2.4 Inverses, one-to-one maps, isomorphism

Linear maps can be added, scaled and chained. A map can be undone exactly when no two inputs land on the same output, and an undoable linear map between two spaces makes them the same space under different names.

Chapter 2.3 asked what a linear map destroys and what it reaches. This chapter asks two more questions. First, how do you chain maps, doing one after the other, and what rules do chains obey? Second, when can a map be undone, and when are two vector spaces really the same space under different names? The first question comes up whenever operations are stacked (rotate, then shift). The second comes up whenever you solve an equation $T(x) = b$ , because solving it means undoing $T$ .

Before you start

2.3: linear map, null space, range, one-to-one and onto, and the rank–nullity theorem. 2.2: basis, dimension and coordinates. 2.1: vector spaces and the subspace test.

Quick checkA linear map

T: \mathbb R^3 \to \mathbb R^2

has rank 2. What is its nullity?

Quick checkWhich condition shows that a LINEAR map

T

is one-to-one?

Quick checkIn a basis

(e_1, e_2, e_3)

V

, what are the coordinates of

x = 2e_1 - e_3

The picture

Two maps of the plane. Start with “Rotate 90°” and press “Apply the map”. The whole grid turns a quarter turn. Every point lands somewhere, and no two points land in the same place. Given where a point ended up, you can say where it came from: turn back a quarter. That is what it means for a map to be invertible.

Now pick “Squash to a line” from the menu. Everything lands on the green line. The whole red line lands on the single point $O$ , and more generally every line parallel to the red one lands on a single point of the green line. If I tell you the output is a point on the green line, you cannot say which input it came from. There are infinitely many candidates, all on one parallel line. Nothing can undo this map.

One warning about the “Undo the map” button. It plays the animation backwards, so it works for any map, even the squash. The film remembers where every point started. The real question is different: given only the final picture, can you recover the start? For the rotation yes. For the squash no.

The idea, in words

Three ideas, in order.

Chaining. Doing map $T$ and then map $S$ is a new map. If $T$ and $S$ are linear, so is the chain. Order matters: rotating a book a quarter turn about one axis and then a quarter turn about another axis gives a different result from doing them in the other order.

Undoing. To undo $T$ , you need a map $S$ that sends each output back to its input. That works exactly when no two inputs share an output, which is the condition “one-to-one” from 2.3. If two inputs share an output, that output cannot remember which input it came from. For linear maps there is a shortcut: $T$ is one-to-one exactly when the only input sent to $O$ is $O$ itself, so you never need to compare all pairs of inputs.

Sameness. A linear map that can be undone, between two vector spaces, is a perfect translation dictionary. Every sum, multiple, basis and dimension in one space matches one in the other. We then call the two spaces isomorphic, and for practical purposes they are the same space. The big payoff is that every $n$ -dimensional real vector space is isomorphic to $\mathbb R^n$ , by taking coordinates. That is why $\mathbb R^n$ is enough.

Why· Why one-to-one is the right condition

Think of $T$ as a machine that takes an input and prints an output. To undo it, you hold the printout and ask ‘what went in?’. If two different inputs print the same page, you cannot answer. If every page was printed by exactly one input, you can. So ‘undoable’ means ‘one-to-one’. Onto is a separate matter: pages that no input prints do not stop you from undoing the pages that do exist, because nobody ever hands you those pages.

Definitions and theorems

Adding and scaling linear maps

Spagnolie writes linearity as one line, $T(ax + y) = aT(x) + T(y)$ for all $x, y$ and every scalar $a$ . It is equivalent to the two conditions of 2.3. If $T$ satisfies the two conditions, then $T(ax + y) = T(ax) + T(y) = aT(x) + T(y)$ . Conversely, suppose the one-line version holds. Taking $a = 1$ gives $T(x + y) = T(x) + T(y)$ . Then $T(O) = T(O + O) = T(O) + T(O)$ , so $T(O) = O$ . Taking $y = O$ gives $T(ax) = aT(x) + T(O) = aT(x)$ . We use the one-line version below because it is shorter.

Let $S, T: V \to W$ be two maps. Define their sum $S + T$ and the multiple $cT$ by

(S + T)(x) = S(x) + T(x), \qquad (cT)(x) = c\,T(x).

This is the pointwise addition of functions from 2.1: add the outputs, input by input. The set of all linear maps from $V$ to $W$ is written $L(V, W)$ (Apostol writes $\mathscr L(V, W)$ ).

Theorem T Apostol Thm 2.4; Spagnolie Lecture 8

Linear maps form a vector space

For vector spaces $V$ and $W$ , the set $L(V, W)$ with the operations above is a vector space. The zero vector is the zero map $O(x) = O$ , and the negative of $T$ is $(-1)T$ .

In one lineThe set $L(V, W)$ of all linear maps from $V$ to $W$ is a vector space under pointwise addition and scaling.

Why it is trueIt is a subspace of the space of all functions from $V$ to $W$ , and a sum or multiple of linear maps is linear.

ProofApostol 2.5; Spagnolie Lecture 8 axiom check

The idea: Use the subspace test inside the space of ALL functions from V to W. Every axiom holds there because it holds in W at each input. Then check that the zero map, sums and multiples of linear maps are still linear.

The set

F

of all functions

V \to W

, with pointwise operations, is a vector space. For example,

S + T = T + S

because

S(x) + T(x) = T(x) + S(x)

W

for every

x

Same argument as the function-space example in 2.1. To check an axiom, evaluate both sides at an input

x

. Each side becomes an expression in

W

, where the axiom holds.

The zero map is linear:

O(ax + y) = O = a\,O + O

, using

aO = O

(a problem in 2.1) and axiom 5 of

W

First part of the subspace test: the zero vector of

F

is the zero function, which must be in

L(V, W)

S, T

are linear:

(S + T)(ax + y) = S(ax + y) + T(ax + y) = aS(x) + S(y) + aT(x) + T(y) = a\,(S(x) + T(x)) + (S(y) + T(y)) = a\,(S + T)(x) + (S + T)(y)

Closure under addition. Apply the definition of

S + T

, then linearity of

S

and of

T

, then rearrange using axioms 3, 4 and 8.

(cT)(ax + y) = c\,(aT(x) + T(y)) = a\,(cT)(x) + (cT)(y)

Closure under scaling. Axiom 8 splits

c\,(aT(x) + T(y))

into two terms. Axiom 7 and the fact that the numbers

c

and

a

can be multiplied in either order move

c

past

a

L(V, W)

is a vector space.

All three parts of the subspace test hold, so

L(V, W)

is a subspace of

F

, hence a vector space.

In 2.5 we will find its dimension: if $\dim V = n$ and $\dim W = m$ , then $\dim L(V, W) = mn$ .

Chaining maps: composition

Definition T

Composition

Let $T: U \to V$ and $S: V \to W$ be functions. The composition $ST: U \to W$ is defined by

(ST)(x) = S\big(T(x)\big) \quad \text{for every } x \in U.

It is defined only when the outputs of $T$ lie in the input space of $S$ . We read it from right to left: $T$ acts first. (In JEE you wrote $(g \circ f)(x) = g(f(x))$ . Apostol and Spagnolie drop the circle, so $gf$ means $f$ first.)

In one lineThe map $ST$ that does $T$ first and then $S$ : $(ST)(x) = S(T(x))$ .

Why it existsIt is how operations are stacked. Rotating then shifting, or one layer of a network followed by the next, are compositions.

Two bits of vocabulary. The identity map $I_V: V \to V$ is $I_V(x) = x$ . For a map $T: V \to V$ from a space to itself we define powers by $T^0 = I$ and $T^n = T\,T^{n-1}$ for $n \ge 1$ .

Everyday example from school. For $f(x) = x^2$ and $g(x) = x + 1$ , we get $(gf)(x) = x^2 + 1$ but $(fg)(x) = (x + 1)^2$ . The order changed the answer.

Theorem T Apostol Thm 2.6; Spagnolie Lecture 8

Composition of linear maps is linear

If $T: U \to V$ and $S: V \to W$ are linear maps, then $ST: U \to W$ is linear.

In one lineIf $T: U \to V$ and $S: V \to W$ are linear, then $ST: U \to W$ is linear.

Why it is true $T$ carries sums and multiples through, and then $S$ does too.

ProofSpagnolie Lecture 8

The idea: Push the combination ax + y through T, then through S, one use of linearity each.

(ST)(ax + y) = S\big(T(ax + y)\big)

Definition of composition.

= S\big(aT(x) + T(y)\big)

T

is linear.

= a\,S(T(x)) + S(T(y))

S

is linear, applied to the vectors

T(x)

and

T(y)

V

= a\,(ST)(x) + (ST)(y)

. So

ST

satisfies the one-line test.

Definition of composition again, read backwards.

Theorem T Apostol Thm 2.5; Spagnolie Lecture 8

Composition is associative

If $T: U \to V$ , $S: V \to W$ and $R: W \to X$ are functions (linear or not), then $R(ST) = (RS)T$ .

In one lineFor any three functions that can be chained, $R(ST) = (RS)T$ .

Why it is trueBoth sides do the same three steps in the same order, so the brackets only say which two steps you named first.

ProofApostol 2.5

The idea: Evaluate both sides at an arbitrary input and see that each is R(S(T(x))).

[R(ST)](x) = R\big((ST)(x)\big)

Definition of composition, applied to the pair

R

and

ST

= R\big(S(T(x))\big)

Definition applied to

ST

[(RS)T](x) = (RS)\big(T(x)\big)

Definition applied to the pair

RS

and

T

= R\big(S(T(x))\big)

. The two sides agree at every

x

, so they are the same function.

Definition applied to

RS

Associativity gives the law of exponents $T^m T^n = T^{m+n}$ , and it lets us write $RST$ with no brackets. It is the reason that, in 2.5, matrix multiplication is associative without anyone expanding a single sum.

Adding and composing mix in the way you hope.

Proposition A Apostol Thm 2.7

Composition distributes over addition

Let $S, T$ be linear maps from $V$ to $W$ , and $c$ a scalar. For any function $R$ with outputs in $V$ : $(S + T)R = SR + TR$ and $(cS)R = c(SR)$ . For any linear map $R: W \to X$ : $R(S + T) = RS + RT$ and $R(cS) = c(RS)$ .

In one line $(S + T)R = SR + TR$ and $R(S + T) = RS + RT$ (the second needs $R$ linear); scalars pass through too.

Why it is trueThe first is just the definition of $S + T$ . The second is where $R$ must respect addition.

ProofApostol 2.5

The idea: Evaluate at an input. The first law uses only the definition of S + T. The second law needs R to carry a sum through.

