 1} 𝑎) 𝑇^{- 1} 𝑇 =

{id}_{𝑉}

, and similarly. (d)

𝐴^{𝖳} {(𝐴^{- 1})}^{𝖳} =

{(𝐴^{- 1} 𝐴)}^{𝖳} =

𝐼_{𝑛}

and

{(𝐴^{- 1})}^{𝖳} 𝐴^{𝖳} =

{(𝐴 𝐴^{- 1})}^{𝖳} =

𝐼_{𝑚}

In each part we exhibit a two-sided inverse; by Definition 2.5.12 the map or matrix is then invertible, and by Theorem 2.5.10 the exhibited inverse is the inverse.

(a) The equations $𝑇^{- 1} 𝑇 = {id}_{𝑉}$ and $𝑇 𝑇^{- 1} = {id}_{𝑊}$ , read with $𝑇^{- 1}$ as the map being inverted, say that $𝑇$ is a left and a right inverse of $𝑇^{- 1}$ . So $𝑇^{- 1}$ is invertible with inverse $𝑇$ .

(b) By associativity and the identity laws (Proposition 2.1.24),

\begin{aligned} (𝑇 𝑆) (𝑆^{- 1} 𝑇^{- 1}) & = 𝑇 (𝑆 𝑆^{- 1}) 𝑇^{- 1} = 𝑇 {id}_{𝑉} 𝑇^{- 1} \\ = 𝑇 𝑇^{- 1} = {id}_{𝑊}, \\ (𝑆^{- 1} 𝑇^{- 1}) (𝑇 𝑆) & = 𝑆^{- 1} (𝑇^{- 1} 𝑇) 𝑆 = 𝑆^{- 1} {id}_{𝑉} 𝑆 \\ = 𝑆^{- 1} 𝑆 = {id}_{𝑈} . \end{aligned}

(c) By Proposition 2.1.24(d), $(𝑎^{- 1} 𝑇^{- 1}) (𝑎 𝑇) =$ $(𝑎^{- 1} 𝑎) (𝑇^{- 1} 𝑇) =$ ${id}_{𝑉}$ , and likewise $(𝑎 𝑇) (𝑎^{- 1} 𝑇^{- 1}) =$ ${id}_{𝑊}$ .

(d) For matrices, the computations in (a)–(c) work verbatim, with Proposition 2.4.14 in place of Proposition 2.1.24 and identity matrices in place of identity maps. For the transpose, $𝐴^{- 1} \in 𝐹^{𝑛 \times 𝑚}$ , with $𝐴^{- 1} 𝐴 = 𝐼_{𝑛}$ and $𝐴 𝐴^{- 1} = 𝐼_{𝑚}$ . By Proposition 2.4.23 and Proposition 2.4.22,

\begin{aligned} 𝐴^{𝖳} {(𝐴^{- 1})}^{𝖳} & = {(𝐴^{- 1} 𝐴)}^{𝖳} = 𝐼_{𝑛}^{𝖳} = 𝐼_{𝑛}, \\ {(𝐴^{- 1})}^{𝖳} 𝐴^{𝖳} & = {(𝐴 𝐴^{- 1})}^{𝖳} = 𝐼_{𝑚}^{𝖳} = 𝐼_{𝑚}, \end{aligned}

so ${(𝐴^{- 1})}^{𝖳}$ is a two-sided inverse of $𝐴^{𝖳}$ . □

By induction on $𝑘$ , part (b) gives ${(𝑇_{1} 𝑇_{2} \dots 𝑇_{𝑘})}^{- 1} =$ $𝑇_{𝑘}^{- 1} \dots 𝑇_{2}^{- 1} 𝑇_{1}^{- 1}$ for invertible $𝑇_{1}, \dots, 𝑇_{𝑘}$ that can be composed, and in particular ${(𝑇^{𝑘})}^{- 1} = {(𝑇^{- 1})}^{𝑘}$ for invertible $𝑇 \in L (𝑉)$ . We write $𝑇^{- 𝑘}$ for this map, and $𝐴^{- 𝑘}$ for ${(𝐴^{- 1})}^{𝑘}$ .

Part (b) also runs backwards: if two of $𝑆$ , $𝑇$ , and $𝑇 𝑆$ are invertible, the third can be written as a product of invertible maps.

Corollary 2.5.18 (Two out of three). Let

𝑆 \in L (𝑈, 𝑉)

and

𝑇 \in L (𝑉, 𝑊)

. If two of

𝑆

𝑇

, and

𝑇 𝑆

are invertible, then so is the third. The same holds for matrices.

Proof.routine

Uses. Proposition 2.5.17 (Algebra of inverses), Proposition 2.1.24 (Algebra of composition)

Hint 1. If

𝑆

and

𝑇

are invertible, this is Proposition 2.5.17(b). In the other two cases, express the unknown factor as a product of two invertible maps.

Hint 2.

𝑆 = 𝑇^{- 1} (𝑇 𝑆)

and

𝑇 = (𝑇 𝑆) 𝑆^{- 1}

Sketch. The case where

𝑆

and

𝑇

are invertible is Proposition 2.5.17(b). If

𝑇

and

𝑇 𝑆

are invertible, then

𝑆 = 𝑇^{- 1} (𝑇 𝑆)

is a product of invertible maps; if

𝑆

and

𝑇 𝑆

are, then

𝑇 = (𝑇 𝑆) 𝑆^{- 1}

is. Apply Proposition 2.5.17(a) and (b).

If $𝑆$ and $𝑇$ are invertible, then $𝑇 𝑆$ is invertible by Proposition 2.5.17(b).

If $𝑇$ and $𝑇 𝑆$ are invertible, then by associativity $𝑇^{- 1} (𝑇 𝑆) =$ $(𝑇^{- 1} 𝑇) 𝑆 =$ $𝑆$ . The map $𝑇^{- 1}$ is invertible by Proposition 2.5.17(a), so $𝑆$ is a composite of two invertible maps, which is invertible by Proposition 2.5.17(b). Explicitly, $𝑆^{- 1} = {(𝑇 𝑆)}^{- 1} 𝑇$ .

If $𝑆$ and $𝑇 𝑆$ are invertible, then $(𝑇 𝑆) 𝑆^{- 1} =$ $𝑇 (𝑆 𝑆^{- 1}) =$ $𝑇$ is invertible in the same way, with $𝑇^{- 1} = 𝑆 {(𝑇 𝑆)}^{- 1}$ . The matrix case is identical. □

If only $𝑇 𝑆$ is known to be invertible, neither factor need be (Example 2.5.27).

Exercise 2.5.19.routine Let $𝐴 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ and $𝐵 = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ in $ℚ^{2 \times 2}$ .

(a)Check that $𝐴^{- 1} = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix})$ and $𝐵^{- 1} = (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix})$ .
(b)Compute ${(𝐴 𝐵)}^{- 1}$ , $𝐴^{- 1} 𝐵^{- 1}$ , and $𝐵^{- 1} 𝐴^{- 1}$ . Which two agree?

Solution.

Uses. Definition 2.5.12 (Invertible), Proposition 2.5.17 (Algebra of inverses), Definition 2.4.1 (Matrix product)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. Multiply out both products in (a).

Hint 2. By Proposition 2.5.17(b),

{(𝐴 𝐵)}^{- 1} = 𝐵^{- 1} 𝐴^{- 1}

, so it is enough to compute the two products of inverses and identify which one it is.

Sketch. (a) Both products are

𝐼_{2}

in each case. (b)

𝐵^{- 1} 𝐴^{- 1} =

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

{(𝐴 𝐵)}^{- 1}

, with

𝐴 𝐵 = (\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})

, while

𝐴^{- 1} 𝐵^{- 1} = (\begin{matrix} 2 & - 1 \\ - 1 & 1 \end{matrix})

is the inverse of

𝐵 𝐴 = (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})

(a) $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) =$ $(\begin{matrix} 1 & - 1 + 1 \\ 0 & 1 \end{matrix}) =$ $𝐼_{2}$ and $(\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) =$ $(\begin{matrix} 1 & 1 - 1 \\ 0 & 1 \end{matrix}) =$ $𝐼_{2}$ . Similarly $(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) =$ $(\begin{matrix} 1 & 0 \\ 1 - 1 & 1 \end{matrix}) =$ $𝐼_{2}$ , and the product in the other order is $𝐼_{2}$ as well.

(b) We have $𝐴 𝐵 = (\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})$ and

\begin{aligned} 𝐵^{- 1} 𝐴^{- 1} & = (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) \\ = (\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}), \\ 𝐴^{- 1} 𝐵^{- 1} & = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) \\ = (\begin{matrix} 2 & - 1 \\ - 1 & 1 \end{matrix}) . \end{aligned}

By Proposition 2.5.17(b), ${(𝐴 𝐵)}^{- 1} =$ $𝐵^{- 1} 𝐴^{- 1} =$ $(\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix})$ ; as a check, $(\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}) =$ $(\begin{matrix} 2 - 1 & - 2 + 2 \\ 1 - 1 & - 1 + 2 \end{matrix}) =$ $𝐼_{2}$ . The product $𝐴^{- 1} 𝐵^{- 1}$ is different; it is ${(𝐵 𝐴)}^{- 1}$ , the inverse of $𝐵 𝐴 = (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})$ .

Inverses also translate between maps and matrices, in every pair of bases. The proof is a direct application of Technique 2.3.27.

Proposition 2.5.20 (Inverses and matrices). Let

ℬ︀

be a basis of

𝑉

of length

𝑛

and

𝒞︀

a basis of

𝑊

of length

𝑚

, and let

𝑇 \in L (𝑉, 𝑊)

and

𝑆 \in L (𝑊, 𝑉)

(a) $𝑆$ is a left inverse of $𝑇$ if and only if ${[𝑆]}_{ℬ︀ 𝒞︀}$ is a left inverse of ${[𝑇]}_{𝒞︀ ℬ︀}$ , and $𝑆$ is a right inverse of $𝑇$ if and only if ${[𝑆]}_{ℬ︀ 𝒞︀}$ is a right inverse of ${[𝑇]}_{𝒞︀ ℬ︀}$ .
(b) $𝑇$ is left-invertible, right-invertible, or invertible if and only if ${[𝑇]}_{𝒞︀ ℬ︀}$ is. If $𝑇$ is invertible, then
${[𝑇^{- 1}]}_{ℬ︀ 𝒞︀} = {({[𝑇]}_{𝒞︀ ℬ︀})}^{- 1} .$
(c)In particular, $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ is left-invertible, right-invertible, or invertible if and only if $𝑇_{𝐴}$ is, and if $𝐴$ is invertible, then ${(𝑇_{𝐴})}^{- 1} = 𝑇_{𝐴^{- 1}}$ .

Proof.standard

Uses. Theorem 2.4.2 (The matrix of a composite), Example 2.3.6 (The zero map and the identity map), Theorem 2.3.20 (Linear maps correspond to matrices), Proposition 2.3.16 (The matrix of $𝑇_{𝐴}$ ), Proposition 2.3.17 (Every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is some $𝑇_{𝐴}$ ), Definition 2.5.12 (Invertible)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. By Theorem 2.4.2,

{[𝑆]}_{ℬ︀ 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} =

{[𝑆 𝑇]}_{ℬ︀ ℬ︀}

. What is the matrix of

{id}_{𝑉}

in the bases

ℬ︀, ℬ︀

Hint 2. Two maps

𝑉 \to 𝑉

are equal exactly when their matrices in

ℬ︀, ℬ︀

are equal (Theorem 2.3.20(a)).

Hint 3. For the “if” in (b), you are given a matrix. Use Theorem 2.3.20(b) to find a map with that matrix, then apply (a).

Sketch. (a)

{[𝑆]}_{ℬ︀ 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} =

{[𝑆 𝑇]}_{ℬ︀ ℬ︀}

and

𝐼_{𝑛} = {[{id}_{𝑉}]}_{ℬ︀ ℬ︀}

, and maps

𝑉 \to 𝑉

are determined by their matrices, so

𝑆 𝑇 = {id}_{𝑉}

iff the matrix product is

𝐼_{𝑛}

; the right side is the same with

𝑇 𝑆

and

𝐼_{𝑚}

. (b) “Only if” is (a). For “if”, realize a one-sided inverse

𝑋

[𝑇]

[𝑆]

by Theorem 2.3.20(b) and apply (a). The inverse formula is (a) plus uniqueness. (c) is (b) for the standard bases, where

{[𝑇_{𝐴}]}_{ℰ︀ ℰ︀} = 𝐴

(a) By Theorem 2.4.2, ${[𝑆]}_{ℬ︀ 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} =$ ${[𝑆 𝑇]}_{ℬ︀ ℬ︀}$ , and by Example 2.3.6, ${[{id}_{𝑉}]}_{ℬ︀ ℬ︀} = 𝐼_{𝑛}$ . Both $𝑆 𝑇$ and ${id}_{𝑉}$ lie in $L (𝑉)$ , and by Theorem 2.3.20(a) (with the basis $ℬ︀$ on both sides) they are equal if and only if their matrices are. So

\begin{aligned} 𝑆 𝑇 = {id}_{𝑉} & ⟺ {[𝑆 𝑇]}_{ℬ︀ ℬ︀} = {[{id}_{𝑉}]}_{ℬ︀ ℬ︀} \\ ⟺ {[𝑆]}_{ℬ︀ 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} = 𝐼_{𝑛} . \end{aligned}

In the same way, $𝑇 𝑆 = {id}_{𝑊}$ if and only if ${[𝑇]}_{𝒞︀ ℬ︀} {[𝑆]}_{ℬ︀ 𝒞︀} =$ ${[𝑇 𝑆]}_{𝒞︀ 𝒞︀} =$ ${[{id}_{𝑊}]}_{𝒞︀ 𝒞︀} =$ $𝐼_{𝑚}$ .

(b) If $𝑇$ has a left inverse $𝑆$ , then ${[𝑆]}_{ℬ︀ 𝒞︀}$ is a left inverse of ${[𝑇]}_{𝒞︀ ℬ︀}$ by (a). Conversely, suppose $𝑋 \in 𝐹^{𝑛 \times 𝑚}$ is a left inverse of ${[𝑇]}_{𝒞︀ ℬ︀}$ . By Theorem 2.3.20(b) there is $𝑆 \in L (𝑊, 𝑉)$ with ${[𝑆]}_{ℬ︀ 𝒞︀} = 𝑋$ , and by (a) this $𝑆$ is a left inverse of $𝑇$ . The same argument works for right inverses, and invertibility means having both (Definition 2.5.12).

If $𝑇$ is invertible, then by (a) the matrix ${[𝑇^{- 1}]}_{ℬ︀ 𝒞︀}$ is both a left and a right inverse of ${[𝑇]}_{𝒞︀ ℬ︀}$ . By the uniqueness in Definition 2.5.12, it is ${({[𝑇]}_{𝒞︀ ℬ︀})}^{- 1}$ .

(c) Apply (b) to $𝑇_{𝐴} \in L (𝐹^{𝑛}, 𝐹^{𝑚})$ with the standard bases, where ${[𝑇_{𝐴}]}_{ℰ︀ ℰ︀} = 𝐴$ (Proposition 2.3.16). If $𝐴$ is invertible, (b) gives ${[{(𝑇_{𝐴})}^{- 1}]}_{ℰ︀ ℰ︀} =$ $𝐴^{- 1}$ , so ${(𝑇_{𝐴})}^{- 1} = 𝑇_{𝐴^{- 1}}$ by Proposition 2.3.17. □

Example 2.5.21 (Change of coordinates is invertible). Let $ℬ︀$ and $𝒞︀$ be bases of $𝑉$ . The identity map ${id}_{𝑉}$ is its own inverse, so by Proposition 2.5.20(b), applied with the basis $ℬ︀$ on the domain and $𝒞︀$ on the target,

{({[{id}_{𝑉}]}_{𝒞︀ ℬ︀})}^{- 1} = {[{id}_{𝑉}]}_{ℬ︀ 𝒞︀} .

The matrix converting $ℬ︀$ -coordinates into $𝒞︀$ -coordinates is invertible, and its inverse converts back. Example 2.4.5(c) checked one instance by multiplying out. These matrices are the subject of Section 4.5. Once we know that invertible matrices are square (Corollary 3.5.15), this also shows that any two bases of the same space have the same length, which Section 4.1 proves by other means.

Exercise 2.5.22.routine Let

𝑇 \in L ({𝐹 [𝑥]}_{\leq 2})

be the shift

𝑇 (𝑝) = 𝑝 (𝑥 + 1)

, whose matrix in

ℬ︀ = 1, 𝑥, 𝑥^{2}

{[𝑇]}_{ℬ︀ ℬ︀} =

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

(Example 2.4.13). Let

𝑆 \in L ({𝐹 [𝑥]}_{\leq 2})

be the shift

𝑆 (𝑝) = 𝑝 (𝑥 - 1)

. Compute

{[𝑆]}_{ℬ︀ ℬ︀}

, and use Proposition 2.5.20 to show that

𝑇

is invertible with

𝑇^{- 1} = 𝑆

Solution.

