Linear-Algebra-Done-Right · Unit 7 · Video 1 · Interactive Practice

Invertible Linear Maps: Injective and Surjective

IKey Formulas

Formula	Name	What it says
$T$ invertible $\iff \exists\, S \in \mathcal{L}(W, V)$ with $ST = I$ (on $V$ ) and $TS = I$ (on $W$ )	Invertible, inverse	Both equations are required, and $S$ itself must be linear
$S_1 = S_1 I = S_1(TS_2) = (S_1 T)S_2 = I S_2 = S_2$	The inverse is unique	Uses $TS_2 = I$ and $S_1 T = I$ , one equation from each inverse; the inverse is written $T^{-1}$
$T$ invertible $\iff$ $T$ injective and surjective	Characterization	Injective: $Tu = Tv \Rightarrow u = v$ ; surjective: $\text{range}\, T = W$
$T^{-1}(w)$ = the unique $v \in V$ with $Tv = w$	Building the inverse	Such $v$ exists because $T$ is surjective and is unique because $T$ is injective; this $T^{-1}$ is automatically linear

Key Insight: For a linear map from an infinite-dimensional space to itself, one property alone is not enough: $Tp = x^2 p$ on $\mathcal{P}(\mathbf{R})$ is injective but not surjective, and the backward shift on $\mathbf{F}^\infty$ is surjective but not injective.

IIVisualization 1 — One Equation Is Not Enough

$T(t) = (2t, t)$ from $\mathbf{R}$ to $\mathbf{R}^2$ has many linear $S$ with $ST = I$ , but none with $TS = I$ .

Product Candidate S

💡 All three choices of $S$ satisfy $ST = I$ , so a one-sided inverse need not be unique; the uniqueness proof needs $S_1 T = I$ and $TS_2 = I$ , one equation from each side.

IIIVisualization 2 — The Inverse Function Is Additive

For $T(x, y) = (x + y,\ x + 2y)$ , defining $S(w)$ as the unique $v$ with $Tv = w$ already gives a linear map.

IVVisualization 3 — Infinite Dimensions: One Property Alone Fails

For a linear map from an infinite-dimensional space to itself, injective alone or surjective alone does not make it invertible.

Map Target

a₀ 1 a₁ 0 a₂ 0 a₃ 0

VQuiz Questions

Problem 1 · Find the Inverse

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (2x,\ x + y)$ — find $T^{-1}(x, y)$ .

✅ Correct!

T(\tfrac12 x,\ y - \tfrac12 x) = (x,\ \tfrac12 x + y - \tfrac12 x) = (x, y)

, and

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

❌ Not quite. Undoing each entry separately ignores that the second entry of

T

mixes both inputs:

T(\tfrac12 x,\ y) = (x,\ \tfrac12 x + y) \neq (x, y)

❌ Check the sign.

T(\tfrac12 x,\ y + \tfrac12 x) = (x,\ x + y)

: the

x

added by

T

must be subtracted, not added again.

❌ Close.

T(\tfrac12 x,\ y - x) = (x,\ y - \tfrac12 x)

. The first entry of the preimage is

\tfrac12 x

, so that is what must be subtracted.

❌ Not quite. Solve

T(a, b) = (x, y)

for

a

and

b

Show solution

Find the unique $(a, b)$ with $T(a, b) = (x, y)$ :

2a = x,\qquad a + b = y \quad\Longrightarrow\quad a = \tfrac12 x,\qquad b = y - \tfrac12 x.

So $T^{-1}(x, y) = (\tfrac12 x,\ y - \tfrac12 x)$ . Check both equations of the definition:

T\big(T^{-1}(x, y)\big) = \big(x,\ \tfrac12 x + y - \tfrac12 x\big) = (x, y),

T^{-1}\big(T(x, y)\big) = T^{-1}(2x,\ x + y) = \big(x,\ x + y - x\big) = (x, y).

Both products are the identity on $\mathbf{R}^2$ , so $T$ is invertible with this inverse.

Problem 2 · Is One Equation Enough?

Given: $T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ , $T(x, y) = (x,\ x - y,\ y)$ , and $S \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $S(x, y, z) = (x, z)$ — which statement is true?

✅ Correct!

S

undoes

T

, but

T

cannot undo

S

(0, 1, 0) \notin \text{range}\, T

, so

T

is not surjective and has no inverse at all.

❌ Only one equation holds.