[(S + T)R](x) = S(R(x)) + T(R(x)) = [SR + TR](x)

Definition of

S + T

at the input

R(x)

, then definition of composition twice.

[R(S + T)](x) = R\big(S(x) + T(x)\big)

Definition of composition, then definition of

S + T

= R(S(x)) + R(T(x)) = [RS + RT](x)

R

is linear, so it respects addition.

So all four laws hold.

The scalar laws are identical:

[(cS)R](x) = c\,S(R(x))

, and

R(c\,S(x)) = c\,R(S(x))

because

R

respects scaling.

Composition does not commute

Spagnolie’s example (Lecture 8, which he called ‘Proof by example’) uses two quarter turns of space. Let

S(x, y, z) = (-y,\ x,\ z), \qquad T(x, y, z) = (x,\ -z,\ y).

$S$ turns space a quarter turn about the $z$ -axis: it sends $(1,0,0)$ to $(0,1,0)$ . $T$ turns it a quarter turn about the $x$ -axis: it sends $(0,1,0)$ to $(0,0,1)$ . Both are linear. Composing,

ST(x, y, z) = S(x, -z, y) = (z,\ x,\ y), \qquad TS(x, y, z) = T(-y, x, z) = (-y,\ -z,\ x).

At $(1, 2, 3)$ these give $ST(1,2,3) = (3, 1, 2)$ and $TS(1,2,3) = (-2, -3, 1)$ . Different. Hold a book and try it: a quarter turn about a vertical axis then a quarter turn about a left-right axis leaves the book in a different position from the same two turns in the other order.

Your answerWith the same

S(x, y, z) = (-y, x, z)

and

T(x, y, z) = (x, -z, y)

, compute

TS(4, 5, 6)

. What is its FIRST entry?

Sometimes $ST$ does not even make sense, because the outputs of $T$ are not inputs of $S$ . When both do make sense and $ST = TS$ , we say $S$ and $T$ commute. This happens rarely, so an identity like $(S + T)^2 = S^2 + 2ST + T^2$ is false in general. By distributivity, $(S + T)^2 = S^2 + ST + TS + T^2$ , and the middle two terms add up to $2ST$ only when the maps commute.

Worked example new, in the style of Apostol 2.8 #26

Compute

ST

and

TS

for

S(x, y) = (2x,\ x + y)

and

T(x, y) = (y,\ x - y)

, and decide whether they commute

Solution.

Write $ST$ as ‘apply $T$ , then $S$ ’. $ST(x, y) = S\big(T(x, y)\big) = S(y,\ x - y)$ . The map on the right acts first, so $T$ eats $(x, y)$ and produces the pair that $S$ then receives.
Feed that pair into $S$ . $S(a, b) = (2a,\ a + b)$ with $a = y$ and $b = x - y$ gives $(2y,\ y + x - y) = (2y,\ x)$ . We replace $x$ by $a$ and $y$ by $b$ in the formula for $S$ .
Write $TS$ as ‘apply $S$ , then $T$ ’. $TS(x, y) = T(2x,\ x + y)$ . Now $S$ acts first.
Feed that pair into $T$ . $T(a, b) = (b,\ a - b)$ with $a = 2x$ and $b = x + y$ gives $(x + y,\ 2x - x - y) = (x + y,\ x - y)$ . Same substitution, with the roles reversed.
Compare at one point. At $(1, 0)$ : $ST(1,0) = (0, 1)$ and $TS(1, 0) = (1, 1)$ . One input where the outputs differ is enough to show two maps are different.

Answer. $ST(x,y) = (2y, x)$ and $TS(x,y) = (x + y, x - y)$ . They do not commute.

Your answerFor the same

S(x, y) = (2x, x + y)

and

T(x, y) = (y, x - y)

, what is the SECOND entry of

TS(3, 7)

Undoing a map: inverses

Spagnolie’s phrase for an inverse is ‘Where did you come from?’. The map $T$ sends $x$ to $w = T(x)$ . An inverse takes $w$ and names its source.

Definition T

Inverse

Let $T: V \to W$ be a function. A function $S: T(V) \to V$ is a left inverse of $T$ if

S\big(T(x)\big) = x \quad \text{for all } x \in V, \qquad \text{that is, } ST = I_V.

A function $R: T(V) \to V$ is a right inverse of $T$ if

T\big(R(w)\big) = w \quad \text{for all } w \in T(V), \qquad \text{that is, } TR = I_{T(V)}.

The domain of both is the range $T(V)$ , because only outputs of $T$ can be asked ‘where did you come from?’. (Some books allow $S$ and $R$ to be defined on all of $W$ . We follow Apostol and Spagnolie. The first stretch problem below shows that, for linear maps, a left inverse defined on all of $W$ can always be built.)

In one lineA map that undoes $T$ : a left inverse satisfies $S(T(x)) = x$ for every $x$ , a right inverse satisfies $T(R(w)) = w$ for every $w$ in the range.

Why it existsSolving $T(x) = b$ is finding a source for $b$ . A left inverse answers 'what went in?' for every output, a right inverse produces some input for every output.

Right inverses are easy to find. For every $w$ in the range, pick any one $x$ with $T(x) = w$ and set $R(w) = x$ . Then $T(R(w)) = w$ . So every function has at least one right inverse. (When the range is infinite, “pick one for each $w$ ” uses a basic set-theory principle called the axiom of choice, which we accept.) The right inverse is not unique if some $w$ has several sources.

Left inverses are not so easy. Spagnolie’s example (Lecture 9): $T: \mathbb R^2 \to \mathbb R$ , $T(x_1, x_2) = 3x_1 - x_2$ . A right inverse is $R(w) = (w/3,\ 0)$ , because $T(w/3, 0) = w$ . Another is $R'(w) = (w,\ 2w)$ , because $3w - 2w = w$ . In fact, for any number $a$ , the map $R_a(w) = (aw,\ (3a - 1)w)$ is a right inverse, since $3aw - (3a - 1)w = w$ . So there are infinitely many. But there is no left inverse. The inputs $(0, 0)$ and $(1, 3)$ both give output $0$ , because $T(1, 3) = 3 - 3 = 0$ . A left inverse $S$ would need $S(0) = (0, 0)$ and $S(0) = (1, 3)$ at once. In terms of 2.3, the culprit is the null space: $T$ is onto $\mathbb R$ , so its rank is 1, and rank–nullity gives $\dim N(T) = 2 - 1 = 1$ .

The smallest example (Apostol 2.6) has $V = \{1, 2\}$ and $W = \{0\}$ with $T(1) = T(2) = 0$ . It has two right inverses, $R(0) = 1$ and $R'(0) = 2$ , and no left inverse.

Theorem T Apostol Thm 2.9; Spagnolie Lectures 8 and 9

Left inverse exists exactly when one-to-one

A function $T: V \to W$ has a left inverse if and only if it is one-to-one: $T(x) = T(y)$ implies $x = y$ .

In one lineA function $T$ has a left inverse if and only if different inputs always give different outputs.

Why it is trueA left inverse reads off the source of each output. That is possible exactly when each output has only one source.

ProofSpagnolie Lecture 8 page 6; Apostol 2.6

The idea: (⇒) Apply the left inverse to both sides of T(x) = T(y). (⇐) Build the left inverse: send each output to its one and only source.

(

\Rightarrow

) Suppose

T(x) = T(y)

Assume a left inverse

S

exists and that

T(x) = T(y)

. We must show

x = y

S(T(x)) = S(T(y))

Apply the function

S

to both sides of an equation. Equal inputs give equal outputs.

x = y

. So

T

is one-to-one.

S

is a left inverse, so

S(T(\cdot))

returns its input.

(

\Leftarrow

) Let

w \in T(V)

. There is some

x

with

T(x) = w

Now assume

T

is one-to-one. Each

w

in the range has at least one source

x

by the definition of range.

There is exactly one such

x

Two sources

x, x'

of the same

w

would have

T(x) = T(x')

, and one-to-one forces

x = x'

Define

S(w)

to be that

x

This is a well-defined function on

T(V)

, because each

w

has exactly one source to name.

Then

S(T(x)) = x

for every

x \in V

. So

S

is a left inverse.

For any

x \in V

, the output

T(x)

has source

x

, so

S

returns

x

In the picture. Go back to the two visualizers at the top. In the rotation, every output point has exactly one source, so ‘send each point back to its source’ is a well-defined map: that is the step ‘Define $S(w)$ to be that $x$ ’. In the squash, the output point has a whole red-parallel line of sources, so the step ‘There is exactly one such $x$ ’ fails, and with it any hope of a left inverse. The first half of the proof is the same picture seen from the other side: if two points land together, no map applied afterwards can pull them apart.

Quick checkDoes

f(x) = x^2

, as a function from

\mathbb R

\mathbb R

, have a left inverse?

Worked example new, in the style of Apostol 2.8 #6

Show

T(x, y) = (x - y,\ 2x - 2y)

is not one-to-one, find its range, and find a right inverse

Solution.

Find two inputs with the same output. $T(1, 1) = (0, 0) = T(0, 0)$ . ‘One-to-one’ says different inputs give different outputs, so one pair that collides shows it fails.
Describe all collisions through the null space. $T(x, y) = O$ means $x - y = 0$ , so $N(T) = \{(t, t)\}$ , the line $y = x$ . $T$ is linear, so it merges $x$ and $x'$ exactly when $x - x'$ is in the null space (the worked example in 2.3 proves this, and it is proved again below as part of ‘Equivalent forms’).
Find the range. Every output is $(x - y,\ 2(x - y)) = s\,(1, 2)$ with $s = x - y$ , and every $s$ occurs. So $T(V) = \{s\,(1, 2)\}$ , a line in the plane. Outputs depend only on the single number $x - y$ , so only one dimension survives (rank 1, nullity 1, and $1 + 1 = 2$ ).
Build a right inverse. For $w = (s, 2s)$ in the range choose the source $(s, 0)$ , so $R(s, 2s) = (s, 0)$ . A right inverse picks ONE source for each output. Any source with $x - y = s$ would do, which is why right inverses are not unique.
Check it. $T\big(R(s, 2s)\big) = T(s, 0) = (s,\ 2s)$ . That is the defining equation $TR = I$ on the range.

Answer. Not one-to-one, so no left inverse. Range: the line through $(1, 2)$ . One right inverse: $R(s, 2s) = (s, 0)$ .

Theorem T Apostol Thm 2.8; Spagnolie Lecture 8 page 5

A left inverse is unique and two-sided

A function $T: V \to W$ has at most one left inverse. If $S$ is a left inverse of $T$ , then $S$ is also a right inverse of $T$ .