Uses. Definition 2.3.1 (Matrix of a linear map), Proposition 2.5.20 (Inverses and matrices), Definition 2.4.1 (Matrix product)

Techniques. Translate between maps and matrices (Technique 2.3.27), Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. The columns of

{[𝑆]}_{ℬ︀ ℬ︀}

are the coordinates of

𝑆 (1) = 1

𝑆 (𝑥) = 𝑥 - 1

, and

𝑆 (𝑥^{2}) =

{(𝑥 - 1)}^{2} =

𝑥^{2} -

2 𝑥 +

1

Hint 2. Multiply

[𝑇] [𝑆]

and

[𝑆] [𝑇]

. If both are

𝐼_{3}

, then by Proposition 2.5.20(a),

𝑆

is a two-sided inverse of

𝑇

Sketch.

{[𝑆]}_{ℬ︀ ℬ︀} =

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

. Both products

[𝑇] [𝑆]

and

[𝑆] [𝑇]

are

𝐼_{3}

, so by Proposition 2.5.20(a),

𝑆

is a left and a right inverse of

𝑇

From $𝑆 (1) = 1$ , $𝑆 (𝑥) = - 1 + 𝑥$ , and $𝑆 (𝑥^{2}) = 1 - 2 𝑥 + 𝑥^{2}$ ,

{[𝑆]}_{ℬ︀ ℬ︀} = (\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}) .

By the entry formula,

\begin{aligned} [𝑇] [𝑆] & = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}) \\ = (\begin{matrix} 1 & - 1 + 1 & 1 - 2 + 1 \\ 0 & 1 & - 2 + 2 \\ 0 & 0 & 1 \end{matrix}) = 𝐼_{3}, \end{aligned}

and

\begin{aligned} [𝑆] [𝑇] & = (\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix}) \\ = (\begin{matrix} 1 & 1 - 1 & 1 - 2 + 1 \\ 0 & 1 & 2 - 2 \\ 0 & 0 & 1 \end{matrix}) = 𝐼_{3}, \end{aligned}

where $[𝑇]$ and $[𝑆]$ abbreviate the matrices in the basis $ℬ︀$ . So ${[𝑆]}_{ℬ︀ ℬ︀}$ is a left and a right inverse of ${[𝑇]}_{ℬ︀ ℬ︀}$ . By Proposition 2.5.20(a), $𝑆$ is a left and a right inverse of $𝑇$ , so $𝑇$ is invertible with $𝑇^{- 1} = 𝑆$ . Shifting by $1$ and then by $- 1$ changes nothing, as expected; the matrices let us confirm it without composing substitutions.

2.5.4One-sided inverses that are not two-sided

Here are the main sources of maps and matrices that have an inverse on one side but not on the other. In each example below there are many inverses on the one side, and by Corollary 2.5.11 having two of them already rules out an inverse on the other side.

Matrices that are not square

Example 2.5.23 (One-sided inverses of a tall matrix). Let $𝑝, 𝑞 \geq 1$ and $𝑋 \in 𝐹^{𝑞 \times 𝑝}$ , and consider the $(𝑝 + 𝑞) \times 𝑝$ matrix

𝐴 = (\begin{matrix} 𝐼_{𝑝} \\ 𝑋 \end{matrix}) .

It sends $𝐱 \in 𝐹^{𝑝}$ to $(𝐱, 𝑋 𝐱)$ , the vector $𝐱$ followed by $𝑞$ more entries. For every $𝑌 \in 𝐹^{𝑝 \times 𝑞}$ , block multiplication (Theorem 2.4.34) gives

\begin{matrix} (\begin{matrix} 𝐼_{𝑝} - 𝑌 𝑋 & 𝑌 \end{matrix}) (\begin{matrix} 𝐼_{𝑝} \\ 𝑋 \end{matrix}) \\ = (𝐼_{𝑝} - 𝑌 𝑋) + 𝑌 𝑋 = 𝐼_{𝑝}, \end{matrix}

so $𝐵_{𝑌} =$ $(\begin{matrix} 𝐼_{𝑝} - 𝑌 𝑋 & 𝑌 \end{matrix})$ is a left inverse of $𝐴$ . These are all of them: if $𝐵 = (\begin{matrix} 𝐵_{1} & 𝐵_{2} \end{matrix})$ with $𝐵_{1} \in 𝐹^{𝑝 \times 𝑝}$ and $𝐵_{2} \in 𝐹^{𝑝 \times 𝑞}$ satisfies $𝐵 𝐴 =$ $𝐵_{1} +$ $𝐵_{2} 𝑋 =$ $𝐼_{𝑝}$ , then $𝐵_{1} = 𝐼_{𝑝} - 𝐵_{2} 𝑋$ , so $𝐵 = 𝐵_{𝐵_{2}}$ . Different $𝑌$ give different $𝐵_{𝑌}$ , since $𝑌$ is the right block of $𝐵_{𝑌}$ , so $𝐴$ has more than one left inverse, and therefore no right inverse (Corollary 2.5.11).

For instance, take $𝑝 = 2$ , $𝑞 = 1$ , and $𝑋 = (\begin{matrix} 2 & - 1 \end{matrix})$ . With $𝑌 = (\begin{matrix} 𝑠 \\ 𝑡 \end{matrix})$ , the left inverses of

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 1 \\ 2 & - 1 \end{matrix})

are the matrices

𝐵_{𝑠, 𝑡} = (\begin{matrix} 1 - 2 𝑠 & 𝑠 & 𝑠 \\ - 2 𝑡 & 1 + 𝑡 & 𝑡 \end{matrix}) (𝑠, 𝑡 \in 𝐹) .

The simplest, $𝐵_{0, 0} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})$ , reads off the first two entries. The others also use the third entry $2 𝑥_{1} - 𝑥_{2}$ , which is redundant information, so it can be mixed in with any weight.

Transposing turns this into right inverses of wide matrices: ${(𝐵 𝐴)}^{𝖳} = 𝐴^{𝖳} 𝐵^{𝖳}$ , so $𝐵$ is a left inverse of $𝐴$ exactly when $𝐵^{𝖳}$ is a right inverse of $𝐴^{𝖳}$ (Exercise 2.5.45). The pattern is general. A left-invertible $𝑚 \times 𝑛$ matrix has $𝑛 \leq 𝑚$ , and a right-invertible one has $𝑛 \geq 𝑚$ (Theorem 3.5.12, Theorem 3.5.13). So a matrix that is not square can have an inverse on at most one side.

Differentiation and integration

Example 2.5.24 (Differentiation has a right inverse but no left inverse). Let $𝐹$ be a field of characteristic $0$ (Definition 1.1.28), such as $ℚ$ , $ℝ$ , or $ℂ$ , so that $𝑘 * 1 = 1 + \dots + 1$ ( $𝑘$ terms) is nonzero for every $𝑘 \geq 1$ ; we write $1 / 𝑘$ for its inverse. Let $𝐷 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be differentiation (Example 2.1.9), and let $𝐽 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be the antiderivative with constant term $0$ ,

𝐽 (\sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑥^{𝑘}) = \sum_{𝑘 = 0}^{𝑛} \frac{𝑎_{𝑘}}{𝑘 + 1} 𝑥^{𝑘 + 1} .

Like $𝐷$ , the map $𝐽$ moves each coefficient to a fixed place and multiplies it by a fixed scalar, so it is linear.

$𝐷 𝐽 = {id}_{𝐹 [𝑥]}$ . Both sides are linear, and $𝐷 (𝐽 (𝑥^{𝑘})) =$ $𝐷 (𝑥^{𝑘 + 1} / (𝑘 + 1)) =$ $𝑥^{𝑘}$ for every $𝑘 \geq 0$ . The monomials span $𝐹 [𝑥]$ , so $𝐷 𝐽 = id$ by Corollary 2.2.6. So $𝐽$ is a right inverse of $𝐷$ , and $𝐷$ is a left inverse of $𝐽$ . This is the fundamental theorem of calculus, in its polynomial form.
$𝐽 𝐷 \neq {id}_{𝐹 [𝑥]}$ , since $𝐽 (𝐷 (1)) = 𝐽 (0) = 0$ . In general $𝐽 (𝐷 (𝑝)) = 𝑝 - 𝑎_{0}$ : differentiating loses the constant term, and $𝐽$ cannot recover it.
$𝐷$ has no left inverse, since $𝐷 (1) = 0 = 𝐷 (0)$ shows that $𝐷$ is not injective (Proposition 2.5.4). $𝐽$ has no right inverse, since every value of $𝐽$ has constant term $0$ , so $1$ is not a value of $𝐽$ (Proposition 2.5.5).
$𝐷$ has many right inverses. If $𝐸 \in L (𝐹 [𝑥])$ takes only constant values, then $𝐷 𝐸 = 0$ and $𝐷 (𝐽 + 𝐸) = id$ . For instance, for $𝑡 \in 𝐹$ the map $𝐽_{𝑡} (𝑝) =$ $𝐽 (𝑝) -$ $(𝐽 (𝑝)) (𝑡)$ , where the scalar $(𝐽 (𝑝)) (𝑡)$ is subtracted as a constant polynomial, is a right inverse of $𝐷$ . It sends $𝑝$ to its antiderivative that vanishes at $𝑡$ , and $𝐽_{0} = 𝐽$ .

The same happens between finite-dimensional spaces of different dimensions. For $𝑛 \geq 1$ , differentiation $𝐷 : {𝐹 [𝑥]}_{\leq 𝑛} \to$ ${𝐹 [𝑥]}_{\leq 𝑛 - 1}$ and the antiderivative $𝐽 : {𝐹 [𝑥]}_{\leq 𝑛 - 1} \to$ ${𝐹 [𝑥]}_{\leq 𝑛}$ still satisfy $𝐷 𝐽 = id$ and $𝐽 𝐷 \neq id$ . For $𝑛 = 2$ , in the bases $1, 𝑥, 𝑥^{2}$ and $1, 𝑥$ ,

\begin{aligned} [𝐷] [𝐽] & = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 / 2 \end{matrix}) = 𝐼_{2}, \\ [𝐽] [𝐷] & = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \neq 𝐼_{3} . \end{aligned}

The finite-dimensional version is an instance of the previous subsection: a $2 \times 3$ matrix with a right inverse. The version on $𝐹 [𝑥]$ is new. There $𝐷$ and $𝐽$ map a space to itself, and still $𝐷 𝐽 = id$ while $𝐽 𝐷 \neq id$ . For a linear map from a finite-dimensional space to itself this cannot happen: a one-sided inverse of a square matrix is automatically two-sided (Theorem 3.5.23), and correspondingly an operator on a finite-dimensional space is injective if and only if it is surjective (Corollary 4.2.18). $𝐹 [𝑥]$ escapes because no finite list spans it, so $𝐷$ can be onto without being one-to-one: it shifts the monomials down by one step, and the bottom one, $1$ , falls off. Over $𝔽_{𝑝}$ even the right inverse is lost (Exercise 2.5.42).

Remark 2.5.25 (One-sided inverses in infinite dimensions).Beyond 21-242 The converses of Remark 2.5.7 hold in every vector space: an injective linear map has a left inverse and a surjective one has a right inverse. Both proofs need a basis of

𝑊

(in the injective case, one that extends a basis of

𝑇 (𝑉)

), and in infinite dimensions bases are known to exist only by Zorn’s lemma (Section 15.1).

Products

If $𝑆$ and $𝑇$ are invertible, so is $𝑇 𝑆$ (Proposition 2.5.17). The converse fails, but a one-sided version survives.

Proposition 2.5.26 (One-sided inverses of composites). Let

𝑆 \in L (𝑈, 𝑉)

and

𝑇 \in L (𝑉, 𝑊)

(a)If $𝑆$ and $𝑇$ are left-invertible, so is $𝑇 𝑆$ . If $𝑆$ and $𝑇$ are right-invertible, so is $𝑇 𝑆$ .
(b)If $𝑇 𝑆$ is left-invertible, then so is $𝑆$ . If $𝑇 𝑆$ is right-invertible, then so is $𝑇$ .
(c)In particular, if $𝑇 𝑆$ is invertible, then ${(𝑇 𝑆)}^{- 1} 𝑇$ is a left inverse of $𝑆$ , and $𝑆 {(𝑇 𝑆)}^{- 1}$ is a right inverse of $𝑇$ .

The same holds for matrices: for instance, if

𝐴 \in 𝐹^{𝑚 \times 𝑛}

and

𝐵 \in 𝐹^{𝑛 \times 𝑚}

and

𝐴 𝐵

is invertible, then

𝐵 {(𝐴 𝐵)}^{- 1}

is a right inverse of

𝐴

and

{(𝐴 𝐵)}^{- 1} 𝐴

is a left inverse of

𝐵

Proof.routine

Uses. Definition 2.5.1 (Left inverse), Definition 2.5.2 (Right inverse), Proposition 2.1.24 (Algebra of composition), Proposition 2.4.14 (Algebra of matrices)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. For (a), the candidate is the composite of the one-sided inverses in reverse order, as in Proposition 2.5.17(b).

Hint 2. For (b), if

𝑅 (𝑇 𝑆) = {id}_{𝑈}

, regroup:

(𝑅 𝑇) 𝑆 = {id}_{𝑈}

Sketch. (a) If

𝑆^{'} 𝑆 = {id}_{𝑈}

and

𝑇^{'} 𝑇 = {id}_{𝑉}

, then

(𝑆^{'} 𝑇^{'}) (𝑇 𝑆) =

𝑆^{'} (𝑇^{'} 𝑇) 𝑆 =

𝑆^{'} 𝑆 =

{id}_{𝑈}

; right inverses are symmetric. (b)

𝑅 (𝑇 𝑆) = {id}_{𝑈}

gives

(𝑅 𝑇) 𝑆 = {id}_{𝑈}

, and

(𝑇 𝑆) 𝑅 = {id}_{𝑊}

gives

𝑇 (𝑆 𝑅) = {id}_{𝑊}

. (c) is (b) with

𝑅 = {(𝑇 𝑆)}^{- 1}

(a) Let $𝑆^{'} 𝑆 = {id}_{𝑈}$ and $𝑇^{'} 𝑇 = {id}_{𝑉}$ . By Proposition 2.1.24,

(𝑆^{'} 𝑇^{'}) (𝑇 𝑆) = 𝑆^{'} (𝑇^{'} 𝑇) 𝑆 = 𝑆^{'} {id}_{𝑉} 𝑆 = 𝑆^{'} 𝑆 = {id}_{𝑈},

so $𝑆^{'} 𝑇^{'}$ is a left inverse of $𝑇 𝑆$ . If $𝑆 𝑆^{″} = {id}_{𝑉}$ and $𝑇 𝑇^{″} = {id}_{𝑊}$ , then $(𝑇 𝑆) (𝑆^{″} 𝑇^{″}) =$ $𝑇 (𝑆 𝑆^{″}) 𝑇^{″} =$ $𝑇 𝑇^{″} =$ ${id}_{𝑊}$ , so $𝑆^{″} 𝑇^{″}$ is a right inverse of $𝑇 𝑆$ .

(b) If $𝑅 (𝑇 𝑆) = {id}_{𝑈}$ , then $(𝑅 𝑇) 𝑆 = {id}_{𝑈}$ , so $𝑅 𝑇$ is a left inverse of $𝑆$ . If $(𝑇 𝑆) 𝑅 = {id}_{𝑊}$ , then $𝑇 (𝑆 𝑅) = {id}_{𝑊}$ , so $𝑆 𝑅$ is a right inverse of $𝑇$ .

(c) Apply (b) with $𝑅 = {(𝑇 𝑆)}^{- 1}$ , which is both a left and a right inverse of $𝑇 𝑆$ . The matrix statements are the same computations with Proposition 2.4.14. □

Example 2.5.27 (An invertible product of non-invertible factors). Let $𝐴 = (\begin{matrix} 1 & 0 \end{matrix}) \in 𝐹^{1 \times 2}$ and $𝐵 = (\begin{matrix} 1 \\ 0 \end{matrix}) \in 𝐹^{2 \times 1}$ . Then $𝐴 𝐵 = (1) = 𝐼_{1}$ is invertible. But $𝐴$ is not: it has the two right inverses $(\begin{matrix} 1 \\ 0 \end{matrix})$ and $(\begin{matrix} 1 \\ 1 \end{matrix})$ , so by Corollary 2.5.11 it has no left inverse. Similarly $𝐵$ has the two left inverses $(\begin{matrix} 1 & 0 \end{matrix})$ and $(\begin{matrix} 1 & 1 \end{matrix})$ , so it has no right inverse. In the other order, $𝐵 𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ has a zero row and is not even right-invertible (Exercise 2.5.9).