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

, so

ST = I

; but

TS(0, 1, 0) = T(0, 0) = (0, 0, 0) \neq (0, 1, 0)

❌ Compute both products.

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

, while

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

Show solution

$ST$ (on $\mathbf{R}^2$ ): $ST(x, y) = S(x,\ x - y,\ y) = (x, y)$ , so $ST = I$ .

$TS$ (on $\mathbf{R}^3$ ): $TS(x, y, z) = T(x, z) = (x,\ x - z,\ z)$ . This equals $(x, y, z)$ only when $y = x - z$ ; for example $TS(0, 1, 0) = (0, 0, 0)$ . So $TS \neq I$ and $S$ is not an inverse of $T$ .

No other $S$ works either: every output of $T$ has middle entry equal to the first entry minus the third, so $(0, 1, 0) \notin \text{range}\, T$ . Thus $T$ is not surjective, and by the theorem $T$ is not invertible.

The definition demands both equations; $ST = I$ alone only shows that $T$ is injective.

Problem 3 · Classify Two Infinite-Dimensional Maps

Given: differentiation $D \in \mathcal{L}(\mathcal{P}(\mathbf{R}))$ , $Dp = p'$ , and the forward shift $R \in \mathcal{L}(\mathbf{F}^\infty)$ , $R(x_1, x_2, x_3, \dots) = (0, x_1, x_2, \dots)$ — classify each map.

Differentiation D on 𝒫(ℝ)

Forward shift R on 𝔽^∞

✅ Correct!

D

is surjective but not injective,

R

is injective but not surjective, and neither is invertible.

❌ Check $D$ .

D(1) = 0 = D(0)

, so the constant

1

lies in

\text{null}\, D

. Is every polynomial the derivative of some polynomial?

❌ Check $R$ . If

Rx = 0

then

x_1 = x_2 = \cdots = 0

. Can an output of

R

start with

1

Show solution

$D$ is not injective: $D1 = 0 = D0$ although $1 \neq 0$ ; equivalently $\text{null}\, D$ contains the constants.

$D$ is surjective: for $q = c_0 + c_1 x + \cdots + c_m x^m$ take

p = c_0 x + \tfrac{c_1}{2} x^2 + \cdots + \tfrac{c_m}{m + 1} x^{m+1}, \qquad Dp = q.

$R$ is injective: $R(x_1, x_2, \dots) = (0, x_1, x_2, \dots)$ is the zero sequence only when every $x_k = 0$ , so $\text{null}\, R = \{0\}$ .

$R$ is not surjective: every output has first entry $0$ , so $(1, 0, 0, \dots) \notin \text{range}\, R$ .

Like the two maps in the video, each has exactly one of the two properties, so neither is invertible.

Problem 4 · A Left Inverse and a Right Inverse

Given: $T \in \mathcal{L}(V, W)$ and $S_1, S_2 \in \mathcal{L}(W, V)$ with $S_1 T = I$ (on $V$ ) and $T S_2 = I$ (on $W$ ); nothing else is assumed — what follows?

✅ Correct! The uniqueness chain uses exactly one equation from each map, so it applies verbatim:

S_1 = S_1(TS_2) = (S_1 T)S_2 = S_2

❌ Not here. Visualization 1 had many maps with

ST = I

only because no map satisfied

TS = I

. Run the chain

S_1 = S_1 I = S_1(TS_2)

❌ Not quite. The chain

S_1 = S_1(TS_2) = (S_1 T)S_2 = S_2

never uses

V = W

; it only needs the products to make sense.

❌ Look at the proof again. It uses only

TS_2 = I

S_1 T = I

and associativity, never that

T

is invertible.

Show solution

Repeat the uniqueness proof, using $TS_2 = I$ (on $W$ ), associativity, then $S_1 T = I$ (on $V$ ):

S_1 = S_1 I = S_1 (T S_2) = (S_1 T) S_2 = I S_2 = S_2.

Now $S_1$ satisfies both equations: $S_1 T = I$ by hypothesis, and $T S_1 = T S_2 = I$ . Since $S_1 \in \mathcal{L}(W, V)$ , it is an inverse of $T$ , so $T$ is invertible and $T^{-1} = S_1 = S_2$ .

Compare Visualization 1: there $T(t) = (2t, t)$ had several maps $S$ with $ST = I$ , which is possible only because no map satisfies $TS = I$ ( $T$ is not surjective).

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