In one lineA function has at most one left inverse, and a left inverse is also a right inverse.

Why it is trueThe left inverse on an output $w$ must name the source of $w$ , and there is only one source.

ProofApostol 2.6

The idea: Uniqueness: two left inverses both name the source of w, so they agree. Two-sidedness: apply T to the equation x = S(T(x)).

Suppose

S

and

S'

are left inverses. Let

w \in T(V)

, say

w = T(x)

Take two left inverses

S

and

S'

. We show they give the same value at every point of their common domain

T(V)

S(w) = S(T(x)) = x

and

S'(w) = S'(T(x)) = x

. So

S(w) = S'(w)

for all

w

Both are left inverses, so each returns

x

from

T(x)

Since

S

is a left inverse,

x = S(T(x)) = S(w)

Now let

S

be a left inverse and

w \in T(V)

, again

w = T(x)

. We must show

T(S(w)) = w

T(x) = T(S(w))

, that is,

w = T(S(w))

. So

S

is a right inverse.

Apply

T

to both sides of

x = S(w)

Definition T

Invertible

A function $T: V \to W$ is invertible if it is one-to-one. By the two theorems, it then has exactly one left inverse, which is also a right inverse. That map $T^{-1}: T(V) \to V$ is the inverse of $T$ . It satisfies

T^{-1}\big(T(x)\big) = x \ \text{ for all } x \in V, \qquad T\big(T^{-1}(w)\big) = w \ \text{ for all } w \in T(V).

In one lineA function is invertible when it is one-to-one. Its unique left inverse, which is also a right inverse, is the inverse $T^{-1}$ .

Why it existsIt gives a single name, $T^{-1}$ , to the map that undoes $T$ , so we can write equations with it.

Do not read $T^{-1}$ as $1/T$ . It is the function that undoes $T$ . The $-1$ is borrowed from numbers, where $x^{-1}$ undoes multiplication by $x$ .

When $T: V \to V$ goes from a space to itself and is also onto, $T^{-1}$ is defined on all of $V$ and both equations in the definition hold on all of $V$ : $T^{-1}T = I_V = TT^{-1}$ . Later parts of this chapter and chapter 2.5 use “invertible” in this stronger sense (one-to-one and onto) for maps $V \to V$ , and say so when they do.

Inverses of linear maps

For general functions, to check one-to-one you must compare all pairs of inputs. For linear maps it is enough to look at one vector, and the inverse is automatically linear.

Theorem T Apostol Thm 2.10; Spagnolie Lecture 9 page 3

Equivalent forms of one-to-one for linear maps

Let $T: V \to W$ be linear. The following statements are equivalent:

(a) $T$ is one-to-one.

(b) $T$ is invertible and its inverse $T^{-1}: T(V) \to V$ is linear.

In one lineFor a linear map $T$ , these are equivalent: one-to-one; invertible with a linear inverse; and $N(T) = \{O\}$ .

Why it is trueLinearity turns 'T(x) = T(y)' into 'T(x − y) = O', so only the vectors sent to $O$ matter.

ProofApostol 2.7; Spagnolie Lecture 9

The idea: Prove a ⇒ b ⇒ c ⇒ a, a circle. The only new work is that the inverse is linear: pull a combination back through T.

Assume (a). Let

u, v \in T(V)

and let

a

be a scalar.

(a ⇒ b) By 'Left inverse exists exactly when one-to-one',

T

is invertible. It remains to show

T^{-1}

is linear.

Write

u = T(x)

and

v = T(y)

Every element of the range has a source. Name the sources.

au + v = aT(x) + T(y) = T(ax + y)

T

is linear, so it carries the combination of sources to the same combination of outputs.

T^{-1}(au + v) = ax + y = a\,T^{-1}(u) + T^{-1}(v)

. So

T^{-1}

is linear.

Apply

T^{-1}

and use that it undoes

T

on the source

ax + y

. Then replace

x = T^{-1}(u)

and

y = T^{-1}(v)

Assume (b). If

T(x) = O

, then

x = T^{-1}(T(x)) = T^{-1}(O)

(b ⇒ c) Let

T(x) = O

. Apply

T^{-1}

T^{-1}(O) = O

, so

x = O

. This is (c).

A linear map sends

O

O

(the first consequence in 2.3), and

T^{-1}

is linear by (b).

Assume (c) and

T(x) = T(y)

. Then

T(x - y) = T(x) - T(y) = O

(c ⇒ a) Assume

T(x) = T(y)

. Move everything to one side using linearity:

T(x - y) = T(x + (-1)y) = T(x) + (-1)T(y) = T(x) - T(y)

, as in 2.3.

x - y = O

, that is

x = y

. This is (a).

By (c), the only vector sent to

O

O

In the picture. Statement (c) is the red line in the ‘Squash to a line’ visualizer, the vectors sent to $O$ . When the red line is just the origin, as in the rotation, no two points can land together, because two points $x$ and $y$ landing together means $x - y$ is a vector on the red line. That is the step $T(x - y) = O$ . In the squash the red line is a whole line, and any two points differing by a vector on it land together.

Proof workshopGive the reason for each line: the inverse of a linear map is linear

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $u, v \in T(V)$ and let $a$ be a scalar. Write $u = T(x)$ and $v = T(y)$ .
$au + v = aT(x) + T(y)$
$aT(x) + T(y) = T(ax + y)$
$T^{-1}(au + v) = T^{-1}\big(T(ax + y)\big)$
$T^{-1}\big(T(ax + y)\big) = ax + y$
$ax + y = a\,T^{-1}(u) + T^{-1}(v)$
So $T^{-1}(au + v) = a\,T^{-1}(u) + T^{-1}(v)$ , and $T^{-1}$ is linear.

Quick checkA linear map

T

has

T(1, 2) = O

. Is

T

one-to-one?

Worked example Apostol 2.8 #20

Show

T(x, y, z) = (x + y,\ y + z,\ x + z)

is one-to-one and find

T^{-1}

Solution.

Test the null space. Suppose $T(x, y, z) = (0, 0, 0)$ : $x + y = 0$ , $y + z = 0$ , $x + z = 0$ . T is linear, so ‘Equivalent forms’ says one-to-one is the same as only $O$ going to $O$ .
Solve. From the first two, $x = -y$ and $z = -y$ . The third gives $x + z = -2y = 0$ , so $y = 0$ , and then $x = z = 0$ . So $N(T) = \{O\}$ . Each equation lets us express one unknown through another, and the third closes the loop.
Conclude one-to-one. Since $N(T) = \{O\}$ , $T$ is one-to-one. ‘Equivalent forms of one-to-one for linear maps’ (c ⇒ a). Its part (b) also tells us that $T^{-1}$ exists and is linear.
Find the inverse by solving for $x, y, z$ . Let $u = x + y$ , $v = y + z$ , $w = x + z$ . Adding $u$ and $w$ and subtracting $v$ gives $u + w - v = 2x$ , so $x = \tfrac{u - v + w}{2}$ . In the same way $y = \tfrac{u + v - w}{2}$ and $z = \tfrac{-u + v + w}{2}$ . Each unknown is isolated by adding the two equations that contain it and subtracting the one that does not.
Check one output. $T(1, 2, 3) = (3, 5, 4)$ . Then $x = (3 - 5 + 4)/2 = 1$ , $y = (3 + 5 - 4)/2 = 2$ , $z = (-3 + 5 + 4)/2 = 3$ . The formula must return the input we started with.

Answer. One-to-one, and $T^{-1}(u, v, w) = \big(\tfrac{u - v + w}{2},\ \tfrac{u + v - w}{2},\ \tfrac{-u + v + w}{2}\big)$ .

Your answerFor

T(x, y, z) = (x + y,\ y + z,\ x + z)

, the inverse sends

(4, 6, 8)

(x, y, z)

. What is

z

Worked example Apostol 2.8 #10, #17 (similar)

T(x, y) = (x + 2,\ y - 1)

linear? Is it one-to-one? Find

T^{-1}

if it exists

Solution.

Test linearity with the zero vector. $T(0, 0) = (2, -1) \ne (0, 0)$ . Every linear map sends $O$ to $O$ (2.3), so one failure settles it.
Do not use the null-space test. ‘Equivalent forms’ is a theorem about linear maps only, so with $T$ not linear it cannot decide one-to-one.
Test one-to-one by hand. If $T(x, y) = T(x', y')$ then $x + 2 = x' + 2$ and $y - 1 = y' - 1$ , so $(x, y) = (x', y')$ . That is the definition: equal outputs force equal inputs.
Solve for the inverse. If $(u, v) = (x + 2, y - 1)$ then $x = u - 2$ and $y = v + 1$ . The inverse sends each output back to its one source.

Answer. Not linear. One-to-one, with $T^{-1}(u, v) = (u - 2,\ v + 1)$ (also not linear).

When $V$ is finite-dimensional, one-to-one can be read off from bases.

Theorem T Apostol Thm 2.11; Spagnolie Lecture 9 page 4

One-to-one in finite dimension

Let $T: V \to W$ be linear with $\dim V = n$ . The following are equivalent:

(a) $T$ is one-to-one.

(b) If $e_1, \dots, e_p$ are independent in $V$ , then $T(e_1), \dots, T(e_p)$ are independent in $T(V)$ .

(d) If $e_1, \dots, e_n$ is a basis of $V$ , then $T(e_1), \dots, T(e_n)$ is a basis of $T(V)$ .

In one lineFor a linear $T$ on an $n$ -dimensional $V$ , these are equivalent: $T$ is one-to-one; independent sets stay independent; $\dim T(V) = n$ ; bases go to bases.

Why it is trueOne-to-one means nothing is destroyed. Independent directions therefore stay independent, and a basis of $V$ lands on a basis of the range.

ProofApostol 2.7; Spagnolie Lecture 9 page 4

The idea: Again a circle, a ⇒ b ⇒ c ⇒ d ⇒ a. Spagnolie left the last two arrows 'to the reader'; they are written out here.

Assume (a). Suppose

\sum_{i=1}^p c_i T(e_i) = O

(a ⇒ b) To test independence of the images, take a combination of them equal to

O

and show the coefficients vanish.

T\!\left(\sum c_i e_i\right) = O = T(O)

Linearity: a combination of images is the image of the combination.

\sum c_i e_i = O

T

is one-to-one, so equal outputs come from equal inputs.

Every

c_i = 0

. So the images are independent.

The

e_i

are independent.