For a $2 \times 2$ example of the same kind, take $𝐴 = (\begin{matrix} 𝐼_{2} & 0 \end{matrix}) \in 𝐹^{2 \times 3}$ and $𝐵 = (\begin{matrix} 𝐼_{2} \\ 0 \end{matrix}) \in 𝐹^{3 \times 2}$ , so that $𝐴 𝐵 = 𝐼_{2}$ by block multiplication. For an example with maps from a space to itself, take $𝐷 𝐽 = {id}_{𝐹 [𝑥]}$ from Example 2.5.24, where neither $𝐷$ nor $𝐽$ is invertible.

These examples use matrices that are not square, or a space that is not finite-dimensional, and they have to. For square matrices, $𝐴 𝐵$ invertible implies that $𝐴$ and $𝐵$ are both invertible (Theorem 3.5.23).

2.5.5Inverting $2 \times 2$ matrices

For $2 \times 2$ matrices there is a formula for the inverse, and a single number decides whether it exists. The candidate comes from the following computation. With $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ , swap the diagonal entries and negate the other two:

\begin{aligned} (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix}) \\ = (\begin{matrix} 𝑎 𝑑 - 𝑏 𝑐 & 0 \\ 0 & 𝑎 𝑑 - 𝑏 𝑐 \end{matrix}) \\ = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix}) (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) . \end{aligned}

So the second matrix is an inverse of $𝐴$ up to the scalar $𝑎 𝑑 - 𝑏 𝑐$ .

Proposition 2.5.28 (Inverse of a

2 \times 2

matrix). Let

𝐴 =

(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in

𝐹^{2 \times 2}

, and let

Δ = 𝑎 𝑑 - 𝑏 𝑐

(a)If $Δ \neq 0$ , then $𝐴$ is invertible, and
$𝐴^{- 1} = \frac{1}{Δ} (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix}) .$
(b)If $Δ = 0$ , then $𝐴$ has neither a left inverse nor a right inverse.

Proof.routine

Uses. Definition 2.5.12 (Invertible), Proposition 2.4.14 (Algebra of matrices), Proposition 2.4.8 (Four views of the product)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. Let

𝐴^{'} = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})

. Compute

𝐴 𝐴^{'}

and

𝐴^{'} 𝐴

Hint 2. Both products are

Δ 𝐼_{2}

. For (a), divide by

Δ

Hint 3. For (b), suppose

𝐵 𝐴 = 𝐼_{2}

. Then

𝐴^{'} =

(𝐵 𝐴) 𝐴^{'} =

𝐵 (𝐴 𝐴^{'})

. What is

𝐴 𝐴^{'}

when

Δ = 0

, and what does that say about

𝐴^{'}

, and then about

𝐴

Sketch. With

𝐴^{'} = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})

, the entry formula gives

𝐴 𝐴^{'} = 𝐴^{'} 𝐴 = Δ 𝐼_{2}

. (a)

Δ^{- 1} 𝐴^{'}

is a two-sided inverse. (b) If

Δ = 0

, then

𝐴 𝐴^{'} = 𝐴^{'} 𝐴 = 0

. A left inverse

𝐵

would give

𝐴^{'} = 𝐵 𝐴 𝐴^{'} = 0

, so

𝑎 = 𝑏 = 𝑐 = 𝑑 = 0

and

𝐴 = 0

, which has no left inverse. A right inverse

𝐶

gives

𝐴^{'} = 𝐴^{'} 𝐴 𝐶 = 0

in the same way.

Let $𝐴^{'} = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})$ . By the entry formula,

\begin{aligned} 𝐴 𝐴^{'} & = (\begin{matrix} 𝑎 𝑑 - 𝑏 𝑐 & - 𝑎 𝑏 + 𝑏 𝑎 \\ 𝑐 𝑑 - 𝑑 𝑐 & - 𝑐 𝑏 + 𝑑 𝑎 \end{matrix}) = Δ 𝐼_{2}, \\ 𝐴^{'} 𝐴 & = (\begin{matrix} 𝑑 𝑎 - 𝑏 𝑐 & 𝑑 𝑏 - 𝑏 𝑑 \\ - 𝑐 𝑎 + 𝑎 𝑐 & - 𝑐 𝑏 + 𝑎 𝑑 \end{matrix}) = Δ 𝐼_{2} . \end{aligned}

(a) If $Δ \neq 0$ , then by Proposition 2.4.14(d),

𝐴 (Δ^{- 1} 𝐴^{'}) = Δ^{- 1} (𝐴 𝐴^{'}) = Δ^{- 1} Δ 𝐼_{2} = 𝐼_{2},

and likewise $(Δ^{- 1} 𝐴^{'}) 𝐴 = 𝐼_{2}$ . So $Δ^{- 1} 𝐴^{'}$ is a two-sided inverse of $𝐴$ .

(b) Suppose $Δ = 0$ , so that $𝐴 𝐴^{'} = 𝐴^{'} 𝐴 = 0$ . If $𝐵$ is a left inverse of $𝐴$ , then

𝐴^{'} = 𝐼_{2} 𝐴^{'} = (𝐵 𝐴) 𝐴^{'} = 𝐵 (𝐴 𝐴^{'}) = 𝐵 0 = 0 .

So $𝑎 = 𝑏 = 𝑐 = 𝑑 = 0$ and $𝐴 = 0$ . But the zero matrix has no left inverse, since $𝐵 0 = 0 \neq 𝐼_{2}$ , a contradiction. If $𝐶$ is a right inverse of $𝐴$ , then $𝐴^{'} =$ $𝐴^{'} (𝐴 𝐶) =$ $(𝐴^{'} 𝐴) 𝐶 =$ $0$ , and we reach the same contradiction. □

The number $Δ = 𝑎 𝑑 - 𝑏 𝑐$ is the determinant of $𝐴$ , studied in Chapter 5, and $𝐴^{'}$ is the $2 \times 2$ case of the adjugate matrix, which inverts matrices of every size (Theorem 5.5.9). For a rotation, $Δ = {cos}^{2} 𝜃 + {sin}^{2} 𝜃 = 1$ and the formula returns $[𝑅_{- 𝜃}]$ , as in Example 2.5.14(b).

Geometrically, $| Δ |$ is the area of the parallelogram spanned by the columns of $𝐴$ , which is the image of the unit square under $𝑇_{𝐴}$ (Section 5.1). When $Δ = 0$ the columns are parallel, and $𝑇_{𝐴}$ flattens the plane onto a line through $𝟎$ , or onto ${𝟎}$ . A flattened plane cannot be unflattened: two points that land on the same point cannot both be recovered. Drag the columns below until they line up, and watch the area reach $0$ .

The map of

𝐴 = (\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})

, with

Δ = 1

. Move

𝐴 𝐞_{2}

onto the line through

𝐴 𝐞_{1}

(for example to

(1, 1 / 2)

) and the grid collapses onto that line; the signed area

𝑎 𝑑 - 𝑏 𝑐

is then

0

, and by Proposition 2.5.28 the matrix has no inverse on either side.

Example 2.5.29 (One matrix over several fields). Let $𝐴 = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ , with $Δ = 4 - 6 = - 2$ . Whether $𝐴$ is invertible depends on the field in which we read its entries.

Over $ℚ$ , $Δ = - 2 \neq 0$ and $𝐴^{- 1} =$ $- \frac{1}{2} (\begin{matrix} 4 & - 2 \\ - 3 & 1 \end{matrix}) =$ $(\begin{matrix} - 2 & 1 \\ 3 / 2 & - 1 / 2 \end{matrix})$ .
Over $𝔽_{5}$ , $Δ = - 2 = 3$ and $3^{- 1} = 2$ , so $𝐴^{- 1} =$ $2 (\begin{matrix} 4 & - 2 \\ - 3 & 1 \end{matrix}) =$ $(\begin{matrix} 8 & - 4 \\ - 6 & 2 \end{matrix}) =$ $(\begin{matrix} 3 & 1 \\ 4 & 2 \end{matrix})$ . As a check, $(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) (\begin{matrix} 3 & 1 \\ 4 & 2 \end{matrix}) =$ $(\begin{matrix} 11 & 5 \\ 25 & 11 \end{matrix})$ , which is $𝐼_{2}$ modulo $5$ .
Over $𝔽_{3}$ , $Δ = - 2 = 1$ and $𝐴 = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , so $𝐴^{- 1} =$ $(\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}) =$ $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .
Over $𝔽_{2}$ , $Δ = 0$ , so $𝐴$ is not invertible. Indeed $𝐴 = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix})$ there, with a zero column.

So a matrix with integer entries can be invertible over $ℚ$ and over some $𝔽_{𝑝}$ but not others; here it fails exactly for $𝑝 = 2$ , the only prime dividing $Δ = - 2$ .

Exercises

Exercise 2.5.30.routine Let $𝐚 = (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}) \in ℚ^{3 \times 1}$ .

(a)Find all left inverses of $𝐚$ .
(b)Show that $𝐚$ has no right inverse, in two ways: directly from Definition 2.5.2, and from (a).

Solution.

Uses. Definition 2.5.1 (Left inverse), Definition 2.5.2 (Right inverse), Corollary 2.5.11 (Many one-sided inverses), Proposition 2.4.8 (Four views of the product)

Hint 1. A left inverse is a row

(\begin{matrix} 𝑏_{1} & 𝑏_{2} & 𝑏_{3} \end{matrix})

with

𝑏_{1} + 2 𝑏_{2} - 𝑏_{3} = 1

. Solve for

𝑏_{3}

Hint 2. For a right inverse

(\begin{matrix} 𝑐_{1} & 𝑐_{2} & 𝑐_{3} \end{matrix})

, the

3 \times 3

product

𝐚 (\begin{matrix} 𝑐_{1} & 𝑐_{2} & 𝑐_{3} \end{matrix})

has

(𝑖, 𝑗)

entry

𝑎_{𝑖} 𝑐_{𝑗}

. Compare the entries in positions

(1, 2)

and

(2, 2)

with those of

𝐼_{3}

Sketch. (a) The rows

(\begin{matrix} 𝑠 & 𝑡 & 𝑠 + 2 𝑡 - 1 \end{matrix})

for

𝑠, 𝑡 \in ℚ

. (b)

𝐚 (\begin{matrix} 𝑐_{1} & 𝑐_{2} & 𝑐_{3} \end{matrix}) = 𝐼_{3}

would need

𝑐_{2} = 0

(position

(1, 2)

) and

2 𝑐_{2} = 1

(position

(2, 2)

). Alternatively,

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

and

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

are two different left inverses, so Corollary 2.5.11 applies.

(a) A left inverse is a matrix $𝐵 =$ $(\begin{matrix} 𝑏_{1} & 𝑏_{2} & 𝑏_{3} \end{matrix}) \in$ $ℚ^{1 \times 3}$ with $𝐵 𝐚 = 𝐼_{1}$ , that is, $𝑏_{1} + 2 𝑏_{2} - 𝑏_{3} = 1$ . For any $𝑏_{1} = 𝑠$ and $𝑏_{2} = 𝑡$ this determines $𝑏_{3} = 𝑠 + 2 𝑡 - 1$ . So the left inverses of $𝐚$ are the rows

(\begin{matrix} 𝑠 & 𝑡 & 𝑠 + 2 𝑡 - 1 \end{matrix}) (𝑠, 𝑡 \in ℚ) .

For instance $𝑠 = 1, 𝑡 = 0$ gives $(\begin{matrix} 1 & 0 & 0 \end{matrix})$ , $𝑠 = 𝑡 = 0$ gives $(\begin{matrix} 0 & 0 & - 1 \end{matrix})$ , and $𝑠 = 1 / 6$ , $𝑡 = 1 / 3$ gives $(\begin{matrix} 1 / 6 & 1 / 3 & - 1 / 6 \end{matrix})$ .

(b) A right inverse would be $𝐶 =$ $(\begin{matrix} 𝑐_{1} & 𝑐_{2} & 𝑐_{3} \end{matrix}) \in$ $ℚ^{1 \times 3}$ with $𝐚 𝐶 = 𝐼_{3}$ . The product $𝐚 𝐶$ has $(𝑖, 𝑗)$ entry $𝑎_{𝑖} 𝑐_{𝑗}$ (Proposition 2.4.8(a)). Position $(1, 2)$ gives $1 \cdot 𝑐_{2} = 0$ , and position $(2, 2)$ gives $2 𝑐_{2} = 1$ , which contradict each other. So there is no right inverse.

Alternatively, by (a), $(\begin{matrix} 1 & 0 & 0 \end{matrix})$ and $(\begin{matrix} 0 & 0 & - 1 \end{matrix})$ are two different left inverses of $𝐚$ , and by Corollary 2.5.11 a matrix with two different left inverses has no right inverse.

Exercise 2.5.31.routine

(a)Over $ℂ$ , find the inverse of $(\begin{matrix} 1 & 𝑖 \\ 𝑖 & 2 \end{matrix})$ .
(b)Over $𝔽_{7}$ , find the inverse of $(\begin{matrix} 3 & 5 \\ 2 & 6 \end{matrix})$ , and check your answer.
(c)For which $𝑡 \in ℝ$ is $(\begin{matrix} 𝑡 & 1 \\ 4 & 𝑡 \end{matrix})$ invertible? For which $𝑡 \in 𝔽_{5}$ ?

Solution.

Uses. Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix)

Hint 1. Compute

Δ = 𝑎 𝑑 - 𝑏 𝑐

in each case and apply Proposition 2.5.28.

Hint 2. In (b),

Δ = 18 - 10 = 8

, which is

1

𝔽_{7}

. In (c),

Δ = 𝑡^{2} - 4

Sketch. (a)

Δ = 2 - 𝑖^{2} = 3

, so the inverse is

\frac{1}{3} (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 1 \end{matrix})

. (b)

Δ = 1

, so the inverse is

(\begin{matrix} 6 & - 5 \\ - 2 & 3 \end{matrix}) = (\begin{matrix} 6 & 2 \\ 5 & 3 \end{matrix})

. (c) Over

ℝ

𝑡 \neq \pm 2

. Over

𝔽_{5}

𝑡^{2} \neq 4

, so

𝑡 \in {0, 1, 4}

(a) $Δ = 1 \cdot 2 - 𝑖 \cdot 𝑖 = 2 + 1 = 3$ , so by Proposition 2.5.28

{(\begin{matrix} 1 & 𝑖 \\ 𝑖 & 2 \end{matrix})}^{- 1} = \frac{1}{3} (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 1 \end{matrix}) .

Check: $(\begin{matrix} 1 & 𝑖 \\ 𝑖 & 2 \end{matrix}) (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 1 \end{matrix}) =$ $(\begin{matrix} 2 - 𝑖^{2} & - 𝑖 + 𝑖 \\ 2 𝑖 - 2 𝑖 & - 𝑖^{2} + 2 \end{matrix}) =$ $(\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix})$ .

(b) $Δ = 3 \cdot 6 - 5 \cdot 2 = 8 = 1$ in $𝔽_{7}$ , so the inverse is $(\begin{matrix} 6 & - 5 \\ - 2 & 3 \end{matrix}) = (\begin{matrix} 6 & 2 \\ 5 & 3 \end{matrix})$ . Check:

\begin{aligned} (\begin{matrix} 3 & 5 \\ 2 & 6 \end{matrix}) (\begin{matrix} 6 & 2 \\ 5 & 3 \end{matrix}) & = (\begin{matrix} 18 + 25 & 6 + 15 \\ 12 + 30 & 4 + 18 \end{matrix}) \\ = (\begin{matrix} 43 & 21 \\ 42 & 22 \end{matrix}), \end{aligned}

which is $𝐼_{2}$ modulo $7$ since $43 = 42 + 1$ and $22 = 21 + 1$ .

(c) Here $Δ = 𝑡^{2} - 4$ . By Proposition 2.5.28 the matrix is invertible exactly when $𝑡^{2} \neq 4$ . Over $ℝ$ this means $𝑡 \neq \pm 2$ , and then the inverse is ${(𝑡^{2} - 4)}^{- 1} (\begin{matrix} 𝑡 & - 1 \\ - 4 & 𝑡 \end{matrix})$ . In $𝔽_{5}$ the squares of $0, 1, 2, 3, 4$ are $0, 1, 4, 4, 1$ , so $𝑡^{2} = 4$ exactly for $𝑡 \in {2, 3}$ , and the matrix is invertible for $𝑡 \in {0, 1, 4}$ .

Exercise 2.5.32.routine Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ with $𝑛 \geq 1$ .