Assume (b). Then

T(V)

contains

n

independent vectors, so

\dim T(V) \ge n

(b ⇒ c) Apply (b) to a basis

e_1, \dots, e_n

V

. Then

T(V)

contains

n

independent vectors. By the

k+1

theorem of 2.2, a space spanned by fewer than

n

vectors cannot contain

n

independent ones. So

T(V)

cannot have dimension below

n

\dim T(V) \le \dim V = n

. So

\dim T(V) = n

Rank–nullity: rank plus nullity is

n

, and the nullity is not negative.

Assume (c) and let

e_1, \dots, e_n

be a basis of

V

. Then

T(e_1), \dots, T(e_n)

span

T(V)

(c ⇒ d) The images of a basis span the range: write any

T(x)

\sum x_i T(e_i)

They are

n

vectors spanning

T(V)

, which has dimension

n

by (c). So they form a basis.

n

vectors that span an

n

-dimensional space form a basis (a problem in 2.2).

Assume (d). Let

T(x) = O

and

x = \sum c_i e_i

. Then

\sum c_i T(e_i) = O

(d ⇒ a) By the previous theorem it is enough to show

N(T) = \{O\}

. Take

x

with

T(x) = O

and expand

x

in a basis.

Every

c_i = 0

, so

x = O

. Hence

N(T) = \{O\}

, and

T

is one-to-one by ‘Equivalent forms of one-to-one for linear maps’ (c ⇒ a).

The

T(e_i)

are a basis by (d), hence independent.

Inverse of a composition: socks and shoes

You put on socks and then shoes. To undo this you take off the shoes first and then the socks. The same reversal holds for every pair of invertible maps. It was Apostol 2.8 #23 on HW5.

Theorem T Apostol 2.8 #23; MATH 375 HW5

Inverse of a composition reverses the order

Let $T: U \to V$ and $S: V \to W$ be one-to-one functions. Then $ST$ is one-to-one, and for every $w$ in its range,

(ST)^{-1}(w) = T^{-1}\big(S^{-1}(w)\big).

In short, $(ST)^{-1} = T^{-1}S^{-1}$ .

In one lineIf $S$ and $T$ are invertible, so is $ST$ , and $(ST)^{-1} = T^{-1}S^{-1}$ .

Why it is true $ST$ does $T$ then $S$ , so to undo it you must undo $S$ first, then $T$ .

ProofApostol 2.8 #23

The idea: Show T⁻¹S⁻¹ is a left inverse of ST. Left inverses are unique, so it is THE inverse.

First,

ST

is one-to-one.

Check one-to-one directly. Suppose

ST(x) = ST(y)

, that is,

S(T(x)) = S(T(y))

T(x) = T(y)

S

is one-to-one, applied to the two inputs

T(x)

and

T(y)

x = y

. So

ST

is one-to-one and has an inverse.

T

is one-to-one.

Define

G(w) = T^{-1}\big(S^{-1}(w)\big)

for

w

in the range of

ST

Let

w = S(T(x))

be in the range of

ST

. Then

S^{-1}(w)

is defined, since

w

is in the range of

S

G(ST(x)) = T^{-1}\big(S^{-1}(S(T(x)))\big) = T^{-1}\big(T(x)\big)

S^{-1}

undoes

S

on the input

T(x)

= x

. So

G

is a left inverse of

ST

T^{-1}

undoes

T

G = (ST)^{-1}

, that is

(ST)^{-1} = T^{-1}S^{-1}

'A left inverse is unique and two-sided': there is only one left inverse, and it is called

(ST)^{-1}

Your answerA sprite point

p

is rotated by

T(x, y) = (-y, x)

and then shifted by

S(x, y) = (x + 5, y + 2)

, and it ends on screen at

(8, 6)

. Undo the two steps in the right order. What was the

x

-coordinate of

p

The step-through above is the picture for this theorem. Steps 4 and 5 apply $T^{-1}S^{-1}$ to $ST(p)$ and get $p$ back, which is the written proof’s check that $G(ST(x)) = x$ , at one point. Step 6 shows the other order fails.

Proof workshopRebuild the proof of socks and shoes

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick check

S

and

T

are invertible. Which is the inverse of

ST

Worked example new, in the style of Apostol 2.8 #24

Prove: if

S, T: V \to V

are linear,

T

is one-to-one and onto, and

ST = TS

, then

ST^{-1} = T^{-1}S

Solution.

Plan. Because $T$ is one-to-one and onto, $T^{-1}$ is defined on all of $V$ , so $TT^{-1} = I = T^{-1}T$ . We want $T^{-1}$ moved from the right of $S$ to the left. Multiply the given equation by $T^{-1}$ on both sides. The hypothesis $ST = TS$ is the only information we have, so we use it and then cancel $T$ with $T^{-1}$ .
Multiply on the left and right. $T^{-1}(ST)T^{-1} = T^{-1}(TS)T^{-1}$ . Both sides of an equation of maps stay equal if we compose both with the same map on the same side.
Regroup the left side. $T^{-1}(ST)T^{-1} = (T^{-1}S)(T\,T^{-1}) = T^{-1}S\,I = T^{-1}S$ . Composition is associative, so brackets can move, and $TT^{-1} = I$ .
Regroup the right side. $T^{-1}(TS)T^{-1} = (T^{-1}T)(S\,T^{-1}) = I\,S\,T^{-1} = ST^{-1}$ . Same associativity, and now $T^{-1}T = I$ .
Equate. $T^{-1}S = ST^{-1}$ . Steps 3 and 4 simplified two expressions that step 2 said were equal.

Answer. $ST^{-1} = T^{-1}S$ , so $S$ commutes with $T^{-1}$ too.

Isomorphism: the same space under different names

Definition T

Isomorphism

A linear map $T: V \to W$ is an isomorphism if it is one-to-one and onto ( $T(V) = W$ ). If an isomorphism $V \to W$ exists, $V$ and $W$ are isomorphic, written $V \cong W$ .

In one lineA linear map between two vector spaces that is one-to-one and onto. Two spaces joined by one are isomorphic: the same space with the vectors renamed.

Why it existsIt is the exact meaning of 'these two spaces are the same as far as linear algebra can tell'. Anything proved about one holds for the other.

The word comes from Greek: isos, equal, and morphe, shape (Spagnolie’s remark in Lecture 11). If $T: V \to W$ is an isomorphism, then $T^{-1}: W \to V$ exists (it is defined on all of $W$ , since $T$ is onto), and it is linear by ‘Equivalent forms of one-to-one for linear maps’. It is one-to-one and onto too, so it is also an isomorphism. An isomorphism matches vectors, sums, multiples, bases and subspaces on the two sides.

Where this came from· Where 'isomorphism' came from

The word began in chemistry. In 1819 the German chemist Eilhard Mitscherlich reported that corresponding phosphates and arsenates crystallize in the same form. Different substances, same shape: he called them isomorphic, from the Greek for “equal form”. Mathematicians later borrowed the word for two structures that can be matched exactly, with the matching respecting the operations. The habit of studying a structure only up to such a matching became standard through Emmy Noether’s work on homomorphisms of rings and modules, where her isomorphism theorems appeared in 1927. This chapter’s theorem is that habit at its cleanest. For finite-dimensional real vector spaces, the single number $\dim V$ settles which spaces are the same.

Being isomorphic is an equivalence relation: it behaves like ‘has the same value as’. That means three things hold, and each is short to check.

Theorem B new

Isomorphic is an equivalence relation

For vector spaces $V, W, X$ : (reflexive) $V \cong V$ ; (symmetric) $V \cong W$ implies $W \cong V$ ; (transitive) $V \cong W$ and $W \cong X$ imply $V \cong X$ .

In one lineEvery space is isomorphic to itself, if $V \cong W$ then $W \cong V$ , and if $V \cong W$ and $W \cong X$ then $V \cong X$ .

Why it is trueThe identity, the inverse and the composition of isomorphisms are isomorphisms.

Proofnew

The idea: Supply the isomorphism in each case: the identity, the inverse, the composition.

Reflexive:

I_V: V \to V

is an isomorphism.

The identity map respects sums and multiples, and it is one-to-one and onto.

Symmetric: if

T: V \to W

is an isomorphism, so is

T^{-1}: W \to V

Shown just above:

T^{-1}

is linear (Equivalent forms), one-to-one (it has the left inverse

T

) and onto (every

x \in V

equals

T^{-1}(T(x))

Transitive: if

T: V \to W

and

S: W \to X

are isomorphisms, then

ST: V \to X

is linear and one-to-one.

'Composition of linear maps is linear' and 'Inverse of a composition' (one-to-one part).

ST

is onto, so it is an isomorphism.

ST

has range

S(T(V)) = S(W) = X

, because

T

is onto

W

and

S

is onto

X

Now the payoff promised in 2.1 and 2.2. A space with a basis is isomorphic to $\mathbb R^n$ , and the isomorphism is the act of taking coordinates.

Theorem B uses Apostol 1.9; Spagnolie Lecture 11 (isomorphism)

Every n-dimensional space is isomorphic to R^n

Let $\dim V = n$ with ordered basis $(e_1, \dots, e_n)$ . The coordinate map $C: V \to \mathbb R^n$ , $C(\sum c_i e_i) = (c_1, \dots, c_n)$ , is an isomorphism. In particular $V \cong \mathbb R^n$ .

In one lineIf $\dim V = n$ and $(e_1, \dots, e_n)$ is an ordered basis, then $x \mapsto (c_1, \dots, c_n)$ , the coordinates of $x$ , is an isomorphism $V \to \mathbb R^n$ .

Why it is trueCoordinates are unique (so the map is a function and is one-to-one), every list of numbers is the coordinate list of some vector (onto), and coordinates of a sum are the sums of coordinates (linear).

ProofApostol 1.9 and 2.9

The idea: Linear because of uniqueness of coordinates; one-to-one because the zero vector is the only one with all coordinates zero; onto by building a vector from any list.

C

is well defined.

'Unique coordinates' from 2.2 makes

C

a function: each

x

has exactly one coordinate list.

Then

ax + y = \sum (ac_i + d_i)\, e_i

Write

x = \sum c_i e_i

and

y = \sum d_i e_i

. Collect terms using the axioms.

C(ax + y) = (ac_1 + d_1, \dots, ac_n + d_n) = a\,C(x) + C(y)

C

is linear.

By uniqueness of coordinates, this IS the coordinate list of

ax + y

C(x) = (0, \dots, 0)

, then

x = 0e_1 + \dots + 0e_n = O

. So

C

is one-to-one.

'Equivalent forms of one-to-one for linear maps' (c ⇒ a): only

N(C)

needs checking.

C

is onto. So

C

is an isomorphism.

Given any list

(c_1, \dots, c_n)

, the vector

\sum c_i e_i

is in

V

by the closure axioms, and its coordinates are that list.

In the picture. This is the coordinate grid from 2.2. The plane drawn with the slanted grid below is $V$ , and the pair $(c_1, c_2)$ displayed is $C(x)$ in $\mathbb R^2$ . Drag $x$ and watch the pair change: each point has exactly one pair (that is ‘well defined’, and ‘one-to-one’), and every pair belongs to some point (that is ‘onto’). Move $b_1$ onto $b_2$ ‘s line and the grid collapses, which is the case where the $e_i$ are not a basis and the coordinate map stops existing.

Where this came from· Descartes and the first coordinate map

In La Géométrie (1637), René Descartes brought algebra into geometry: a curve is described by an equation linking lengths measured from a fixed line. Pierre de Fermat found the same idea independently. Both used a single reference line. The now-familiar pair of axes came a little later, as Frans van Schooten and his students translated and explained La Géométrie (a Latin edition appeared in 1649). Giving each point of the plane a pair of numbers is the case $n = 2$ of this chapter’s theorem: the plane, with a chosen pair of axes, is matched with $\mathbb R^2$ . The general statement, that any $n$ -dimensional space is $\mathbb R^n$ once a basis is chosen, belongs to the later study of abstract vector spaces (see the history box in 2.1). A different pair of axes gives a different matching, which is exactly the freedom that 2.5 studies.

Quick checkWhich pair of spaces is NOT isomorphic?

Your answerThe space of

3 \times 4

real matrices is isomorphic to

\mathbb R^n

for which

n

Worked example Axler 3.D style

Show that the symmetric

2 \times 2

matrices are isomorphic to

\mathbb R^3

, by writing down an isomorphism

Solution.

Name the set. A $2 \times 2$ matrix is symmetric if its two off-diagonal entries are equal. So $V = \left\{\begin{bmatrix}a&b\\b&c\end{bmatrix} : a, b, c \in \mathbb R\right\}$ . We need the objects before we can map them.
Check $V$ is a vector space. It is closed under addition, since $\begin{bmatrix}a&b\\b&c\end{bmatrix} + \begin{bmatrix}a'&b'\\b'&c'\end{bmatrix} = \begin{bmatrix}a+a'&b+b'\\b+b'&c+c'\end{bmatrix}$ , and under scaling, since $k\begin{bmatrix}a&b\\b&c\end{bmatrix} = \begin{bmatrix}ka&kb\\kb&kc\end{bmatrix}$ is symmetric, and it contains the zero matrix. By the subspace test it is a subspace of the $2 \times 2$ matrices.
Define the map. $C\begin{bmatrix}a&b\\b&c\end{bmatrix} = (a, b, c)$ . This is ‘take coordinates’ in the basis $\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}$ , because the symmetric matrix above equals $a$ times the first plus $b$ times the second plus $c$ times the third.
Check linear. The sum of two matrices has entries $(a + a', b + b', c + c')$ , which is $C(\text{first}) + C(\text{second})$ . Scaling by $k$ multiplies all three entries by $k$ . Both conditions of a linear map.
Check one-to-one. If $C = (0, 0, 0)$ then $a = b = c = 0$ , so the matrix is the zero matrix. ‘Equivalent forms’ (c ⇒ a): the null space is only $O$ .
Check onto. Any $(a, b, c)$ is $C$ of $\begin{bmatrix}a&b\\b&c\end{bmatrix}$ . Every output is attained.

Answer. $C$ is an isomorphism, so the symmetric $2 \times 2$ matrices are isomorphic to $\mathbb R^3$ and have dimension 3.

Theorem B Apostol 2.7 and 2.14, assembled

Same dimension exactly when isomorphic

Let $V$ and $W$ be finite-dimensional. Then $V \cong W$ if and only if $\dim V = \dim W$ .

In one lineTwo finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

Why it is trueAn isomorphism sends a basis to a basis, so it preserves the count. Conversely, two spaces of dimension $n$ are both isomorphic to $\mathbb R^n$ , and isomorphic is transitive.

Proofnew

The idea: Forward: an isomorphism is one-to-one, so (d) of the finite-dimensional theorem gives a basis of the range. Backward: go through R^n.

(

\Rightarrow

)

T(e_1), \dots, T(e_n)

is a basis of

T(V)

Let

T: V \to W

be an isomorphism and

e_1, \dots, e_n

a basis of

V

. 'One-to-one in finite dimension' (a ⇒ d).

W

has a basis of

n

vectors, and

\dim W = n = \dim V

T

is onto, so

T(V) = W

\dim V = \dim W = n

, then

V \cong \mathbb R^n

and

W \cong \mathbb R^n

(

\Leftarrow

) Both spaces have dimension

n

. The previous theorem gives

V \cong \mathbb R^n

and

W \cong \mathbb R^n

V \cong W

Symmetric:

\mathbb R^n \cong W

. Transitive: chain

V \cong \mathbb R^n \cong W

So dimension is the one number that classifies finite-dimensional real vector spaces. The plane is not isomorphic to 3-space, and $P_3$ (cubic polynomials, dimension 4) is isomorphic to the $2 \times 2$ matrices (dimension 4). Isomorphic does not mean equal, and the isomorphism is not unique: each ordered basis gives a different one. This is why 2.5 will work hard on what happens when you change the basis.

Not every vector space has finite dimension. In infinite dimensions the picture changes (the second stretch problem shows the polynomials are isomorphic to a smaller subspace of themselves).

Prescribed values on a basis

2.3 showed that a linear map is determined by its values on a basis. The converse is the useful half: you may choose those values freely.

Theorem A Apostol Thm 2.12; Spagnolie Lecture 10

Prescribed values on a basis

Let $e_1, \dots, e_n$ be a basis of $V$ and $u_1, \dots, u_n$ any $n$ vectors in $W$ (repeats allowed). There is one and only one linear map $T: V \to W$ with $T(e_k) = u_k$ for $k = 1, \dots, n$ . It is given by

T\!\left(\sum_{k=1}^n x_k e_k\right) = \sum_{k=1}^n x_k u_k.

In one lineGiven a basis $e_1, \dots, e_n$ of $V$ and ANY vectors $u_1, \dots, u_n$ in $W$ , there is exactly one linear map $T$ with $T(e_k) = u_k$ .

Why it is trueA basis gives each vector unique coordinates, so $T(x) = \sum x_k u_k$ is well defined, and it is the only possible linear map.

ProofApostol 2.9

The idea: Existence: define T by the formula. Uniqueness: a linear map is forced to use the formula.

The formula defines a function

T: V \to W

'Unique coordinates' (2.2): each

x

has exactly one list

(x_1, \dots, x_n)

, so the right-hand side has exactly one value for each

x

T(ax + y) = \sum (ax_k + y_k)\,u_k = a \sum x_k u_k + \sum y_k u_k = aT(x) + T(y)

. So

T

is linear.

x = \sum x_k e_k

and

y = \sum y_k e_k

, then

ax + y

has coordinates

ax_k + y_k

(proved in the isomorphism theorem above).

T(e_k) = 0u_1 + \dots + 1u_k + \dots + 0u_n = u_k

The vector

e_k

has coordinates

0, \dots, 0, 1, 0, \dots, 0

, with the

1

in place

k

T'(x) = \sum x_k T'(e_k) = \sum x_k u_k = T(x)

, using linearity of

T'

Uniqueness. Let

T'

be another linear map with

T'(e_k) = u_k

. It is enough to compare them at an arbitrary

x = \sum x_k e_k

T' = T

Same values at every input.

In the picture. Drag the two arrow tips in the visualizer below. The arrows $\hat\imath$ and $\hat\jmath$ are the basis $e_1, e_2$ , and their landing spots are the values $u_1, u_2$ . You may drop them anywhere you like (existence: the formula always produces a map), and once you have dropped them, the whole grid is forced into place (uniqueness: no other linear map has those landing spots).

Apostol’s example: define $T: \mathbb R^2 \to \mathbb R^2$ by $T(\hat\imath) = \hat\imath + \hat\jmath$ and $T(\hat\jmath) = 2\hat\imath - \hat\jmath$ . Then $T(x_1, x_2) = x_1(1, 1) + x_2(2, -1) = (x_1 + 2x_2,\ x_1 - x_2)$ .

Another, with a basis that is not the standard one. Let $b_1 = (1, 1)$ , $b_2 = (1, -1)$ , and require $T(b_1) = (3, 1)$ , $T(b_2) = (1, 5)$ . Writing $(x, y) = \tfrac{x+y}{2}\,b_1 + \tfrac{x-y}{2}\,b_2$ ,

T(x, y) = \tfrac{x+y}{2}(3, 1) + \tfrac{x-y}{2}(1, 5) = (2x + y,\ 3x - 2y).

Check: $T(1, 1) = (3, 1)$ and $T(1, -1) = (1, 5)$ .

Your answerFor this map (

T(1, 1) = (3, 1)

and

T(1, -1) = (1, 5)

), what is the SECOND entry of

T(5, 1)

Quick checkCan you prescribe

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

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

and

T(1, 1) = (5, 5)

for a linear map

T

Worked example Apostol 2.12 style; new numbers

Find the linear map

T

from

\mathbb R^2

\mathbb R^3

with

T(1, 1) = (1, 0, 2)

and

T(1, 2) = (0, 1, 1)

. Is it one-to-one? Onto?

Solution.

Check the inputs form a basis. $(1, 1)$ and $(1, 2)$ are not multiples of each other, so they are independent, and two independent vectors form a basis of $\mathbb R^2$ . ‘Prescribed values on a basis’ needs a basis, and ‘Extending to a basis’ (b) from 2.2 lets us skip checking they span.
Write $(x, y)$ in this basis. Solve $a(1,1) + b(1,2) = (x, y)$ : $a + b = x$ and $a + 2b = y$ . Subtracting, $b = y - x$ , and then $a = x - b = 2x - y$ . We need the coordinates $a, b$ , because $T$ is defined by $T(x, y) = a\,T(1,1) + b\,T(1,2)$ .
Apply the formula. $T(x, y) = (2x - y)(1, 0, 2) + (y - x)(0, 1, 1) = (2x - y,\ y - x,\ 4x - 2y + y - x) = (2x - y,\ y - x,\ 3x - y)$ . Linearity: the values on the basis are combined with the coordinates as weights.
Check the prescribed values. $T(1, 1) = (1, 0, 2)$ and $T(1, 2) = (0, 1, 1)$ . The formula must reproduce the data we started with.
Decide one-to-one. $T(x, y) = O$ gives $y - x = 0$ from the second entry and $2x - y = 0$ from the first, so $x = y$ and then $x = 0$ . Hence $N(T) = \{O\}$ and $T$ is one-to-one. ‘Equivalent forms of one-to-one for linear maps’ (c ⇒ a).
Decide onto. The range is spanned by the two values $(1, 0, 2)$ and $(0, 1, 1)$ , so it is at most a plane in $\mathbb R^3$ . It is not all of $\mathbb R^3$ . The range is the set of combinations of the prescribed values, since $T(x, y) = a\,T(1,1) + b\,T(1,2)$ . Two vectors cannot span 3 dimensions. The corollary ‘Building a linear map from the images of a basis’, just after this section, proves both answers in general.