(a)Show that if $𝐴$ is invertible and $𝐴 𝐵 = 0$ for some $𝐵 \in 𝐹^{𝑛 \times 𝑝}$ , then $𝐵 = 0$ .
(b)Show that if $𝐴^{𝑘} = 0$ for some $𝑘 \geq 1$ , then $𝐴$ is not invertible.
(c)Show that if $𝐴$ is invertible and $𝐴^{2} = 𝐴$ , then $𝐴 = 𝐼_{𝑛}$ .
(d)Show that if $𝐴^{𝑘} = 𝐼_{𝑛}$ for some $𝑘 \geq 1$ , then $𝐴$ is invertible with $𝐴^{- 1} = 𝐴^{𝑘 - 1}$ .

Solution.

Uses. Corollary 2.5.6 (One-sided cancellation), Proposition 2.5.17 (Algebra of inverses), Exercise 2.5.9, Notation 2.4.16 (Powers of a matrix)

Hint 1. (a) and (c) are cancellations: write the hypothesis as

𝐴 𝑋 = 𝐴 𝑌

Hint 2. (b) Powers of an invertible matrix are invertible. Is the zero matrix invertible?

Hint 3. (d) Compute

𝐴 𝐴^{𝑘 - 1}

and

𝐴^{𝑘 - 1} 𝐴

Sketch. (a)

𝐴 𝐵 = 𝐴 0

, and

𝐴

can be cancelled. (b) If

𝐴

were invertible, so would be

𝐴^{𝑘} = 0

, but

0

has a zero column. (c)

𝐴 𝐴 = 𝐴 𝐼_{𝑛}

; cancel. (d)

𝐴 𝐴^{𝑘 - 1} =

𝐴^{𝑘 - 1} 𝐴 =

𝐴^{𝑘} =

𝐼_{𝑛}

(a) $𝐴 𝐵 = 0 = 𝐴 0$ , and $𝐴$ is left-invertible, so $𝐵 = 0$ by Corollary 2.5.6.

(b) If $𝐴$ were invertible, then $𝐴^{𝑘}$ would be invertible by Proposition 2.5.17(b) and induction. But $𝐴^{𝑘} = 0$ , whose columns are all zero, and a matrix with a zero column has no left inverse (Exercise 2.5.9). So $𝐴$ is not invertible.

(c) $𝐴 𝐴 = 𝐴 = 𝐴 𝐼_{𝑛}$ , so $𝐴 = 𝐼_{𝑛}$ by Corollary 2.5.6.

(d) By Notation 2.4.16, $𝐴 𝐴^{𝑘 - 1} =$ $𝐴^{𝑘} =$ $𝐼_{𝑛}$ and $𝐴^{𝑘 - 1} 𝐴 =$ $𝐴^{𝑘} =$ $𝐼_{𝑛}$ (for $𝑘 = 1$ , $𝐴^{0} = 𝐼_{𝑛}$ and $𝐴 = 𝐼_{𝑛}$ ). So $𝐴^{𝑘 - 1}$ is a two-sided inverse of $𝐴$ .

Exercise 2.5.33.routine Let $𝑇 \in L (𝑉, 𝑊)$ and $𝑆 \in L (𝑊, 𝑉)$ .

(a)Suppose $𝑆 𝑇 = {id}_{𝑉}$ and $𝑇$ is surjective. Show that $𝑇 𝑆 = {id}_{𝑊}$ .
(b)Suppose $𝑇 𝑆 = {id}_{𝑊}$ and $𝑇$ is injective. Show that $𝑆 𝑇 = {id}_{𝑉}$ .
(c)Deduce that a left-invertible linear map is invertible if and only if it is surjective, and a right-invertible one is invertible if and only if it is injective.

Solution.

Uses. Definition 2.5.12 (Invertible), Proposition 2.5.4 (Left inverses give uniqueness), Proposition 2.5.5 (Right inverses give existence)

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

Hint 1. (a) Fix

𝐰 \in 𝑊

and write it as

𝑇 (𝐯)

. Compute

𝑇 (𝑆 (𝐰))

Hint 2. (b) Fix

𝐯

. Show that

𝑇 (𝑆 (𝑇 (𝐯))) =

𝑇 (𝐯)

, and use injectivity.

Sketch. (a) For

𝐰 = 𝑇 (𝐯)

𝑇 𝑆 (𝐰) =

𝑇 (𝑆 𝑇 (𝐯)) =

𝑇 (𝐯) =

𝐰

. (b)

𝑇 (𝑆 𝑇 (𝐯)) =

(𝑇 𝑆) (𝑇 (𝐯)) =

𝑇 (𝐯)

, so

𝑆 𝑇 (𝐯) = 𝐯

by injectivity. (c) One direction is (a) or (b); the other is Proposition 2.5.5 or Proposition 2.5.4.

(a) Let $𝐰 \in 𝑊$ . Since $𝑇$ is surjective, $𝐰 = 𝑇 (𝐯)$ for some $𝐯 \in 𝑉$ . Then

\begin{aligned} 𝑇 (𝑆 (𝐰)) & = 𝑇 (𝑆 (𝑇 (𝐯))) \\ = 𝑇 ((𝑆 𝑇) (𝐯)) = 𝑇 (𝐯) = 𝐰 . \end{aligned}

As $𝐰$ was arbitrary, $𝑇 𝑆 = {id}_{𝑊}$ .

(b) Let $𝐯 \in 𝑉$ . Then $𝑇 (𝑆 (𝑇 (𝐯))) =$ $(𝑇 𝑆) (𝑇 (𝐯)) =$ $𝑇 (𝐯)$ . Since $𝑇$ is injective, $𝑆 (𝑇 (𝐯)) = 𝐯$ . As $𝐯$ was arbitrary, $𝑆 𝑇 = {id}_{𝑉}$ .

(c) Let $𝑆$ be a left inverse of $𝑇$ . If $𝑇$ is surjective, then by (a) $𝑆$ is also a right inverse, so $𝑇$ is invertible. Conversely, an invertible map has a right inverse and is therefore surjective (Proposition 2.5.5). The second statement follows in the same way from (b) and Proposition 2.5.4.

Exercise 2.5.34.standard Find the error in the following argument, and show that its conclusion is false. Then add a hypothesis on $𝐴$ under which the conclusion holds.

Claim. Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ . If $𝐴 𝐵 = 𝐼_{𝑚}$ , then $𝐵 𝐴 = 𝐼_{𝑛}$ .

Proof. Since $𝐴 𝐵 = 𝐼_{𝑚}$ , the matrix $𝐵$ is the inverse of $𝐴$ , that is, $𝐵 = 𝐴^{- 1}$ . Hence $𝐵 𝐴 = 𝐴^{- 1} 𝐴 = 𝐼_{𝑛}$ .

Solution.

Uses. Definition 2.5.12 (Invertible), Definition 2.5.2 (Right inverse), Theorem 2.5.10 (Left and right inverses agree), Example 2.5.27 (An invertible product of non-invertible factors)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. What does

𝐴 𝐵 = 𝐼_{𝑚}

say about

𝐵

, according to the definitions? What does “

𝐵 = 𝐴^{- 1}

” require?

Hint 2. Try

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

and

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

Hint 3. If

𝐴

also has some left inverse, what does Theorem 2.5.10 say?

Sketch.

𝐴 𝐵 = 𝐼_{𝑚}

only says that

𝐵

is a right inverse of

𝐴

; the step “

𝐵 = 𝐴^{- 1}

” assumes that

𝐴

is invertible, which is what the claim was supposed to deliver. With

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

and

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

𝐴 𝐵 = 𝐼_{1}

but

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

. If

𝐴

also has a left inverse

𝐶

, then

𝐶 = 𝐵

by Theorem 2.5.10, so

𝐵 𝐴 = 𝐶 𝐴 = 𝐼_{𝑛}

The error is the step “ $𝐵 = 𝐴^{- 1}$ ”. By Definition 2.5.2, $𝐴 𝐵 = 𝐼_{𝑚}$ says only that $𝐵$ is a right inverse of $𝐴$ . The inverse $𝐴^{- 1}$ is a matrix that is a left and a right inverse at the same time (Definition 2.5.12), and nothing in the hypothesis says that $𝐴$ has an inverse on the left. The argument assumes, in the notation $𝐴^{- 1}$ , the very fact $𝐵 𝐴 = 𝐼_{𝑛}$ that it set out to prove.

The conclusion is false: as in Example 2.5.27, $𝐴 = (\begin{matrix} 1 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 1 \\ 0 \end{matrix})$ satisfy $𝐴 𝐵 = (1) = 𝐼_{1}$ , but $𝐵 𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) \neq 𝐼_{2}$ .

It becomes true if $𝐴$ is also assumed to be left-invertible. If $𝐶 𝐴 = 𝐼_{𝑛}$ , then $𝐶 = 𝐵$ by Theorem 2.5.10, so $𝐵 𝐴 = 𝐶 𝐴 = 𝐼_{𝑛}$ . For square matrices no extra hypothesis is needed (Theorem 3.5.23), but that is a deeper fact, and the argument above does not prove it. For maps from a space to itself it fails without finite dimension: $𝐷 𝐽 = id$ but $𝐽 𝐷 \neq id$ on $ℚ [𝑥]$ (Example 2.5.24).

Exercise 2.5.35.standard Prove or disprove each statement, where $𝑚, 𝑛 \geq 1$ .

(a)If $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ are invertible, then $𝐴 + 𝐵$ is invertible.
(b)If $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ and $𝐴^{2}$ is invertible, then $𝐴$ is invertible.
(c)If $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ is invertible and symmetric, then $𝐴^{- 1}$ is symmetric.
(d)If $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐴^{𝖳} 𝐴$ is invertible, then $𝐴$ is left-invertible.
(e)If $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ is left-invertible, then $𝐴^{𝖳} 𝐴$ is invertible.

Solution.

Uses. Definition 2.5.12 (Invertible), Theorem 2.5.10 (Left and right inverses agree), Proposition 2.5.17 (Algebra of inverses), Exercise 2.5.9

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. (a) Try

𝐵 = - 𝐴

. (b) From an inverse

𝑋

𝐴^{2}

, build a left and a right inverse of

𝐴

separately.

Hint 2. (c) Use Proposition 2.5.17(d). (d) Write down a candidate left inverse that involves

{(𝐴^{𝖳} 𝐴)}^{- 1}

Hint 3. (e) The field matters. Try a column

𝐴 \in 𝔽_{2}^{2 \times 1}

𝐴 \in ℂ^{2 \times 1}

with

𝐴^{𝖳} 𝐴 = 0

Sketch. (a) False:

𝐼_{𝑛} + (- 𝐼_{𝑛}) = 0

. (b) True: if

𝑋 𝐴^{2} = 𝐴^{2} 𝑋 = 𝐼_{𝑛}

, then

𝑋 𝐴

is a left inverse and

𝐴 𝑋

a right inverse of

𝐴

. (c) True:

{(𝐴^{- 1})}^{𝖳} =

{(𝐴^{𝖳})}^{- 1} =

𝐴^{- 1}

. (d) True:

{(𝐴^{𝖳} 𝐴)}^{- 1} 𝐴^{𝖳}

is a left inverse. (e) False:

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

over

𝔽_{2}

, or

(\begin{matrix} 1 \\ 𝑖 \end{matrix})

over

ℂ

, is left-invertible with

𝐴^{𝖳} 𝐴 = 0

(a) False. $𝐼_{𝑛}$ and $- 𝐼_{𝑛}$ are invertible (each is its own inverse), but $𝐼_{𝑛} + (- 𝐼_{𝑛}) = 0$ has a zero column and is not invertible (Exercise 2.5.9). This works over every field, including $𝔽_{2}$ , where $- 𝐼_{𝑛} = 𝐼_{𝑛}$ .

(b) True. Let $𝑋 = {(𝐴^{2})}^{- 1}$ , so $𝑋 𝐴^{2} = 𝐴^{2} 𝑋 = 𝐼_{𝑛}$ . Then $(𝑋 𝐴) 𝐴 =$ $𝑋 𝐴^{2} =$ $𝐼_{𝑛}$ , so $𝑋 𝐴$ is a left inverse of $𝐴$ , and $𝐴 (𝐴 𝑋) =$ $𝐴^{2} 𝑋 =$ $𝐼_{𝑛}$ , so $𝐴 𝑋$ is a right inverse of $𝐴$ . By Theorem 2.5.10, $𝐴$ is invertible (and $𝑋 𝐴 = 𝐴 𝑋 = 𝐴^{- 1}$ ).

(c) True. By Proposition 2.5.17(d) and $𝐴^{𝖳} = 𝐴$ , ${(𝐴^{- 1})}^{𝖳} =$ ${(𝐴^{𝖳})}^{- 1} =$ $𝐴^{- 1}$ .

(d) True. Let $𝑋 =$ ${(𝐴^{𝖳} 𝐴)}^{- 1} \in$ $𝐹^{𝑛 \times 𝑛}$ . Then $(𝑋 𝐴^{𝖳}) 𝐴 =$ $𝑋 (𝐴^{𝖳} 𝐴) =$ $𝐼_{𝑛}$ , so $𝑋 𝐴^{𝖳} \in 𝐹^{𝑛 \times 𝑚}$ is a left inverse of $𝐴$ .

(e) False. Over $𝔽_{2}$ , let $𝐴 = (\begin{matrix} 1 \\ 1 \end{matrix})$ . It is left-invertible, with left inverse $(\begin{matrix} 1 & 0 \end{matrix})$ , but $𝐴^{𝖳} 𝐴 = 1 + 1 = 0$ is not invertible. Over $ℂ$ , $𝐴 = (\begin{matrix} 1 \\ 𝑖 \end{matrix})$ has the left inverse $(\begin{matrix} 1 & 0 \end{matrix})$ and $𝐴^{𝖳} 𝐴 = 1 + 𝑖^{2} = 0$ . Over $ℝ$ the statement is true. By positivity of sums of squares, $𝐴^{𝖳} 𝐴 𝐱 = 𝟎$ forces $𝐴 𝐱 = 𝟎$ (Section 6.5), hence $𝐱 = 𝟎$ when $𝐴$ is left-invertible. So the square matrix $𝐴^{𝖳} 𝐴$ defines an injective map, and that makes it invertible (Section 3.5).

Exercise 2.5.36.standard Let $𝑝$ be a prime and $𝐴 =$ $(\begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}) \in$ $𝔽_{𝑝}^{3 \times 2}$ .

(a)How many left inverses does $𝐴$ have?
(b)Does $𝐴$ have a right inverse?
(c)How many right inverses does $𝐴^{𝖳}$ have?

Solution.

Uses. Definition 2.5.1 (Left inverse), Proposition 2.4.8 (Four views of the product), Corollary 2.5.11 (Many one-sided inverses), Proposition 2.4.23 (Transpose of a product)

Hint 1. By the row view,

𝐵 𝐴 = 𝐼_{2}

means: row

1

𝐵

times

𝐴

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

and row

2

𝐵

times

𝐴

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

. Treat the two rows separately.

Hint 2. A row

(\begin{matrix} 𝑟_{1} & 𝑟_{2} & 𝑟_{3} \end{matrix})

times

𝐴

(\begin{matrix} 𝑟_{1} + 𝑟_{3} & 𝑟_{1} + 𝑟_{2} \end{matrix})

Hint 3. For (c), transpose

𝐵 𝐴 = 𝐼_{2}

Sketch.

(\begin{matrix} 𝑟_{1} & 𝑟_{2} & 𝑟_{3} \end{matrix}) 𝐴 =

(\begin{matrix} 𝑟_{1} + 𝑟_{3} & 𝑟_{1} + 𝑟_{2} \end{matrix})

. Row

1

𝐵

must be

(\begin{matrix} 𝑡 & - 𝑡 & 1 - 𝑡 \end{matrix})

and row

2

must be

(\begin{matrix} 𝑠 & 1 - 𝑠 & - 𝑠 \end{matrix})

, with

𝑠, 𝑡 \in 𝔽_{𝑝}

free:

𝑝^{2}

left inverses. Since

𝑝^{2} \geq 4 > 1

, there is no right inverse.

𝐶

is a right inverse of

𝐴^{𝖳}

iff

𝐶^{𝖳}

is a left inverse of

𝐴

, so

𝐴^{𝖳}

has

𝑝^{2}

right inverses.