Answer. $T(x, y) = (2x - y,\ y - x,\ 3x - y)$ . It is one-to-one and not onto.

Why a basis is the right thing to prescribe on. Try to prescribe on a dependent set: $T(1,0) = (1,0)$ , $T(0,1) = (0,1)$ , $T(1,1) = (5,5)$ . Linearity forces $T(1,1) = T(1,0) + T(0,1) = (1,1)$ , so no linear map exists. Independence is what prevents values from contradicting each other. Spanning is what prevents any vector’s value from being left undecided. A basis has both, which is exactly why it works.

Corollary A uses Apostol Thms 2.11, 2.12

Building a linear map from the images of a basis

Let $e_1, \dots, e_n$ be a basis of $V$ , and let $T: V \to W$ be the linear map with $T(e_k) = u_k$ . Then (a) $T(V) = L(\{u_1, \dots, u_n\})$ ; (b) $T$ is one-to-one if and only if $u_1, \dots, u_n$ are independent; (c) $T$ is an isomorphism if and only if $u_1, \dots, u_n$ is a basis of $W$ .

In one lineIf $T(e_k) = u_k$ on a basis, then $T$ is onto $W$ exactly when the $u_k$ span $W$ , one-to-one exactly when they are independent, and an isomorphism exactly when they are a basis of $W$ .

Why it is trueThe range is the span of the $u_k$ , and $T(x) = O$ is the equation $\sum x_k u_k = O$ .

Proofnew

The idea: The range is everything of the form sum x_k u_k, which is the span. The null space question is the independence question word for word.

(a)

T(V)

is exactly the set of all combinations of the

u_k

, that is, their span.

Every output is

T(x) = \sum x_k u_k

, a combination of the

u_k

. Conversely, each combination

\sum x_k u_k

equals

T(\sum x_k e_k)

(b)

N(T) = \{O\}

if and only if the only combination

\sum x_k u_k = O

has all

x_k = 0

, which is the definition of independent. And

N(T) = \{O\}

is one-to-one by ‘Equivalent forms’ (c ⇔ a).

T(x) = O

means

\sum x_k u_k = O

. This is a combination equal to

O

whose coefficients are the coordinates of

x

(c)

T

is one-to-one and onto exactly when the

u_k

are independent and span

W

, that is, form a basis.

Onto is (a):

T(V) = W

when the

u_k

span

W

. Together with (b), an isomorphism is a set that spans and is independent.

This is the cleanest way to manufacture isomorphisms. Pick bases, send one to the other. Every isomorphism between finite-dimensional spaces arises this way.

Examples and non-examples

Running sums and differences (Apostol 2.8 #19). $T(x, y, z) = (x,\ x + y,\ x + y + z)$ keeps a running total. If $T(x, y, z) = (u, v, w)$ , then $u = x$ , $v = x + y$ , $w = x + y + z$ . Solve backwards: $x = u$ , $y = v - u$ , $z = w - v$ . So $T$ is one-to-one (its output determines its input), and

T^{-1}(u, v, w) = (u,\ v - u,\ w - v).

The inverse takes differences. This is the discrete version of the fact that integrating and differentiating undo each other, and it is the prefix-sum array from programming: store running totals, and you can recover the list by subtracting neighbours.

Worked example Apostol 2.8 #27; MATH 375 HW6

Let

D

be differentiation and

T(p)(x) = \int_0^x p(t)\,dt

on all real polynomials. Show

DT = I

but

TD \ne I

, and find the null space and range of

TD

Solution.

Check both are linear maps of the polynomials to themselves. $D(af + g) = af' + g'$ and $\int_0^x (af + g) = a\int_0^x f + \int_0^x g$ . Sum and constant-multiple rules from calculus are exactly the two conditions.
Compute $DT$ . $DT(p) = \dfrac{d}{dx}\displaystyle\int_0^x p(t)\,dt = p(x)$ . So $DT = I$ . The fundamental theorem of calculus: differentiating an integral with variable upper limit returns the integrand.
Compute $TD$ . $TD(p) = \displaystyle\int_0^x p'(t)\,dt = p(x) - p(0)$ . The fundamental theorem again, in the other direction, which leaves behind the starting value.
Show $TD \ne I$ with one example. $p = 2 + 3x$ gives $TD(p) = 3x \ne p$ . One input where two maps differ proves the maps differ.
Null space of $TD$ . $p - p(0) = 0$ means $p$ equals its constant term, so $N(TD)$ is the constant polynomials. A polynomial equal to its value at $0$ has no $x$ terms.
Range of $TD$ . Every output $p - p(0)$ has value $0$ at $x = 0$ , and any $q$ with $q(0) = 0$ is hit by $p = q$ . So $R(TD) = \{q : q(0) = 0\}$ . Subtracting $p(0)$ forces the constant term to vanish, and nothing else is lost.

Answer. $DT = I$ , $TD(p) = p - p(0)$ , $N(TD)$ = constants, $R(TD)$ = polynomials vanishing at $0$ .

Why does this not contradict ‘A left inverse is unique and two-sided’? That theorem requires the left inverse to be defined on the range $T(V)$ only. Here $T$ (integration) has range $\{q : q(0) = 0\}$ . On that set, $TD(q) = q - q(0) = q$ , so $D$ restricted to the range is both a left and right inverse of $T$ . Applied to all of $V$ , $D$ is no longer a right inverse. $D$ itself is not one-to-one, since $D(1) = D(2) = 0$ , so $D$ has no left inverse. In infinite dimensions, a linear map can be one-to-one and not onto, as $T$ is here. When the domain and codomain have the same finite dimension, that cannot happen (a prove problem in 2.3).

The six reorderings of three things (Apostol 2.8 #2, HW6). Let $V = \{0, 1, 2\}$ , and write each reordering as the list of images of $0, 1, 2$ : $T_1 = (0,1,2)$ , $T_2 = (0,2,1)$ , $T_3 = (1,0,2)$ , $T_4 = (1,2,0)$ , $T_5 = (2,0,1)$ , $T_6 = (2,1,0)$ . All six are one-to-one and onto. The table gives $T_jT_i$ , with $T_i$ done first (columns) and $T_j$ second (rows):

	$T_1$	$T_2$	$T_3$	$T_4$	$T_5$	$T_6$
$T_1$	$T_1$	$T_2$	$T_3$	$T_4$	$T_5$	$T_6$
$T_2$	$T_2$	$T_1$	$T_5$	$T_6$	$T_3$	$T_4$
$T_3$	$T_3$	$T_4$	$T_1$	$T_2$	$T_6$	$T_5$
$T_4$	$T_4$	$T_3$	$T_6$	$T_5$	$T_1$	$T_2$
$T_5$	$T_5$	$T_6$	$T_2$	$T_1$	$T_4$	$T_3$
$T_6$	$T_6$	$T_5$	$T_4$	$T_3$	$T_2$	$T_1$

Look at the diagonal. For $T_1, T_2, T_3, T_6$ the diagonal entry is $T_1$ , so each of these is its own inverse. But $T_4T_4 = T_5$ and $T_5T_5 = T_4$ , while $T_4T_5 = T_5T_4 = T_1$ . So $T_4$ and $T_5$ are inverses of EACH OTHER. They are the two ‘cyclic shifts’: $T_4$ sends $0 \to 1 \to 2 \to 0$ . The answer key issued for HW6 claimed every $T_i^2 = T_1$ , which is wrong for $T_4$ and $T_5$ . Also $T_2T_3 = T_5$ but $T_3T_2 = T_4$ , so even these simple maps do not commute. (For $\{0, 1\}$ only the identity and the swap are one-to-one, and each is its own inverse: Apostol 2.8 #1.) This little table is a first look at a “group”, the structure of invertible maps under composition that MATH 541 studies.

The positive reals are secretly the real line (Apostol 1.5 #29, from 2.1). Let $P$ be the positive real numbers with ‘sum’ $x \oplus y = xy$ and ‘scaling’ $c \odot x = x^c$ . Define $L: P \to \mathbb R$ by $L(x) = \ln x$ . Then

L(x \oplus y) = \ln(xy) = \ln x + \ln y = L(x) + L(y), \qquad L(c \odot x) = \ln(x^c) = c \ln x = c\,L(x).

So $L$ is linear. It is one-to-one, since $\ln$ is strictly increasing, and onto, since every real $r$ equals $\ln(e^r)$ . So $L$ is an isomorphism and $P \cong \mathbb R$ . The inverse is $\exp$ . The strange ‘vector space’ of 2.1 is the real line in disguise, and its laws of exponents are the vector-space axioms. It has dimension 1. The single number $2$ is a basis: it is not the zero vector of $P$ (that is $1$ ), so it is independent, and every $x > 0$ equals $c \odot 2 = 2^c$ with $c = \log_2 x$ , so it spans. That $c$ is the coordinate of $x$ .

Where it is used

Used inside maths· Inside maths: solving T(x) = b, and polar coordinates

Solving $T(x) = b$ is two questions. Does a solution exist? That is: is $b$ in the range, is $T$ onto? Is it the only one? That is: is $T$ one-to-one? When $\dim V = \dim W$ is finite, 2.3 showed these are the same question, so a square system has exactly one solution for every $b$ or fails both ways. Chapter 2.6 turns this into a procedure and chapter 2.7 into one number.

In multivariable calculus the same idea reaches into non-linear maps. The change to polar coordinates $F(r, \theta) = (r\cos\theta,\ r\sin\theta)$ is not linear. Near a point $(r, \theta)$ , though, it behaves like its derivative (the matrix of partial derivatives from multivariable calculus), which is the linear map

J(a, b) = (a\cos\theta - b\,r\sin\theta,\ a\sin\theta + b\,r\cos\theta).

Factor it as a composition: first $D(a, b) = (a,\ rb)$ (stretch the second coordinate by $r$ ), then rotate by $\theta$ . Write $\mathrm{Rot}_\theta$ for the rotation by $\theta$ . Then $J = \mathrm{Rot}_\theta D$ , since rotating the point $(a, rb)$ by $\theta$ gives exactly the formula for $J$ . For $r \ne 0$ both pieces are invertible, so by socks and shoes $J^{-1} = D^{-1}\,\mathrm{Rot}_{-\theta}$ : rotate back by $-\theta$ , then divide the second coordinate by $r$ . At $r = 0$ , $J(a, b) = (a\cos\theta,\ a\sin\theta)$ sends the whole vertical direction $(0, b)$ to $O$ , so $J$ is not one-to-one. Geometrically, every point $(0, \theta)$ with $r = 0$ lands on the origin, so the polar map cannot be undone there. (B: a theorem of calculus, the inverse function theorem, says that when the derivative is invertible at a point, the non-linear map itself can be undone near that point.)

Used in the real world· Graphics: undoing in reverse order

A 2D game engine places a sprite by rotating it $90^\circ$ and shifting it by $(5, 2)$ . The corner $(1, 0)$ goes to $(5, 3)$ (the step-through above). When you click the screen at $(5, 3)$ , the engine must find the point on the sprite. It applies the inverse: subtract $(5, 2)$ first, then rotate by $-90^\circ$ . Game engines with parent and child objects (a hand attached to an arm attached to a body) do this for every click. They store each object’s transform and apply the parents’ inverses in reverse order to turn a world point into a point in the object’s own coordinates (Unity’s InverseTransformPoint is this). An editor’s undo button works the same way: it keeps a stack of edits and undoes the newest first.

Used in the real world· Encryption: the Hill cipher (B)

The Hill cipher (1929) encrypts letters in pairs using a matrix mod 26. Number the letters $A = 0, \dots, Z = 25$ . Take the key $K = \begin{bmatrix}3&3\\2&5\end{bmatrix}$ and encrypt “HELP”. The pair HE is $(7, 4)$ , and $K(7, 4) = (3\cdot7 + 3\cdot4,\ 2\cdot7 + 5\cdot4) = (33, 34) \equiv (7, 8)$ mod 26, which is “HI”. The pair LP is $(11, 15)$ , and $K(11, 15) = (78, 97) \equiv (0, 19)$ , which is “AT”. The ciphertext is “HIAT”. To decrypt, apply the inverse map. The inverse key is $K^{-1} = \begin{bmatrix}15&17\\20&9\end{bmatrix}$ , and one checks $KK^{-1} = \begin{bmatrix}105&78\\130&79\end{bmatrix} \equiv \begin{bmatrix}1&0\\0&1\end{bmatrix}$ mod 26.

The cipher is usable only if encryption is one-to-one. The key $\begin{bmatrix}2&3\\4&5\end{bmatrix}$ is not: the pairs “NA” $= (13, 0)$ and “AA” $= (0, 0)$ both encrypt to “AA”, because $2\cdot13 = 26 \equiv 0$ and $4 \cdot 13 = 52 \equiv 0$ . Nobody could decrypt “AA”. The theorem ‘Left inverse exists exactly when one-to-one’ holds for any function, so it still applies. The linear-map theorems of this chapter do not apply directly, because the numbers mod 26 do not form a field (26 is not prime), so these pairs are not a vector space in the sense of 2.1. The correct criterion is that the determinant must share no factor with 26. For the working key it is $3\cdot5 - 3\cdot2 = 9$ , and for the broken key it is $2\cdot5 - 3\cdot4 = -2$ , which shares the factor 2 with 26.

Used in the real world· AI: why language-model software works with log-probabilities

A language model gives each next word a probability. For open-ended text these are often small, say around $0.01$ . The probability of a 1000-word passage is the product of 1000 such numbers, about $10^{-2000}$ . A 64-bit floating-point number cannot hold anything much below $10^{-308}$ (the smallest ones, down to about $10^{-324}$ , lose precision), so a computer stores this product as $0$ and loses all information. Software stores $\ln$ of the probability instead. By the isomorphism $L = \ln$ above, the product becomes a sum: $1000 \ln 0.01 = -4605.17$ , an ordinary number. Comparing two passages by log-probability picks the same winner, because $\ln$ is increasing. The product world $(P, \oplus)$ and the sum world $\mathbb R$ are the same vector space, and the computer works in the one that its number format can hold.

Traps

Watch out· Order: ST means T first, and inverses reverse it

$ST(x) = S(T(x))$ , so the map written on the RIGHT acts first. And $(ST)^{-1} = T^{-1}S^{-1}$ , not $S^{-1}T^{-1}$ . The two orders differ whenever the maps do not commute. Piazza on HW5 (2.8 #23): the identity $(ST)^{-1} = T^{-1}S^{-1}$ holds for any invertible $S$ and $T$ , but a proof that quietly uses $ST = TS$ is wrong, since you were not given that.

Watch out· Do not assume commuting

$(S + T)^2 = S^2 + ST + TS + T^2$ always. It equals $S^2 + 2ST + T^2$ only when $ST = TS$ . And $(S + T)^3$ is the sum of all eight orderings of three letters from $S$ and $T$ , not $S^3 + 3S^2T + 3ST^2 + T^3$ unless the maps commute (HW5, 2.8 #25).

Watch out· A one-sided inverse is not two-sided on all of V

The theorem ‘A left inverse is unique and two-sided’ needs the inverse defined only on the range $T(V)$ . The calculus example shows what goes wrong otherwise: $DT = I$ but $TD \ne I$ on the full space. Between spaces of the same finite dimension the trouble disappears (see the prove problem on $ST = I$ ).

Watch out· Isomorphic does not mean equal, and the isomorphism is a choice

$P_2$ (polynomials of degree at most 2) is isomorphic to $\mathbb R^3$ , but a polynomial is not a triple. The isomorphism depends on a basis. With the basis $(1, x, x^2)$ the polynomial $3 + x^2$ becomes $(3, 0, 1)$ , and with the basis $(x^2, x, 1)$ the same polynomial becomes $(1, 0, 3)$ . Both are isomorphisms. Statements that survive every isomorphism are about the vector space itself.

Watch out· One-to-one and onto are different questions

‘One-to-one’ says different inputs give different outputs. ‘Onto’ says every element of the codomain is an output. Neither implies the other: $T(x) = (x, 0)$ from $\mathbb R$ to $\mathbb R^2$ is one-to-one and not onto, while $T(x, y) = x$ from $\mathbb R^2$ to $\mathbb R$ is onto and not one-to-one.

Problems

checkApostol 2.8 #5 and #6

Which of $T(x, y) = (x, 0)$ and $T(x, y) = (x, x)$ are one-to-one? For each, find a right inverse.

Solution

Neither. $T(x, y) = (x, 0)$ has $T(0, 1) = T(0, 2)$ . $T(x, y) = (x, x)$ has $T(1, 0) = T(1, 1)$ . Right inverses: for the first, $R(x, 0) = (x, 0)$ , since $T(x, 0) = (x, 0)$ ; for the second, $R(x, x) = (x, 0)$ , since $T(x, 0) = (x, x)$ . (Many others work: any second coordinate may be chosen.)

checkApostol 2.8 #10

Is $T(x, y) = (x + 1, y + 1)$ linear? Is it one-to-one? What is its inverse?

Solution

Not linear, since $T(0, 0) = (1, 1) \ne (0, 0)$ . It is one-to-one: $(x+1, y+1) = (x'+1, y'+1)$ gives $x = x'$ and $y = y'$ . Inverse: $T^{-1}(u, v) = (u - 1, v - 1)$ .

check

Let $S(x, y) = (y, x)$ and $T(x, y) = (x, 0)$ . Compute $ST$ and $TS$ . Are they equal?

Solution

$ST(x, y) = S(x, 0) = (0, x)$ . $TS(x, y) = T(y, x) = (y, 0)$ . They are different: at $(1, 0)$ , $ST$ gives $(0, 1)$ and $TS$ gives $(0, 0)$ .

check

Which pairs are isomorphic? (a) $\mathbb R^2$ and $\mathbb R^3$ . (b) $P_3$ and $\mathbb R^{2 \times 2}$ . (c) The plane $\{(x, y, z) : x + y + z = 0\}$ and $\mathbb R^2$ .

Solution

(a) No, dimensions 2 and 3 differ. (b) Yes, both have dimension 4. (c) Yes. The plane has dimension 2 (problem in 2.2), the same as $\mathbb R^2$ . An explicit isomorphism sends $(1, 0)$ to $(-1, 1, 0)$ and $(0, 1)$ to $(-1, 0, 1)$ (use ‘Building a linear map from the images of a basis’).

useApostol 2.8 #19

Show $T(x, y, z) = (x,\ x + y,\ x + y + z)$ is one-to-one and find $T^{-1}$ .

Solution

If $T(x, y, z) = O$ , then $x = 0$ , then $y = 0$ from $x + y = 0$ , then $z = 0$ from $x + y + z = 0$ . So $N(T) = \{O\}$ and $T$ is one-to-one by ‘Equivalent forms’ (c ⇒ a). Given $(u, v, w)$ , solve $u = x$ , $v = x + y$ , $w = x + y + z$ to get $T^{-1}(u, v, w) = (u,\ v - u,\ w - v)$ . Check: $T(u, v - u, w - v) = (u, v, w)$ .

useApostol 2.8 #26

Let $S(x, y, z) = (z, y, x)$ and $T(x, y, z) = (x,\ x + y,\ x + y + z)$ . Find $ST$ , $TS$ and $(ST)^{-1}$ , and check $(ST)^{-1} = T^{-1}S^{-1}$ .

Solution

$ST(x, y, z) = S(x,\ x + y,\ x + y + z) = (x + y + z,\ x + y,\ x)$ . $TS(x, y, z) = T(z, y, x) = (z,\ z + y,\ z + y + x)$ . To invert $ST$ , let $u = x + y + z$ , $v = x + y$ , $w = x$ : then $x = w$ , $y = v - w$ , $z = u - v$ , so $(ST)^{-1}(u, v, w) = (w,\ v - w,\ u - v)$ . Check with $T^{-1}S^{-1}$ : $S^{-1} = S$ (swapping first and third entries twice does nothing), so $S^{-1}(u, v, w) = (w, v, u)$ . Then $T^{-1}(w, v, u) = (w,\ v - w,\ u - v)$ by the previous problem. They agree.

useApostol 2.9, example; new numbers

Let $T: \mathbb R^2 \to \mathbb R^2$ be the linear map with $T(1, 2) = (1, 0)$ and $T(3, 4) = (0, 1)$ . Find $T(x, y)$ . Is $T$ an isomorphism?