(a) By the row view (Proposition 2.4.8(c)), $𝐵 𝐴 = 𝐼_{2}$ means that row $1$ of $𝐵$ times $𝐴$ is $(\begin{matrix} 1 & 0 \end{matrix})$ and row $2$ of $𝐵$ times $𝐴$ is $(\begin{matrix} 0 & 1 \end{matrix})$ , and the two rows can be chosen independently. For a row $𝐫 = (\begin{matrix} 𝑟_{1} & 𝑟_{2} & 𝑟_{3} \end{matrix})$ , the row view gives $𝐫 𝐴 =$ $𝑟_{1} (\begin{matrix} 1 & 1 \end{matrix}) +$ $𝑟_{2} (\begin{matrix} 0 & 1 \end{matrix}) +$ $𝑟_{3} (\begin{matrix} 1 & 0 \end{matrix}) =$ $(\begin{matrix} 𝑟_{1} + 𝑟_{3} & 𝑟_{1} + 𝑟_{2} \end{matrix})$ . So $𝐫 𝐴 = (\begin{matrix} 1 & 0 \end{matrix})$ means $𝑟_{3} = 1 - 𝑟_{1}$ and $𝑟_{2} = - 𝑟_{1}$ , giving the $𝑝$ rows $(\begin{matrix} 𝑡 & - 𝑡 & 1 - 𝑡 \end{matrix})$ with $𝑡 \in 𝔽_{𝑝}$ . And $𝐫 𝐴 = (\begin{matrix} 0 & 1 \end{matrix})$ means $𝑟_{3} = - 𝑟_{1}$ and $𝑟_{2} = 1 - 𝑟_{1}$ , giving the $𝑝$ rows $(\begin{matrix} 𝑠 & 1 - 𝑠 & - 𝑠 \end{matrix})$ . Hence the left inverses are

𝐵_{𝑡, 𝑠} = (\begin{matrix} 𝑡 & - 𝑡 & 1 - 𝑡 \\ 𝑠 & 1 - 𝑠 & - 𝑠 \end{matrix}) (𝑠, 𝑡 \in 𝔽_{𝑝}),

and there are $𝑝^{2}$ of them.

(b) No. Since $𝑝^{2} \geq 4$ , $𝐴$ has two different left inverses, so it has no right inverse by Corollary 2.5.11.

(c) By Proposition 2.4.23, ${(𝐵 𝐴)}^{𝖳} = 𝐴^{𝖳} 𝐵^{𝖳}$ , and $𝐼_{2}^{𝖳} = 𝐼_{2}$ . Since transposing is a bijection (Proposition 2.4.22), $𝐵 𝐴 = 𝐼_{2}$ if and only if $𝐴^{𝖳} 𝐵^{𝖳} = 𝐼_{2}$ . So $𝐶$ is a right inverse of $𝐴^{𝖳}$ exactly when $𝐶^{𝖳}$ is a left inverse of $𝐴$ , and $𝐴^{𝖳}$ has $𝑝^{2}$ right inverses.

Exercise 2.5.37.standard Let

𝑇 \in L (𝑉, 𝑊)

, and let

𝑆

be a left inverse of

𝑇

. Show that

𝑇

is invertible if and only if

𝑆

is left-invertible, and that in this case

𝑇^{- 1} = 𝑆

Solution.

Uses. Theorem 2.5.10 (Left and right inverses agree), Definition 2.5.12 (Invertible)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. The equation

𝑆 𝑇 = {id}_{𝑉}

also says something about

𝑆

: on which side is

𝑇

an inverse of

𝑆

Hint 2. If

𝑅 𝑆 = {id}_{𝑊}

, then

𝑆

has a left inverse

𝑅

and a right inverse

𝑇

. Apply Theorem 2.5.10 to

𝑆

Sketch.

𝑆 𝑇 = {id}_{𝑉}

says that

𝑇

is a right inverse of

𝑆

. If

𝑆

also has a left inverse

𝑅

, then

𝑅 = 𝑇

by Theorem 2.5.10, so

𝑇 𝑆 = 𝑅 𝑆 = {id}_{𝑊}

and

𝑆

is a two-sided inverse of

𝑇

. Conversely, if

𝑇

is invertible, then

𝑆 = 𝑇^{- 1}

and

𝑇

is a left inverse of

𝑆

Note first that $𝑆 𝑇 = {id}_{𝑉}$ says that $𝑇$ is a right inverse of $𝑆$ .

Suppose $𝑆$ has a left inverse $𝑅 \in L (𝑉, 𝑊)$ , so $𝑅 𝑆 = {id}_{𝑊}$ . Then $𝑆$ has the left inverse $𝑅$ and the right inverse $𝑇$ , so $𝑅 = 𝑇$ by Theorem 2.5.10, applied to $𝑆$ . Hence $𝑇 𝑆 = 𝑅 𝑆 = {id}_{𝑊}$ . Together with $𝑆 𝑇 = {id}_{𝑉}$ , this shows that $𝑆$ is a two-sided inverse of $𝑇$ , so $𝑇$ is invertible and $𝑇^{- 1} = 𝑆$ .

Conversely, suppose $𝑇$ is invertible. By Theorem 2.5.10, its only left inverse is $𝑇^{- 1}$ , so $𝑆 = 𝑇^{- 1}$ . Then $𝑇 𝑆 = 𝑇 𝑇^{- 1} = {id}_{𝑊}$ , so $𝑇$ is a left inverse of $𝑆$ .

Exercise 2.5.38.standard Let $𝑇 \in L (𝑉)$ , and suppose that $𝑎_{𝑘} 𝑇^{𝑘} +$ $\dots +$ $𝑎_{1} 𝑇 +$ $𝑎_{0} {id}_{𝑉} =$ $0$ for some $𝑎_{0}, \dots, 𝑎_{𝑘} \in 𝐹$ with $𝑎_{0} \neq 0$ .

(a)Show that $𝑇$ is invertible, and that $𝑇^{- 1} =$ $- 𝑎_{0}^{- 1} (𝑎_{𝑘} 𝑇^{𝑘 - 1} + \dots + 𝑎_{2} 𝑇 + 𝑎_{1} {id}_{𝑉})$ .
(b)Let $𝐴 = (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix}) \in ℝ^{2 \times 2}$ . Verify that $𝐴^{2} - 𝐴 + 𝐼_{2} = 0$ , find $𝐴^{- 1}$ from (a), and show that $𝐴^{3} = - 𝐼_{2}$ .

Solution.

Uses. Proposition 2.1.24 (Algebra of composition), Notation 2.1.18, Definition 2.5.12 (Invertible), Proposition 2.4.14 (Algebra of matrices)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. Move

𝑎_{0} {id}_{𝑉}

to the other side and factor

𝑇

out of what remains, on the left and on the right.

Hint 2. With

𝑄 =

𝑎_{𝑘} 𝑇^{𝑘 - 1} +

\dots +

𝑎_{1} {id}_{𝑉}

, both

𝑇 𝑄

and

𝑄 𝑇

equal

𝑎_{𝑘} 𝑇^{𝑘} +

\dots +

𝑎_{1} 𝑇 =

- 𝑎_{0} {id}_{𝑉}

Hint 3. For

𝐴^{3}

, multiply

𝐴^{2} - 𝐴 + 𝐼_{2} = 0

𝐴 + 𝐼_{2}

Sketch. (a) With

𝑄 =

𝑎_{𝑘} 𝑇^{𝑘 - 1} +

\dots +

𝑎_{1} {id}_{𝑉}

, distributivity and

𝑇 𝑇^{𝑗} =

𝑇^{𝑗} 𝑇 =

𝑇^{𝑗 + 1}

give

𝑇 𝑄 =

𝑄 𝑇 =

- 𝑎_{0} {id}_{𝑉}

, so

- 𝑎_{0}^{- 1} 𝑄

is a two-sided inverse. (b)

𝐴^{2} = (\begin{matrix} 0 & 1 \\ - 1 & - 1 \end{matrix})

, so

𝐴^{2} - 𝐴 + 𝐼_{2} = 0

, and (a) gives

𝐴^{- 1} =

𝐼_{2} -

𝐴 =

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

. Then

0 =

(𝐴 + 𝐼_{2}) (𝐴^{2} - 𝐴 + 𝐼_{2}) =

𝐴^{3} +

𝐼_{2}

(a) Let $𝑄 =$ $𝑎_{𝑘} 𝑇^{𝑘 - 1} +$ $\dots +$ $𝑎_{2} 𝑇 +$ $𝑎_{1} {id}_{𝑉}$ . By the distributive laws in Proposition 2.1.24 and $𝑇 𝑇^{𝑗} =$ $𝑇^{𝑗} 𝑇 =$ $𝑇^{𝑗 + 1}$ (Notation 2.1.18),

\begin{aligned} 𝑇 𝑄 = 𝑄 𝑇 & = 𝑎_{𝑘} 𝑇^{𝑘} + \dots + 𝑎_{2} 𝑇^{2} + 𝑎_{1} 𝑇 \\ = - 𝑎_{0} {id}_{𝑉}, \end{aligned}

where the last step is the hypothesis. Multiplying by the scalar $- 𝑎_{0}^{- 1}$ (Proposition 2.1.24(d)) gives

𝑇 (- 𝑎_{0}^{- 1} 𝑄) = (- 𝑎_{0}^{- 1} 𝑄) 𝑇 = {id}_{𝑉} .

So $- 𝑎_{0}^{- 1} 𝑄$ is a two-sided inverse of $𝑇$ , and $𝑇$ is invertible with $𝑇^{- 1} = - 𝑎_{0}^{- 1} 𝑄$ .

(b) By the entry formula, $𝐴^{2} =$ $(\begin{matrix} 1 - 1 & 1 + 0 \\ - 1 + 0 & - 1 + 0 \end{matrix}) =$ $(\begin{matrix} 0 & 1 \\ - 1 & - 1 \end{matrix})$ , so

\begin{aligned} 𝐴^{2} - 𝐴 + 𝐼_{2} & = (\begin{matrix} 0 & 1 \\ - 1 & - 1 \end{matrix}) - (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix}) \\ + (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = 0 . \end{aligned}

The relation has $𝑎_{2} = 1$ , $𝑎_{1} = - 1$ , $𝑎_{0} = 1$ , so by (a) (in its matrix form, with the same proof), $𝐴^{- 1} =$ $- (𝐴 - 𝐼_{2}) =$ $𝐼_{2} -$ $𝐴 =$ $(\begin{matrix} 0 & - 1 \\ 1 & 1 \end{matrix})$ . Check: $𝐴 (𝐼_{2} - 𝐴) =$ $𝐴 -$ $𝐴^{2} =$ $(\begin{matrix} 1 - 0 & 1 - 1 \\ - 1 + 1 & 0 + 1 \end{matrix}) =$ $𝐼_{2}$ . Finally,

\begin{aligned} 0 & = (𝐴 + 𝐼_{2}) (𝐴^{2} - 𝐴 + 𝐼_{2}) \\ = 𝐴^{3} - 𝐴^{2} + 𝐴 + 𝐴^{2} - 𝐴 + 𝐼_{2} \\ = 𝐴^{3} + 𝐼_{2}, \end{aligned}

so $𝐴^{3} = - 𝐼_{2}$ . In particular $𝐴^{6} = 𝐼_{2}$ .

Exercise 2.5.39.standard

(a)Let $𝑁 \in L (𝑉)$ with $𝑁^{𝑘} = 0$ for some $𝑘 \geq 1$ . Show that ${id}_{𝑉} - 𝑁$ and ${id}_{𝑉} + 𝑁$ are invertible, with
${({id}_{𝑉} - 𝑁)}^{- 1} = {id}_{𝑉} + 𝑁 + 𝑁^{2} + \dots + 𝑁^{𝑘 - 1} .$
(b)Use (a) to invert $(\begin{matrix} 1 & 𝑎 & 𝑏 \\ 0 & 1 & 𝑐 \\ 0 & 0 & 1 \end{matrix})$ for $𝑎, 𝑏, 𝑐 \in 𝐹$ .
(c)Show that $id + 𝐷$ is invertible on ${𝐹 [𝑥]}_{\leq 3}$ , where $𝐷$ is differentiation, and compute ${(id + 𝐷)}^{- 1} (𝑥^{3})$ .

Solution.

Uses. Proposition 2.1.24 (Algebra of composition), Notation 2.1.18, Definition 2.5.12 (Invertible), Example 2.1.19 (Composites)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. Multiply

(id - 𝑁) (id + 𝑁 + \dots + 𝑁^{𝑘 - 1})

out. Most terms cancel in pairs.

Hint 2. For

id + 𝑁

, apply the first part to

- 𝑁

. For (b), write the matrix as

𝐼_{3} + 𝑁

with

𝑁

strictly upper triangular, and compute

𝑁^{2}

and

𝑁^{3}

Hint 3. For (c),

𝐷^{4} = 0

{𝐹 [𝑥]}_{\leq 3}

Sketch. (a) The product telescopes to

id - 𝑁^{𝑘} = id

, in either order.

id + 𝑁 = id - (- 𝑁)

and

{(- 𝑁)}^{𝑘} =

{(- 1)}^{𝑘} 𝑁^{𝑘} =

0

. (b) With

𝑁 = (\begin{matrix} 0 & 𝑎 & 𝑏 \\ 0 & 0 & 𝑐 \\ 0 & 0 & 0 \end{matrix})

𝑁^{2} =

(\begin{matrix} 0 & 0 & 𝑎 𝑐 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

and

𝑁^{3} = 0

, so the inverse is

𝐼_{3} -

𝑁 +

𝑁^{2} =

(\begin{matrix} 1 & - 𝑎 & 𝑎 𝑐 - 𝑏 \\ 0 & 1 & - 𝑐 \\ 0 & 0 & 1 \end{matrix})

. (c)

{(id + 𝐷)}^{- 1} =

id -

𝐷 +

𝐷^{2} -

𝐷^{3}

, which sends

𝑥^{3}

𝑥^{3} - 3 𝑥^{2} + 6 𝑥 - 6

(a) Let $𝐺 = id + 𝑁 + \dots + 𝑁^{𝑘 - 1}$ . By the distributive laws in Proposition 2.1.24,

\begin{aligned} (id - 𝑁) 𝐺 & = (id + 𝑁 + \dots + 𝑁^{𝑘 - 1}) \\ - (𝑁 + 𝑁^{2} + \dots + 𝑁^{𝑘}) \\ = id - 𝑁^{𝑘} = id, \end{aligned}

and in the same way $𝐺 (id - 𝑁) =$ $id -$ $𝑁^{𝑘} =$ $id$ . So $id - 𝑁$ is invertible with inverse $𝐺$ . Since ${(- 𝑁)}^{𝑘} =$ ${(- 1)}^{𝑘} 𝑁^{𝑘} =$ $0$ , the same applies to $- 𝑁$ , so $id + 𝑁 = id - (- 𝑁)$ is invertible, with inverse $id -$ $𝑁 +$ $𝑁^{2} -$ $\dots +$ ${(- 1)}^{𝑘 - 1} 𝑁^{𝑘 - 1}$ .

(b) The matrix is $𝐼_{3} + 𝑁$ with $𝑁 = (\begin{matrix} 0 & 𝑎 & 𝑏 \\ 0 & 0 & 𝑐 \\ 0 & 0 & 0 \end{matrix})$ . By the entry formula, $𝑁^{2} =$ $(\begin{matrix} 0 & 0 & 𝑎 𝑐 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$ and $𝑁^{3} = 𝑁^{2} 𝑁 = 0$ . By (a) applied to $- 𝑁$ ,

\begin{aligned} {(𝐼_{3} + 𝑁)}^{- 1} & = 𝐼_{3} - 𝑁 + 𝑁^{2} \\ = (\begin{matrix} 1 & - 𝑎 & 𝑎 𝑐 - 𝑏 \\ 0 & 1 & - 𝑐 \\ 0 & 0 & 1 \end{matrix}) . \end{aligned}

Check, first row of the product of the matrix with this inverse: $(1, - 𝑎 + 𝑎, (𝑎 𝑐 - 𝑏) - 𝑎 𝑐 + 𝑏) =$ $(1, 0, 0)$ ; the other rows are similar.

(c) Each application of $𝐷$ lowers the degree of a nonconstant polynomial and sends constants to $0$ (Example 2.1.19), so $𝐷^{4} = 0$ on ${𝐹 [𝑥]}_{\leq 3}$ . By (a), $id + 𝐷$ is invertible with inverse $id - 𝐷 + 𝐷^{2} - 𝐷^{3}$ . Applied to $𝑥^{3}$ :

{(id + 𝐷)}^{- 1} (𝑥^{3}) = 𝑥^{3} - 3 𝑥^{2} + 6 𝑥 - 6 .

Check: $(id + 𝐷) (𝑥^{3} - 3 𝑥^{2} + 6 𝑥 - 6) =$ $(𝑥^{3} - 3 𝑥^{2} + 6 𝑥 - 6) +$ $(3 𝑥^{2} - 6 𝑥 + 6) =$ $𝑥^{3}$ .

Exercise 2.5.40.standard Let $𝐴 \in 𝐹^{𝑝 \times 𝑝}$ and $𝐷 \in 𝐹^{𝑞 \times 𝑞}$ be invertible, and let $𝑋 \in 𝐹^{𝑝 \times 𝑞}$ .

(a)Show that $𝑀 = (\begin{matrix} 𝐴 & 𝑋 \\ 0 & 𝐷 \end{matrix})$ is invertible, with
$𝑀^{- 1} = (\begin{matrix} 𝐴^{- 1} & - 𝐴^{- 1} 𝑋 𝐷^{- 1} \\ 0 & 𝐷^{- 1} \end{matrix}) .$
(b)Use (a) to invert $(\begin{matrix} 1 & 2 & 1 \\ 1 & 3 & 0 \\ 0 & 0 & 2 \end{matrix}) \in ℚ^{3 \times 3}$ .

Solution.

Uses. Theorem 2.4.34 (Block multiplication), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix), Definition 2.5.12 (Invertible), Proposition 2.4.14 (Algebra of matrices)

Techniques. Guess the inverse, then check both products (Technique 2.5.16), Block decomposition (Technique 2.4.35)

Hint 1. Multiply

𝑀

by the candidate on both sides with Theorem 2.4.34, partitioning with sizes

𝑝, 𝑞

Hint 2. To find the candidate yourself, solve

(\begin{matrix} 𝐴 & 𝑋 \\ 0 & 𝐷 \end{matrix}) (\begin{matrix} 𝑃 & 𝑄 \\ 0 & 𝑆 \end{matrix}) =

𝐼

block by block.

Hint 3. For (b), take

𝑝 = 2

𝑞 = 1

𝐴 = (\begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix})

𝑋 = (\begin{matrix} 1 \\ 0 \end{matrix})

𝐷 = (2)

Sketch. By blocks,

𝑀 𝑁

has diagonal blocks

𝐴 𝐴^{- 1} = 𝐼_{𝑝}

and

𝐷 𝐷^{- 1} = 𝐼_{𝑞}

and top right block

- 𝐴 𝐴^{- 1} 𝑋 𝐷^{- 1} +

𝑋 𝐷^{- 1} =

0

;

𝑁 𝑀

has top right block

𝐴^{- 1} 𝑋 -

𝐴^{- 1} 𝑋 𝐷^{- 1} 𝐷 =

0

. (b)

𝐴^{- 1} = (\begin{matrix} 3 & - 2 \\ - 1 & 1 \end{matrix})

𝐷^{- 1} = (1 / 2)

, and

- 𝐴^{- 1} 𝑋 𝐷^{- 1} =

(\begin{matrix} - 3 / 2 \\ 1 / 2 \end{matrix})

(a) Let $𝑁$ be the proposed inverse, partitioned with sizes $𝑝, 𝑞$ like $𝑀$ . By Theorem 2.4.34, the four blocks of $𝑀 𝑁$ are

\begin{aligned} 𝐴 𝐴^{- 1} + 𝑋 \cdot 0 & = 𝐼_{𝑝}, \\ 𝐴 (- 𝐴^{- 1} 𝑋 𝐷^{- 1}) + 𝑋 𝐷^{- 1} & = 0, \\ 0 \cdot 𝐴^{- 1} + 𝐷 \cdot 0 & = 0, \\ 0 \cdot (- 𝐴^{- 1} 𝑋 𝐷^{- 1}) + 𝐷 𝐷^{- 1} & = 𝐼_{𝑞}, \end{aligned}

so $𝑀 𝑁 = 𝐼_{𝑝 + 𝑞}$ . In the same way, the blocks of $𝑁 𝑀$ are $𝐴^{- 1} 𝐴 = 𝐼_{𝑝}$ , then $𝐴^{- 1} 𝑋 +$ $(- 𝐴^{- 1} 𝑋 𝐷^{- 1}) 𝐷 =$ $𝐴^{- 1} 𝑋 -$ $𝐴^{- 1} 𝑋 =$ $0$ , then $0$ , and $𝐷^{- 1} 𝐷 = 𝐼_{𝑞}$ , so $𝑁 𝑀 = 𝐼_{𝑝 + 𝑞}$ . Hence $𝑁$ is a two-sided inverse of $𝑀$ .

(b) Take $𝐴 = (\begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix})$ , $𝑋 = (\begin{matrix} 1 \\ 0 \end{matrix})$ , and $𝐷 = (2)$ . By Proposition 2.5.28, $Δ = 3 - 2 = 1$ and $𝐴^{- 1} = (\begin{matrix} 3 & - 2 \\ - 1 & 1 \end{matrix})$ , and $𝐷^{- 1} = (1 / 2)$ . Then

\begin{aligned} - 𝐴^{- 1} 𝑋 𝐷^{- 1} & = - \frac{1}{2} (\begin{matrix} 3 & - 2 \\ - 1 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) \\ = (\begin{matrix} - 3 / 2 \\ 1 / 2 \end{matrix}), \end{aligned}

{(\begin{matrix} 1 & 2 & 1 \\ 1 & 3 & 0 \\ 0 & 0 & 2 \end{matrix})}^{- 1} = (\begin{matrix} 3 & - 2 & - 3 / 2 \\ - 1 & 1 & 1 / 2 \\ 0 & 0 & 1 / 2 \end{matrix}) .

Check, row by row against the original: $(3 - 2, - 2 + 2, - 3 / 2 + 1 + 1 / 2) =$ $(1, 0, 0)$ for the first row of the product, and similarly for the others.

Exercise 2.5.41.standard Let $𝑀 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be multiplication by $𝑥$ , $𝑀 (𝑝) = 𝑥 𝑝$ , and let $𝐿 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be

\begin{matrix} 𝐿 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛}) \\ = 𝑎_{1} + 𝑎_{2} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛 - 1} . \end{matrix}

(a)Show that $𝐿 𝑀 = {id}_{𝐹 [𝑥]}$ , compute $𝑀 𝐿$ , and show that $𝑀$ has no right inverse and $𝐿$ no left inverse.
(b)Find all left inverses of $𝑀$ .

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Proposition 2.2.20 (Linear maps out of $𝐹 [𝑥]$ ), Proposition 2.5.4 (Left inverses give uniqueness), Proposition 2.5.5 (Right inverses give existence), Corollary 2.2.6 (Agreeing on a spanning list)

Hint 1.

𝑀

shifts coefficients up by one place, and

𝐿

shifts them down, discarding

𝑎_{0}

Hint 2. A left inverse

𝑆

must satisfy

𝑆 (𝑥^{𝑘 + 1}) =

𝑆 (𝑀 (𝑥^{𝑘})) =

𝑥^{𝑘}

for all

𝑘 \geq 0

. What about

𝑆 (1)

Hint 3. A linear map out of

𝐹 [𝑥]

can take arbitrary values on the monomials (Proposition 2.2.20).

Sketch. (a)

𝐿 (𝑀 (𝑝)) = 𝑝

and

𝑀 (𝐿 (𝑝)) = 𝑝 - 𝑎_{0}

𝑀

has no right inverse since

1

is not a value of

𝑀

, and

𝐿

has no left inverse since

𝐿 (1) = 0 = 𝐿 (0)

. (b) A left inverse is determined by

𝑆 (𝑥^{𝑘 + 1}) = 𝑥^{𝑘}

and the free choice

𝑆 (1) = 𝑞 \in 𝐹 [𝑥]

: the left inverses are

𝑆_{𝑞} (𝑝) =

𝐿 (𝑝) +

𝑎_{0} 𝑞

, one for each

𝑞 \in 𝐹 [𝑥]

(a) For $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛}$ , $𝑀 (𝑝) =$ $𝑎_{0} 𝑥 +$ $𝑎_{1} 𝑥^{2} +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛 + 1}$ , whose coefficients, shifted down by one place, are $𝑎_{0}, \dots, 𝑎_{𝑛}$ . So $𝐿 (𝑀 (𝑝)) = 𝑝$ , that is, $𝐿 𝑀 = id$ . In the other order, $𝑀 (𝐿 (𝑝)) =$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛} =$ $𝑝 -$ $𝑎_{0}$ , so $𝑀 𝐿 \neq id$ : it kills the constant term.

Every value of $𝑀$ has constant term $0$ , so $1$ is not a value of $𝑀$ , and $𝑀$ is not surjective. By Proposition 2.5.5, $𝑀$ has no right inverse. Since $𝐿 (1) = 0 = 𝐿 (0)$ , $𝐿$ is not injective, so it has no left inverse by Proposition 2.5.4.

(b) Let $𝑆 \in L (𝐹 [𝑥])$ with $𝑆 𝑀 = id$ . Then $𝑆 (𝑥^{𝑘 + 1}) =$ $𝑆 (𝑀 (𝑥^{𝑘})) =$ $𝑥^{𝑘}$ for every $𝑘 \geq 0$ , so $𝑆$ agrees with $𝐿$ on $𝑥, 𝑥^{2}, 𝑥^{3}, \dots$ , while $𝑞 = 𝑆 (1)$ is unconstrained. Conversely, for any $𝑞 \in 𝐹 [𝑥]$ , Proposition 2.2.20 gives a linear map $𝑆_{𝑞}$ with $𝑆_{𝑞} (1) = 𝑞$ and $𝑆_{𝑞} (𝑥^{𝑘 + 1}) = 𝑥^{𝑘}$ , namely

𝑆_{𝑞} (𝑝) = 𝑎_{0} 𝑞 + 𝐿 (𝑝),

where $𝑎_{0}$ is the constant term of $𝑝$ . Then $𝑆_{𝑞} 𝑀$ and $id$ agree on every monomial, since $𝑆_{𝑞} (𝑀 (𝑥^{𝑘})) =$ $𝑆_{𝑞} (𝑥^{𝑘 + 1}) =$ $𝑥^{𝑘}$ , so $𝑆_{𝑞} 𝑀 = id$ by Corollary 2.2.6. Conversely, a left inverse $𝑆$ agrees with $𝑆_{𝑞}$ for $𝑞 = 𝑆 (1)$ on every monomial, so $𝑆 = 𝑆_{𝑞}$ , again by Corollary 2.2.6. So the left inverses of $𝑀$ are exactly the maps $𝑆_{𝑞}$ , one for each $𝑞 \in 𝐹 [𝑥]$ , and $𝑆_{0} = 𝐿$ . (Since there are many, Corollary 2.5.11 gives another proof that $𝑀$ has no right inverse.)

Exercise 2.5.42.standard

(a)On $ℚ [𝑥]$ , construct a right inverse $𝑅$ of differentiation $𝐷$ such that $(𝑅 (𝑝)) (1) = 0$ for every $𝑝 \in ℚ [𝑥]$ .
(b)Let $𝑝$ be a prime. Show that $𝑥^{𝑝 - 1}$ is not a value of $𝐷 : 𝔽_{𝑝} [𝑥] \to$ $𝔽_{𝑝} [𝑥]$ , and conclude that over $𝔽_{𝑝}$ , $𝐷$ has no right inverse.

Solution.

Uses. Example 2.5.24 (Differentiation has a right inverse but no left inverse), Example 2.1.9 (Maps of polynomials), Proposition 2.5.5 (Right inverses give existence), Definition 1.2.17 (Evaluation), Proposition 2.1.17 (Compositions of linear maps are linear)

Hint 1. Start from

𝐽

of Example 2.5.24. Adding a constant to

𝐽 (𝑝)

does not change its derivative. Which constant makes the value at

1

vanish?

Hint 2. For (b), the coefficient of

𝑥^{𝑝 - 1}

𝐷 (\sum 𝑎_{𝑘} 𝑥^{𝑘})

𝑝 𝑎_{𝑝}

, computed in

𝔽_{𝑝}

Sketch. (a)

𝑅 (𝑝) =

𝐽 (𝑝) -

(𝐽 (𝑝)) (1)

, the antiderivative vanishing at

1

; it is linear because evaluation at

1

is linear, and

𝐷 𝑅 = 𝐷 𝐽 = id

. (b) The coefficient of

𝑥^{𝑝 - 1}

𝐷 (\sum 𝑎_{𝑘} 𝑥^{𝑘}) =

\sum 𝑘 𝑎_{𝑘} 𝑥^{𝑘 - 1}

𝑝 * 𝑎_{𝑝} = 0

. So

𝐷

is not surjective, and Proposition 2.5.5 rules out a right inverse.

(a) Let $𝐽$ be the antiderivative of Example 2.5.24 and define

𝑅 (𝑝) = 𝐽 (𝑝) - (𝐽 (𝑝)) (1),

where the scalar $(𝐽 (𝑝)) (1)$ is subtracted as a constant polynomial. So $𝑅 (𝑝)$ is the antiderivative of $𝑝$ that vanishes at $1$ . The map $𝑝 \mapsto (𝐽 (𝑝)) (1)$ is the composite of $𝐽$ with evaluation at $1$ , both linear (Example 2.1.9, Proposition 2.1.17), so $𝑅$ is linear. The subtracted term is a constant polynomial, so its derivative is $0$ , and $𝐷 (𝑅 (𝑝)) =$ $𝐷 (𝐽 (𝑝)) =$ $𝑝$ . So $𝑅$ is a right inverse of $𝐷$ , and $(𝑅 (𝑝)) (1) =$ $(𝐽 (𝑝)) (1) -$ $(𝐽 (𝑝)) (1) =$ $0$ . For example, $𝑅 (𝑥) = 𝑥^{2} / 2 - 1 / 2$ .

(b) By Example 2.1.9, $𝐷 (\sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑥^{𝑘}) =$ $\sum_{𝑘 = 1}^{𝑛} (𝑘 * 𝑎_{𝑘}) 𝑥^{𝑘 - 1}$ , so the coefficient of $𝑥^{𝑝 - 1}$ in $𝐷 (𝑓)$ comes only from $𝑘 = 𝑝$ and equals $𝑝 * 𝑎_{𝑝} =$ $(𝑝 * 1) 𝑎_{𝑝} =$ $0$ in $𝔽_{𝑝}$ (it is $0$ outright if $𝑛 < 𝑝$ ). So no polynomial has derivative $𝑥^{𝑝 - 1}$ : $𝐷$ is not surjective, and by Proposition 2.5.5 it has no right inverse. Over $ℚ$ , $𝑥^{𝑝 - 1} =$ $𝐷 (𝑥^{𝑝} / 𝑝)$ , and the division by $𝑝$ is exactly what $𝔽_{𝑝}$ cannot do.

Exercise 2.5.43.standard Let

𝑝

be a prime. Use Proposition 2.5.28 to show that exactly

(𝑝^{2} - 1) (𝑝^{2} - 𝑝)

matrices in

𝔽_{𝑝}^{2 \times 2}

are invertible. List them for

𝑝 = 2

Solution.

Uses. Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix), Proposition 1.1.17 (No zero divisors), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime)

Hint 1. Count the complement: quadruples

(𝑎, 𝑏, 𝑐, 𝑑)

with

𝑎 𝑑 = 𝑏 𝑐

. Group them by the common value

𝑡 = 𝑎 𝑑 = 𝑏 𝑐

Hint 2. The equation

𝑎 𝑑 = 0

has

2 𝑝 - 1

solutions

(𝑎, 𝑑)

, and for

𝑡 \neq 0

the equation

𝑎 𝑑 = 𝑡

has

𝑝 - 1

solutions.

Sketch. By Proposition 2.5.28,

𝐴

is invertible iff

𝑎 𝑑 \neq 𝑏 𝑐

. The number of

(𝑎, 𝑏, 𝑐, 𝑑)

with

𝑎 𝑑 = 𝑏 𝑐 = 𝑡

{(2 𝑝 - 1)}^{2}

for

𝑡 = 0

and

{(𝑝 - 1)}^{2}

for each of the

𝑝 - 1

nonzero

𝑡

. So

{(2 𝑝 - 1)}^{2} + {(𝑝 - 1)}^{3}

matrices are not invertible, and

𝑝^{4}

minus this is

𝑝^{4} -

𝑝^{3} -

𝑝^{2} +

𝑝 =

(𝑝^{2} - 1) (𝑝^{2} - 𝑝)

By Proposition 2.5.28, $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ is invertible if and only if $𝑎 𝑑 \neq 𝑏 𝑐$ . We count the matrices with $𝑎 𝑑 = 𝑏 𝑐$ , grouping them by the common value $𝑡$ .

For $𝑡 = 0$ : by Proposition 1.1.17, $𝑎 𝑑 = 0$ means $𝑎 = 0$ or $𝑑 = 0$ , which holds for $𝑝 + 𝑝 - 1 = 2 𝑝 - 1$ pairs $(𝑎, 𝑑)$ . Likewise there are $2 𝑝 - 1$ pairs $(𝑏, 𝑐)$ with $𝑏 𝑐 = 0$ , so ${(2 𝑝 - 1)}^{2}$ quadruples.

For $𝑡 \neq 0$ : $𝑎 𝑑 = 𝑡$ forces $𝑎 \neq 0$ , and then $𝑑 = 𝑡 𝑎^{- 1}$ is determined, so there are $𝑝 - 1$ pairs $(𝑎, 𝑑)$ , and likewise $𝑝 - 1$ pairs $(𝑏, 𝑐)$ . That gives ${(𝑝 - 1)}^{2}$ quadruples for each of the $𝑝 - 1$ nonzero values of $𝑡$ .

So ${(2 𝑝 - 1)}^{2} + {(𝑝 - 1)}^{3}$ matrices are not invertible. Expanding ${(2 𝑝 - 1)}^{2} =$ $4 𝑝^{2} -$ $4 𝑝 +$ $1$ and ${(𝑝 - 1)}^{3} =$ $𝑝^{3} -$ $3 𝑝^{2} +$ $3 𝑝 -$ $1$ , the number of invertible ones is

\begin{aligned} 𝑝^{4} - {(2 𝑝 - 1)}^{2} - {(𝑝 - 1)}^{3} \\ = 𝑝^{4} - 𝑝^{3} - 𝑝^{2} + 𝑝 \\ = (𝑝^{2} - 1) (𝑝^{2} - 𝑝) . \end{aligned}

For $𝑝 = 2$ this is $3 \cdot 2 = 6$ :

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

The factored form $(𝑝^{2} - 1) (𝑝^{2} - 𝑝)$ suggests another count: $𝑝^{2} - 1$ choices of a nonzero first column, then $𝑝^{2} - 𝑝$ choices of a second column that is not a multiple of the first. That count is justified by the Fundamental Theorems (Theorem 2.7.8), which say that $𝐴$ is invertible exactly when its columns form a basis.

Exercise 2.5.44.standard

(a)Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ be invertible. Show that $𝑚 * 1 = 𝑛 * 1$ in $𝐹$ . Deduce that over a field of characteristic $0$ , such as $ℚ$ , $ℝ$ , or $ℂ$ , every invertible matrix is square.
(b)More generally, let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ with $𝐴 𝐵 = 𝐼_{𝑚}$ , and suppose $𝑚 * 1 \neq 𝑛 * 1$ in $𝐹$ . Show that $𝐵 𝐴 \neq 𝐼_{𝑛}$ .
(c)Explain why this argument does not show that invertible matrices over $𝔽_{2}$ are square.

Solution.

Uses. Proposition 2.4.28 (Cyclicity of the trace), Definition 2.4.26 (Trace), Definition 1.1.28 (Characteristic), Definition 2.5.12 (Invertible)

Techniques. Cyclicity of trace (Technique 2.4.29)

Hint 1. Take the trace of

𝐴 𝐴^{- 1} = 𝐼_{𝑚}

and of

𝐴^{- 1} 𝐴 = 𝐼_{𝑛}

Hint 2. Proposition 2.4.28 applies to an

𝑚 \times 𝑛

matrix times an

𝑛 \times 𝑚

matrix. What is

tr 𝐼_{𝑘}

Sketch. (a)

𝑚 * 1 =

tr 𝐼_{𝑚} =

tr (𝐴 𝐴^{- 1}) =

tr (𝐴^{- 1} 𝐴) =

tr 𝐼_{𝑛} =

𝑛 * 1

. In characteristic

0

𝑚 * 1 = 𝑛 * 1

forces

𝑚 = 𝑛

. (b) If

𝐵 𝐴 = 𝐼_{𝑛}

, then

𝑛 * 1 =

tr (𝐵 𝐴) =

tr (𝐴 𝐵) =

𝑚 * 1

. (c) In

𝔽_{2}

𝑚 * 1 = 𝑛 * 1

whenever

𝑚

and

𝑛

have the same parity, so the trace cannot tell

𝑚 = 1

from

𝑛 = 3

(a) The trace of $𝐼_{𝑘}$ is $1 + \dots + 1 = 𝑘 * 1$ . Since $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐴^{- 1} \in 𝐹^{𝑛 \times 𝑚}$ , Proposition 2.4.28 gives

\begin{aligned} 𝑚 * 1 & = tr 𝐼_{𝑚} = tr (𝐴 𝐴^{- 1}) \\ = tr (𝐴^{- 1} 𝐴) = tr 𝐼_{𝑛} = 𝑛 * 1 . \end{aligned}

Suppose $𝐹$ has characteristic $0$ and, say, $𝑚 > 𝑛$ . Then $(𝑚 - 𝑛) * 1 =$ $𝑚 * 1 -$ $𝑛 * 1 =$ $0$ with $𝑚 - 𝑛 \geq 1$ , contradicting Definition 1.1.28. So $𝑚 = 𝑛$ .

(b) If $𝐵 𝐴 = 𝐼_{𝑛}$ , then $𝑛 * 1 =$ $tr (𝐵 𝐴) =$ $tr (𝐴 𝐵) =$ $tr 𝐼_{𝑚} =$ $𝑚 * 1$ by Proposition 2.4.28, contradicting the hypothesis. So $𝐵 𝐴 \neq 𝐼_{𝑛}$ . For instance, over $ℚ$ , whenever $𝐴 𝐵 = 𝐼_{𝑚}$ with $𝐴$ not square, $𝐵 𝐴$ is not the identity.

(c) In $𝔽_{2}$ , $𝑘 * 1$ is $0$ for even $𝑘$ and $1$ for odd $𝑘$ , so the trace only sees the parity of the size. For $𝑚 = 1$ and $𝑛 = 3$ the argument gives $1 = 1$ , no contradiction. Invertible matrices over $𝔽_{2}$ , and over every field, are nevertheless square (Corollary 3.5.15), but the proof needs rank.

Exercise 2.5.45.routine Let

𝐴 \in 𝐹^{𝑚 \times 𝑛}

and

𝐵 \in 𝐹^{𝑛 \times 𝑚}

. Show that

𝐵

is a left inverse of

𝐴

if and only if

𝐵^{𝖳}

is a right inverse of

𝐴^{𝖳}

. Deduce that

𝐴

is left-invertible if and only if

𝐴^{𝖳}

is right-invertible, and that

𝐴

is invertible if and only if

𝐴^{𝖳}

is.

Solution.

Uses. Proposition 2.4.23 (Transpose of a product), Proposition 2.4.22 (Transpose is a linear involution), Proposition 2.5.17 (Algebra of inverses)

Techniques. Transpose to trade rows for columns (Technique 3.2.27)

Hint 1. Transpose the equation

𝐵 𝐴 = 𝐼_{𝑛}

Hint 2. Transposing is a bijection, so

𝑋 = 𝑌

if and only if

𝑋^{𝖳} = 𝑌^{𝖳}

Sketch.

{(𝐵 𝐴)}^{𝖳} = 𝐴^{𝖳} 𝐵^{𝖳}

and

𝐼_{𝑛}^{𝖳} = 𝐼_{𝑛}

, and transposing is injective, so

𝐵 𝐴 = 𝐼_{𝑛}

iff

𝐴^{𝖳} 𝐵^{𝖳} = 𝐼_{𝑛}

. The second claim follows since

𝐵 \mapsto 𝐵^{𝖳}

is a bijection

𝐹^{𝑛 \times 𝑚} \to 𝐹^{𝑚 \times 𝑛}

. Two-sided: apply both one-sided statements, and use

{(𝐴^{𝖳})}^{𝖳} = 𝐴

By Proposition 2.4.23 and Proposition 2.4.22, ${(𝐵 𝐴)}^{𝖳} = 𝐴^{𝖳} 𝐵^{𝖳}$ and $𝐼_{𝑛}^{𝖳} = 𝐼_{𝑛}$ . Transposing is a bijection (Proposition 2.4.22), so two matrices are equal if and only if their transposes are. Hence

\begin{aligned} 𝐵 𝐴 = 𝐼_{𝑛} & ⟺ {(𝐵 𝐴)}^{𝖳} = 𝐼_{𝑛}^{𝖳} \\ ⟺ 𝐴^{𝖳} 𝐵^{𝖳} = 𝐼_{𝑛}, \end{aligned}

that is, $𝐵$ is a left inverse of $𝐴$ if and only if $𝐵^{𝖳}$ is a right inverse of $𝐴^{𝖳}$ .

If $𝐴$ is left-invertible, with left inverse $𝐵$ , then $𝐵^{𝖳}$ is a right inverse of $𝐴^{𝖳}$ . Conversely, if $𝐶$ is a right inverse of $𝐴^{𝖳}$ , then $𝐵 = 𝐶^{𝖳}$ satisfies $𝐵^{𝖳} = 𝐶$ , so $𝐵$ is a left inverse of $𝐴$ . The same argument with the roles of the two sides exchanged shows that $𝐴$ is right-invertible if and only if $𝐴^{𝖳}$ is left-invertible. Combining the two, $𝐴$ is invertible if and only if $𝐴^{𝖳}$ is; this is Proposition 2.5.17(d) together with its converse, which follows by applying it to $𝐴^{𝖳}$ , since ${(𝐴^{𝖳})}^{𝖳} = 𝐴$ .

Exercise 2.5.46.hard

(a)Show that every $𝐴 =$ $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in$ $𝐹^{2 \times 2}$ satisfies
$𝐴^{2} - (𝑎 + 𝑑) 𝐴 + (𝑎 𝑑 - 𝑏 𝑐) 𝐼_{2} = 0 .$
(b)Use (a) and Exercise 2.5.38 to give a second proof of Proposition 2.5.28(a).
(c)Let $𝑝$ be an odd prime. How many $𝐴 \in 𝔽_{𝑝}^{2 \times 2}$ satisfy $𝐴^{- 1} = 𝐴$ ?

Solution.

Uses. Proposition 2.4.8 (Four views of the product), Exercise 2.5.38, Proposition 1.1.17 (No zero divisors)

Hint 1. (a) is a direct computation. (b) The relation has constant term

𝑎 𝑑 - 𝑏 𝑐

Hint 2. (c)

𝐴^{- 1} = 𝐴

means

𝐴^{2} = 𝐼_{2}

. Combine with (a):

(1 + Δ) 𝐼_{2} =

(𝑎 + 𝑑) 𝐴

, where

Δ = 𝑎 𝑑 - 𝑏 𝑐

. What if

𝑎 + 𝑑 \neq 0

Hint 3. If

𝑎 + 𝑑 \neq 0

, then

𝐴

is a scalar matrix. Otherwise

𝑑 = - 𝑎

and

Δ = - 1

; count the triples

(𝑎, 𝑏, 𝑐)

with

𝑎^{2} + 𝑏 𝑐 = 1

, splitting according to whether

𝑎^{2} = 1

Sketch. (a) Compute

𝐴^{2}

and subtract. (b) With

Δ \neq 0

, Exercise 2.5.38 gives

𝐴^{- 1} =

- Δ^{- 1} (𝐴 - (𝑎 + 𝑑) 𝐼_{2}) =

Δ^{- 1} (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})

. (c) Besides

\pm 𝐼_{2}

, the solutions are exactly the

𝐴

with

𝑑 = - 𝑎

and

𝑎^{2} + 𝑏 𝑐 = 1

. For

𝑎 = \pm 1

𝑏 𝑐 = 0

has

2 𝑝 - 1

solutions; for the other

𝑝 - 2

values,

𝑏 𝑐 = 1 - 𝑎^{2} \neq 0

has

𝑝 - 1

. Total

2 +

2 (2 𝑝 - 1) +

(𝑝 - 2) (𝑝 - 1) =

𝑝^{2} +

𝑝 +

2

(a) By the entry formula,

\begin{aligned} 𝐴^{2} & = (\begin{matrix} 𝑎^{2} + 𝑏 𝑐 & 𝑎 𝑏 + 𝑏 𝑑 \\ 𝑐 𝑎 + 𝑑 𝑐 & 𝑐 𝑏 + 𝑑^{2} \end{matrix}), \\ (𝑎 + 𝑑) 𝐴 & = (\begin{matrix} 𝑎^{2} + 𝑎 𝑑 & 𝑎 𝑏 + 𝑏 𝑑 \\ 𝑎 𝑐 + 𝑐 𝑑 & 𝑎 𝑑 + 𝑑^{2} \end{matrix}), \end{aligned}

so $𝐴^{2} -$ $(𝑎 + 𝑑) 𝐴 =$ $(\begin{matrix} 𝑏 𝑐 - 𝑎 𝑑 & 0 \\ 0 & 𝑏 𝑐 - 𝑎 𝑑 \end{matrix}) =$ $- (𝑎 𝑑 - 𝑏 𝑐) 𝐼_{2}$ , which is the claim.

(b) Let $Δ = 𝑎 𝑑 - 𝑏 𝑐 \neq 0$ and $𝑡 = 𝑎 + 𝑑$ . The relation of (a) reads $𝐴^{2} - 𝑡 𝐴 + Δ 𝐼_{2} = 0$ , with $𝑎_{2} = 1$ , $𝑎_{1} = - 𝑡$ , and $𝑎_{0} = Δ \neq 0$ . By Exercise 2.5.38 (in matrix form),

\begin{aligned} 𝐴^{- 1} & = - Δ^{- 1} (𝐴 - 𝑡 𝐼_{2}) = Δ^{- 1} (𝑡 𝐼_{2} - 𝐴) \\ = Δ^{- 1} (\begin{matrix} 𝑎 + 𝑑 - 𝑎 & - 𝑏 \\ - 𝑐 & 𝑎 + 𝑑 - 𝑑 \end{matrix}) \\ = \frac{1}{Δ} (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix}) . \end{aligned}

(c) $𝐴^{- 1} = 𝐴$ means $𝐴^{2} = 𝐼_{2}$ . Let $𝑡 = 𝑎 + 𝑑$ and $Δ = 𝑎 𝑑 - 𝑏 𝑐$ . By (a), $𝐴^{2} = 𝑡 𝐴 - Δ 𝐼_{2}$ , so $𝐴^{2} = 𝐼_{2}$ if and only if $𝑡 𝐴 = (1 + Δ) 𝐼_{2}$ .

If $𝑡 \neq 0$ , this says $𝐴 = 𝑡^{- 1} (1 + Δ) 𝐼_{2}$ is a scalar matrix $𝜆 𝐼_{2}$ , and $𝐴^{2} = 𝐼_{2}$ means $𝜆^{2} = 1$ , so $𝜆 = \pm 1$ (since $(𝜆 - 1) (𝜆 + 1) = 0$ , Proposition 1.1.17). Both $𝐼_{2}$ and $- 𝐼_{2}$ satisfy $𝐴^{2} = 𝐼_{2}$ , and they are different because $𝑝$ is odd.

If $𝑡 = 0$ , the condition becomes $(1 + Δ) 𝐼_{2} = 0$ , that is, $Δ = - 1$ . With $𝑑 = - 𝑎$ , $Δ = - 𝑎^{2} - 𝑏 𝑐$ , so the condition is $𝑎^{2} + 𝑏 𝑐 = 1$ . Conversely every such $𝐴$ has $𝑡 = 0$ and $Δ = - 1$ , so $𝐴^{2} = 𝑡 𝐴 - Δ 𝐼_{2} = 𝐼_{2}$ . These matrices are not $\pm 𝐼_{2}$ , whose traces are $\pm 2 \neq 0$ .

It remains to count the triples $(𝑎, 𝑏, 𝑐) \in 𝔽_{𝑝}^{3}$ with $𝑏 𝑐 = 1 - 𝑎^{2}$ . The equation $𝑎^{2} = 1$ has exactly the two solutions $𝑎 = \pm 1$ ; for each, $𝑏 𝑐 = 0$ has $2 𝑝 - 1$ solutions. For each of the other $𝑝 - 2$ values of $𝑎$ , $1 - 𝑎^{2} \neq 0$ , so $𝑏 \neq 0$ and $𝑐 = (1 - 𝑎^{2}) 𝑏^{- 1}$ : $𝑝 - 1$ solutions. The total is

\begin{aligned} 2 + 2 (2 𝑝 - 1) + (𝑝 - 2) (𝑝 - 1) \\ = 2 + 4 𝑝 - 2 + 𝑝^{2} - 3 𝑝 + 2 \\ = 𝑝^{2} + 𝑝 + 2 . \end{aligned}

For $𝑝 = 3$ this gives $14$ matrices. Over $𝔽_{2}$ the count is different, because $𝐼_{2} = - 𝐼_{2}$ there: of the six invertible matrices listed in Exercise 2.5.43, exactly four satisfy $𝐴^{2} = 𝐼_{2}$ .

Exercise 2.5.47.hard Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ .

(a)Suppose that $𝐼_{𝑚} - 𝐴 𝐵$ is invertible. Show that $𝐼_{𝑛} - 𝐵 𝐴$ is invertible, with
${(𝐼_{𝑛} - 𝐵 𝐴)}^{- 1} = 𝐼_{𝑛} + 𝐵 {(𝐼_{𝑚} - 𝐴 𝐵)}^{- 1} 𝐴 .$
(b)Deduce: if $𝐮, 𝐯 \in 𝐹^{𝑛}$ and $1 + 𝐯^{𝖳} 𝐮 \neq 0$ , then $𝐼_{𝑛} + 𝐮 𝐯^{𝖳}$ is invertible, with inverse $𝐼_{𝑛} -$ ${(1 + 𝐯^{𝖳} 𝐮)}^{- 1} 𝐮 𝐯^{𝖳}$ .

Solution.

Uses. Proposition 2.4.14 (Algebra of matrices), Definition 2.5.12 (Invertible)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. Where does the formula come from? Pretend the matrices are numbers and expand

1 / (1 - 𝑏 𝑎)

as a geometric series:

1 +

𝑏 𝑎 +

𝑏 𝑎 𝑏 𝑎 +

\dots =

1 +

𝑏 (1 + 𝑎 𝑏 + 𝑎 𝑏 𝑎 𝑏 + \dots) 𝑎

. Then forget the heuristic and verify.

Hint 2. Write

𝐶 = {(𝐼_{𝑚} - 𝐴 𝐵)}^{- 1}

. In

(𝐼_{𝑛} - 𝐵 𝐴) (𝐼_{𝑛} + 𝐵 𝐶 𝐴)

, collect the terms of the form

𝐵 (\dots) 𝐴

Hint 3. The bracket you get is

𝐶 -

𝐼_{𝑚} -

𝐴 𝐵 𝐶 =

(𝐼_{𝑚} - 𝐴 𝐵) 𝐶 -

𝐼_{𝑚}

Sketch. With

𝐶 = {(𝐼_{𝑚} - 𝐴 𝐵)}^{- 1}

(𝐼_{𝑛} - 𝐵 𝐴) (𝐼_{𝑛} + 𝐵 𝐶 𝐴) =

𝐼_{𝑛} +

𝐵 [𝐶 - 𝐼_{𝑚} - 𝐴 𝐵 𝐶] 𝐴 =

𝐼_{𝑛} +

𝐵 [(𝐼_{𝑚} - 𝐴 𝐵) 𝐶 - 𝐼_{𝑚}] 𝐴 =

𝐼_{𝑛}

, and

(𝐼_{𝑛} + 𝐵 𝐶 𝐴) (𝐼_{𝑛} - 𝐵 𝐴) =

𝐼_{𝑛} +

𝐵 [𝐶 - 𝐼_{𝑚} - 𝐶 𝐴 𝐵] 𝐴 =

𝐼_{𝑛} +

𝐵 [𝐶 (𝐼_{𝑚} - 𝐴 𝐵) - 𝐼_{𝑚}] 𝐴 =

𝐼_{𝑛}

. (b) is (a) with

𝑚 = 1

𝐵 = 𝐮

𝐴 = - 𝐯^{𝖳}

(a) Let $𝐶 = {(𝐼_{𝑚} - 𝐴 𝐵)}^{- 1}$ , so $(𝐼_{𝑚} - 𝐴 𝐵) 𝐶 =$ $𝐶 (𝐼_{𝑚} - 𝐴 𝐵) =$ $𝐼_{𝑚}$ . By the distributive and associative laws (Proposition 2.4.14),

\begin{aligned} (𝐼_{𝑛} - 𝐵 𝐴) (𝐼_{𝑛} + 𝐵 𝐶 𝐴) \\ = 𝐼_{𝑛} + 𝐵 𝐶 𝐴 - 𝐵 𝐴 - 𝐵 𝐴 𝐵 𝐶 𝐴 \\ = 𝐼_{𝑛} + 𝐵 (𝐶 - 𝐼_{𝑚} - 𝐴 𝐵 𝐶) 𝐴 \\ = 𝐼_{𝑛} + 𝐵 ((𝐼_{𝑚} - 𝐴 𝐵) 𝐶 - 𝐼_{𝑚}) 𝐴 \\ = 𝐼_{𝑛} + 𝐵 (𝐼_{𝑚} - 𝐼_{𝑚}) 𝐴 = 𝐼_{𝑛}, \end{aligned}

and similarly

\begin{aligned} (𝐼_{𝑛} + 𝐵 𝐶 𝐴) (𝐼_{𝑛} - 𝐵 𝐴) \\ = 𝐼_{𝑛} - 𝐵 𝐴 + 𝐵 𝐶 𝐴 - 𝐵 𝐶 𝐴 𝐵 𝐴 \\ = 𝐼_{𝑛} + 𝐵 (𝐶 - 𝐼_{𝑚} - 𝐶 𝐴 𝐵) 𝐴 \\ = 𝐼_{𝑛} + 𝐵 (𝐶 (𝐼_{𝑚} - 𝐴 𝐵) - 𝐼_{𝑚}) 𝐴 = 𝐼_{𝑛} . \end{aligned}

So $𝐼_{𝑛} + 𝐵 𝐶 𝐴$ is a two-sided inverse of $𝐼_{𝑛} - 𝐵 𝐴$ .

(b) Apply (a) with $𝑚 = 1$ , $𝐵 = 𝐮 \in 𝐹^{𝑛 \times 1}$ , and $𝐴 = - 𝐯^{𝖳} \in 𝐹^{1 \times 𝑛}$ . Then $𝐼_{𝑛} -$ $𝐵 𝐴 =$ $𝐼_{𝑛} +$ $𝐮 𝐯^{𝖳}$ , and $𝐼_{1} - 𝐴 𝐵 = 1 + 𝐯^{𝖳} 𝐮$ , a $1 \times 1$ matrix, which is invertible exactly when it is nonzero. By (a),

\begin{aligned} {(𝐼_{𝑛} + 𝐮 𝐯^{𝖳})}^{- 1} \\ = 𝐼_{𝑛} + 𝐮 {(1 + 𝐯^{𝖳} 𝐮)}^{- 1} (- 𝐯^{𝖳}) \\ = 𝐼_{𝑛} - {(1 + 𝐯^{𝖳} 𝐮)}^{- 1} 𝐮 𝐯^{𝖳} . \end{aligned}

The point of (a) is that inverting the $𝑛 \times 𝑛$ matrix $𝐼_{𝑛} - 𝐵 𝐴$ reduces to inverting the $𝑚 \times 𝑚$ matrix $𝐼_{𝑚} - 𝐴 𝐵$ , which in (b) is a single number.

Exercise 2.5.48.hard Let $𝑆, 𝑇 \in L (𝑉)$ satisfy $𝑆 𝑇 = 𝑆 + 𝑇$ .

(a)Show that $(id - 𝑆) (id - 𝑇) = id$ .
(b)Suppose that $id - 𝑆$ is invertible. Show that $𝑆 𝑇 = 𝑇 𝑆$ . (When $𝑉 = 𝐹^{𝑛}$ , this hypothesis follows from (a) by Theorem 3.5.23.)
(c)Find $𝑆, 𝑇 \in L (ℚ [𝑥])$ with $𝑆 𝑇 = 𝑆 + 𝑇$ but $𝑆 𝑇 \neq 𝑇 𝑆$ .

Solution.

Uses. Proposition 2.1.24 (Algebra of composition), Theorem 2.5.10 (Left and right inverses agree), Example 2.5.24 (Differentiation has a right inverse but no left inverse)

Hint 1. Expand

(id - 𝑆) (id - 𝑇)

and use the hypothesis.

Hint 2. By (a),

id - 𝑇

is a right inverse of

id - 𝑆

. If

id - 𝑆

is invertible, then

id - 𝑇

is its inverse, so the product in the other order is also

id

. Expand that product.

Hint 3. For (c), you need

id - 𝑆

and

id - 𝑇

that are one-sided inverses of each other but not two-sided. Use

𝐷

and

𝐽

Sketch. (a)

(id - 𝑆) (id - 𝑇) =

id -

𝑆 -

𝑇 +

𝑆 𝑇 =

id

. (b) Then

id - 𝑇 = {(id - 𝑆)}^{- 1}

by Theorem 2.5.10, so

(id - 𝑇) (id - 𝑆) = id

, which expands to

𝑇 𝑆 = 𝑆 + 𝑇 = 𝑆 𝑇

. (c)

𝑆 = id - 𝐷

and

𝑇 = id - 𝐽

𝑆 𝑇 =

2 id -

𝐷 -

𝐽 +

𝐷 𝐽 =

𝑆 +

𝑇

, while

𝑇 𝑆 -

𝑆 𝑇 =

𝐽 𝐷 -

𝐷 𝐽 =

𝐽 𝐷 -

id \neq

0

(a) By Proposition 2.1.24 and the hypothesis,

\begin{aligned} (id - 𝑆) (id - 𝑇) & = id - 𝑇 - 𝑆 + 𝑆 𝑇 \\ = id - 𝑇 - 𝑆 + 𝑆 + 𝑇 = id . \end{aligned}

(b) By (a), $id - 𝑇$ is a right inverse of $id - 𝑆$ . If $id - 𝑆$ is invertible, then by Theorem 2.5.10 its right inverse $id - 𝑇$ is its inverse, so also $(id - 𝑇) (id - 𝑆) = id$ . Expanding, $id - 𝑆 - 𝑇 + 𝑇 𝑆 = id$ , so $𝑇 𝑆 = 𝑆 + 𝑇 = 𝑆 𝑇$ .

(c) Let $𝐷, 𝐽 \in L (ℚ [𝑥])$ be as in Example 2.5.24, so $𝐷 𝐽 = id$ and $𝐽 𝐷 \neq id$ . Let $𝑆 = id - 𝐷$ and $𝑇 = id - 𝐽$ . Then

\begin{aligned} 𝑆 𝑇 & = (id - 𝐷) (id - 𝐽) = id - 𝐽 - 𝐷 + 𝐷 𝐽 \\ = 2 id - 𝐷 - 𝐽 = 𝑆 + 𝑇, \end{aligned}

while

\begin{aligned} 𝑇 𝑆 & = (id - 𝐽) (id - 𝐷) = id - 𝐷 - 𝐽 + 𝐽 𝐷 \\ = 𝑆 𝑇 + (𝐽 𝐷 - id) . \end{aligned}

Since $𝐽 𝐷 (1) = 0 \neq 1$ , $𝐽 𝐷 - id \neq 0$ , so $𝑇 𝑆 \neq 𝑆 𝑇$ . Here $id - 𝑆 = 𝐷$ has the right inverse $id - 𝑇 = 𝐽$ but is not invertible, which is exactly the hypothesis of (b) failing.

Exercise 2.5.49.challenge Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ with $𝑛 \geq 1$ and $𝐴 𝐵 = 𝐼_{𝑛}$ . You may use that any $𝑛^{2} + 1$ matrices in $𝐹^{𝑛 \times 𝑛}$ are linearly dependent, because $𝐹^{𝑛 \times 𝑛}$ has dimension $𝑛^{2}$ (Section 4.1).

(a)Show that $𝐴^{𝑟} 𝐵^{𝑘} = 𝐵^{𝑘 - 𝑟}$ whenever $𝑘 \geq 𝑟 \geq 0$ .
(b)Show that $𝐵$ has a right inverse that is a linear combination of $𝐼_{𝑛}, 𝐵, 𝐵^{2}, \dots$ .
(c)Conclude that $𝐵 𝐴 = 𝐼_{𝑛}$ .

Solution.

Uses. Theorem 2.5.10 (Left and right inverses agree), Notation 2.4.16 (Powers of a matrix), Proposition 2.4.14 (Algebra of matrices), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Too many vectors force a dependence (Technique 4.1.30)

Hint 1. (a) First show

𝐴 𝐵^{𝑘} = 𝐵^{𝑘 - 1}

for

𝑘 \geq 1

, then induct on

𝑟

Hint 2. (b) The matrices

𝐼_{𝑛}, 𝐵, \dots, 𝐵^{𝑛^{2}}

are dependent. Take a nontrivial relation

\sum 𝑐_{𝑘} 𝐵^{𝑘} = 0

, let

𝑟

be the smallest index with

𝑐_{𝑟} \neq 0

, and multiply the relation on the left by

𝐴^{𝑟}

Hint 3. After multiplying by

𝐴^{𝑟}

, the relation reads

𝑐_{𝑟} 𝐼_{𝑛} +

𝑐_{𝑟 + 1} 𝐵 +

𝑐_{𝑟 + 2} 𝐵^{2} +

\dots =

0

. Solve for

𝐼_{𝑛}

and factor out

𝐵

Hint 4. (c)

𝐵

now has a left inverse and a right inverse.

Sketch. (a)

𝐴 𝐵^{𝑘} =

(𝐴 𝐵) 𝐵^{𝑘 - 1} =

𝐵^{𝑘 - 1}

, and induction on

𝑟

. (b) Take a nontrivial relation

\sum_{𝑘 = 𝑟}^{𝑁} 𝑐_{𝑘} 𝐵^{𝑘} =

0

with

𝑐_{𝑟} \neq 0

and multiply by

𝐴^{𝑟}

\sum_{𝑘 = 𝑟}^{𝑁} 𝑐_{𝑘} 𝐵^{𝑘 - 𝑟} =

0

, so

𝐼_{𝑛} = 𝐵 𝑄

with

𝑄 =

- 𝑐_{𝑟}^{- 1} \sum_{𝑘 = 𝑟 + 1}^{𝑁} 𝑐_{𝑘} 𝐵^{𝑘 - 𝑟 - 1}

. (c)

𝐴

is a left inverse of

𝐵

and

𝑄

a right inverse, so

𝐴 = 𝑄

by Theorem 2.5.10 and

𝐵 𝐴 = 𝐵 𝑄 = 𝐼_{𝑛}

(a) For $𝑘 \geq 1$ , $𝐴 𝐵^{𝑘} =$ $(𝐴 𝐵) 𝐵^{𝑘 - 1} =$ $𝐵^{𝑘 - 1}$ by associativity. We induct on $𝑟$ ; for $𝑟 = 0$ the claim is $𝐵^{𝑘} = 𝐵^{𝑘}$ . If $𝐴^{𝑟} 𝐵^{𝑘} = 𝐵^{𝑘 - 𝑟}$ for all $𝑘 \geq 𝑟$ , and $𝑘 \geq 𝑟 + 1$ , then

𝐴^{𝑟 + 1} 𝐵^{𝑘} = 𝐴 (𝐴^{𝑟} 𝐵^{𝑘}) = 𝐴 𝐵^{𝑘 - 𝑟} = 𝐵^{𝑘 - 𝑟 - 1},

using $𝑘 - 𝑟 \geq 1$ .

(b) The $𝑛^{2} + 1$ matrices $𝐼_{𝑛},$ $𝐵,$ $𝐵^{2},$ $\dots,$ $𝐵^{𝑛^{2}}$ are linearly dependent, so there are $𝑐_{0}, \dots, 𝑐_{𝑛^{2}} \in 𝐹$ , not all $0$ , with $\sum_{𝑘 = 0}^{𝑛^{2}} 𝑐_{𝑘} 𝐵^{𝑘} =$ $0$ (Theorem 1.5.5). Let $𝑟$ be the smallest index with $𝑐_{𝑟} \neq 0$ , so the relation is $\sum_{𝑘 = 𝑟}^{𝑛^{2}} 𝑐_{𝑘} 𝐵^{𝑘} =$ $0$ . Multiplying on the left by $𝐴^{𝑟}$ and using (a),

𝑐_{𝑟} 𝐼_{𝑛} + 𝑐_{𝑟 + 1} 𝐵 + 𝑐_{𝑟 + 2} 𝐵^{2} + \dots + 𝑐_{𝑛^{2}} 𝐵^{𝑛^{2} - 𝑟} = 0 .

Let $𝑄 =$ $- 𝑐_{𝑟}^{- 1} (𝑐_{𝑟 + 1} 𝐼_{𝑛} + 𝑐_{𝑟 + 2} 𝐵 + \dots + 𝑐_{𝑛^{2}} 𝐵^{𝑛^{2} - 𝑟 - 1})$ , an empty sum being $0$ . Solving for $𝐼_{𝑛}$ gives $𝐼_{𝑛} = 𝐵 𝑄$ . So $𝑄$ is a right inverse of $𝐵$ , and it is a linear combination of powers of $𝐵$ .

(c) $𝐴 𝐵 = 𝐼_{𝑛}$ says that $𝐴$ is a left inverse of $𝐵$ , and by (b), $𝑄$ is a right inverse of $𝐵$ . By Theorem 2.5.10, $𝐴 = 𝑄$ , so $𝐵 𝐴 = 𝐵 𝑄 = 𝐼_{𝑛}$ .

This proves the square case of Theorem 3.5.23 using only that $𝐹^{𝑛 \times 𝑛}$ is finite-dimensional, and finite dimension is essential. On $ℚ [𝑥]$ , with $𝐷 𝐽 = id$ , the argument breaks down at (b), because the powers of $𝐽$ are linearly independent.

More practice

The practice bank has 48 exercises on this section’s topics.