Solution

The vectors $(1,2)$ and $(3,4)$ are independent (neither is a multiple of the other), so they form a basis of $\mathbb R^2$ , and $T$ exists and is unique. Write $(x, y) = a(1, 2) + b(3, 4)$ : $a + 3b = x$ and $2a + 4b = y$ . Subtracting twice the first from the second gives $-2b = y - 2x$ , so $b = (2x - y)/2$ , and $a = x - 3b = (3y - 4x)/2$ . Then $T(x, y) = a(1, 0) + b(0, 1) = \big(\tfrac{3y - 4x}{2},\ \tfrac{2x - y}{2}\big)$ . Check: $T(1, 2) = \big(\tfrac{6 - 4}{2},\ \tfrac{2 - 2}{2}\big) = (1, 0)$ and $T(3, 4) = \big(\tfrac{12 - 12}{2},\ \tfrac{6 - 4}{2}\big) = (0, 1)$ . $T$ sends the basis $(1,2), (3,4)$ to the basis $(1,0), (0,1)$ , so it is an isomorphism by ‘Building a linear map from the images of a basis’.

useApostol 2.8 #31

On the space $V$ of all real polynomials, let $S$ shift down, $S(c_0 + c_1x + c_2x^2 + \cdots) = c_1 + c_2x + \cdots$ , and let $T(p) = x\,p$ (multiply by $x$ ). Compute $ST$ and $TS$ . Which of $S$ , $T$ is one-to-one? Which is onto?

Solution

$ST(p) = S(xp) = p$ , so $ST = I_V$ . $TS(p) = x\,S(p) = c_1x + c_2x^2 + \cdots = p - p(0)$ , so $TS \ne I_V$ (for $p = 1$ , $TS(1) = 0$ ). $T$ is one-to-one, because it has the left inverse $S$ . Its range is the polynomials with $p(0) = 0$ , so $T$ is not onto. $S$ is onto (since $ST = I$ , every $p$ equals $S(xp)$ ) and not one-to-one ( $S(1) = S(2) = 0$ ). So here $S$ is a left inverse of $T$ , defined on all of $V$ , that fails to be a right inverse. By the prove problem on $ST = I_V$ below, this can only happen in infinite dimensions.

proveApostol 2.8 #24; MATH 375 HW6

Let $S, T: V \to V$ be invertible (one-to-one and onto) and $ST = TS$ . Prove that $S^{-1}T^{-1} = T^{-1}S^{-1}$ .

Solution

By socks and shoes, $(ST)^{-1} = T^{-1}S^{-1}$ and $(TS)^{-1} = S^{-1}T^{-1}$ . Since $ST = TS$ are the same map, their inverses are the same map: $T^{-1}S^{-1} = (ST)^{-1} = (TS)^{-1} = S^{-1}T^{-1}$ .

prove

Let $V$ be finite-dimensional and $S, T: V \to V$ linear with $ST = I_V$ . Prove that $TS = I_V$ too, so $S = T^{-1}$ .

Solution

First, $T$ is one-to-one: if $T(x) = T(y)$ then $x = S(T(x)) = S(T(y)) = y$ . A one-to-one linear map from $V$ to $V$ is onto (a prove problem in 2.3: equal finite dimensions). So $T$ is invertible with $T^{-1}$ defined on all of $V$ . Now $S = S\,(T\,T^{-1}) = (S\,T)\,T^{-1} = I\,T^{-1} = T^{-1}$ , using associativity. Hence $TS = T\,T^{-1} = I_V$ . (The polynomial example above shows this fails in infinite dimensions.)

proveApostol 2.8 #21

Let $T: V \to V$ be one-to-one and onto. Prove $T^mT^n = T^{m+n}$ for all $m, n \ge 0$ , and $(T^n)^{-1} = (T^{-1})^n$ .

Solution

For the first, use induction on $m$ . For $m = 0$ : $T^0T^n = I\,T^n = T^n$ . If $T^{m-1}T^n = T^{m-1+n}$ , then $T^mT^n = (T\,T^{m-1})T^n = T\,(T^{m-1}T^n) = T\,T^{m+n-1} = T^{m+n}$ , by associativity and the definition of powers.

For the second, induct on $n$ . $n = 0$ is $I^{-1} = I$ . If $(T^{n-1})^{-1} = (T^{-1})^{n-1}$ , then $T^n = T\,T^{n-1}$ is a composition of invertible maps, so by ‘Inverse of a composition’ $(T^n)^{-1} = (T^{n-1})^{-1}T^{-1} = (T^{-1})^{n-1}\,T^{-1} = (T^{-1})^n$ . The last step is the first part applied to the map $T^{-1}$ . Note that $T^n$ is invertible because a composition of one-to-one onto maps is one-to-one and onto.

proveApostol 2.8 #25; MATH 375 HW5

Let $S, T: V \to V$ be linear. (a) Prove $(S + T)^2 = S^2 + ST + TS + T^2$ . (b) If $ST = TS$ , prove $(S + T)^2 = S^2 + 2ST + T^2$ . (c) Write down $(S + T)^3$ without assuming $ST = TS$ .

Solution

(a) $(S + T)^2 = (S + T)(S + T) = (S + T)S + (S + T)T$ by distributivity (right factor as $R(S + T) = RS + RT$ with $R = S + T$ , which is linear), $= S^2 + TS + ST + T^2$ by distributivity again (left form). Adding in the order $S^2 + ST + TS + T^2$ is the same by commutativity of addition. (b) If $TS = ST$ the middle terms are $ST + ST = 2ST$ . (c) $(S+T)^3 = SSS + SST + STS + STT + TSS + TST + TTS + TTT$ , the sum of all $2^3 = 8$ orderings. If $ST = TS$ they collapse to $S^3 + 3S^2T + 3ST^2 + T^3$ .

prove

Let $T: V \to W$ be linear and onto, with $V$ finite-dimensional. Prove that $T$ has a LINEAR right inverse $R: W \to V$ , meaning $TR = I_W$ .

Solution

$W = T(V)$ is finite-dimensional (rank–nullity), so it has a basis $w_1, \dots, w_m$ . Each $w_i$ is in the range, so pick $x_i \in V$ with $T(x_i) = w_i$ . By ‘Prescribed values on a basis’ there is a linear map $R: W \to V$ with $R(w_i) = x_i$ . Then $TR(w_i) = T(x_i) = w_i = I_W(w_i)$ for every basis vector. Two linear maps that agree on a basis are equal (2.3), so $TR = I_W$ .

stretchAxler 3B

Let $V$ and $W$ be finite-dimensional and $T: V \to W$ linear and one-to-one. Prove there is a linear map $S: W \to V$ , defined on ALL of $W$ , with $ST = I_V$ .

Solution

Take a basis $e_1, \dots, e_n$ of $V$ . By ‘One-to-one in finite dimension’ (a ⇒ b), $T(e_1), \dots, T(e_n)$ are independent in $W$ . By ‘Extending to a basis’ (2.2) extend them to a basis $T(e_1), \dots, T(e_n), w_{n+1}, \dots, w_m$ of $W$ . By ‘Prescribed values on a basis’ there is a linear $S: W \to V$ with $S(T(e_i)) = e_i$ for $i \le n$ and $S(w_j) = O$ for $j > n$ . Then $ST(e_i) = e_i = I_V(e_i)$ for every basis vector of $V$ , so $ST = I_V$ .

This is why linear algebra books define a left inverse on all of $W$ : the apparent loss in restricting to $T(V)$ costs nothing.

stretch

Prove that a finite-dimensional vector space is never isomorphic to a proper subspace of itself (a subspace $U \ne V$ ). Then show the space of all real polynomials IS isomorphic to a proper subspace of itself.

Solution

If $V \cong U$ with $U \subseteq V$ , then $\dim U = \dim V$ by ‘Same dimension exactly when isomorphic’. A subspace of a finite-dimensional $V$ with $\dim U = \dim V$ equals $V$ (a problem in 2.2). So $U$ is not proper.

For the polynomials, let $T(p) = x\,p$ . It is linear, and one-to-one because $xp = 0$ forces $p = 0$ (a nonzero polynomial times $x$ is nonzero). Its range is $U = \{q : q(0) = 0\}$ , a subspace not containing the constant $1$ , so $U$ is proper. As a map onto its range, $T: V \to U$ is one-to-one and onto, so $V \cong U$ . This is possible only because $V$ is infinite-dimensional.

Summary

The linear maps from $V$ to $W$ form a vector space $L(V, W)$ under pointwise addition and scaling.
Composition $ST$ means ‘do $T$ first’. It is linear when $S$ and $T$ are, associative, distributive over addition, and not commutative in general.
A map has a left inverse exactly when it is one-to-one. A left inverse is unique and is also a right inverse. We call a one-to-one map invertible and write $T^{-1}$ for its inverse. Right inverses always exist and are usually not unique.
For a linear map, one-to-one is the same as $N(T) = \{O\}$ , and then $T^{-1}$ is automatically linear. In finite dimension it is also the same as ‘a basis goes to a basis’.
Socks and shoes: $(ST)^{-1} = T^{-1}S^{-1}$ . Undo the last thing first.
An isomorphism is a one-to-one, onto linear map. Taking coordinates shows every $n$ -dimensional real vector space is isomorphic to $\mathbb R^n$ , and two finite-dimensional spaces are isomorphic exactly when their dimensions are equal.
You can prescribe a linear map freely on a basis. It is one-to-one when the chosen vectors are independent, onto when they span, and an isomorphism when they are a basis. This is how every isomorphism is built, and it is the starting point for matrices in 2.5.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

2.4 Inverses, one-to-one maps, isomorphism

The picture

The idea, in words

Definitions and theorems

Adding and scaling linear maps

Linear maps form a vector space

Chaining maps: composition

Composition

Composition of linear maps is linear

Composition is associative

Composition distributes over addition

Composition does not commute

Undoing a map: inverses

Inverse

Left inverse exists exactly when one-to-one

A left inverse is unique and two-sided

Invertible

Inverses of linear maps

Equivalent forms of one-to-one for linear maps

One-to-one in finite dimension

Inverse of a composition: socks and shoes

Inverse of a composition reverses the order

Isomorphism: the same space under different names

Isomorphism

Isomorphic is an equivalence relation

Every n-dimensional space is isomorphic to R^n

Same dimension exactly when isomorphic

Prescribed values on a basis

Prescribed values on a basis

Building a linear map from the images of a basis

Examples and non-examples

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere