Linear-Algebra-Done-Right · Unit 7 · Video 2 · Interactive Practice

Equal Dimensions: Injective or Surjective Means Invertible

IKey Formulas

Formula	Name	Hypotheses
$T$ injective $\iff$ $T$ surjective $\iff$ $T$ invertible	Equal dimensions	$T \in \mathcal{L}(V, W)$ , $\dim V = \dim W < \infty$ (includes $W = V$ )
$\dim \text{null}\, T = \dim V - \dim \text{range}\, T$	Fundamental theorem, rearranged	$V$ finite-dimensional
$ST = I \iff TS = I$	One-sided inverse is two-sided	$S \in \mathcal{L}(W, V)$ , $T \in \mathcal{L}(V, W)$ , $\dim V = \dim W < \infty$
$\text{null}\, T = \{0\} \Rightarrow$ every $q$ is some $Tp$	Existence from injectivity	$T$ maps a finite-dimensional space such as $\mathcal{P}_m(\mathbf{R})$ to itself

Key Insight: When $\dim V = \dim W < \infty$ , the fundamental theorem gives $\dim \text{null}\, T = 0 \iff \dim \text{range}\, T = \dim W$ , so a null-space check settles surjectivity, and with it the existence of solutions to $Tp = q$ .

IIOne Count Controls Both Conditions

Since $\dim \text{null}\, T + \dim \text{range}\, T = \dim V$ , equal dimensions make $\text{null}\, T = \{0\}$ and $\text{range}\, T = W$ the same condition.

dim V dim W

💡 Challenge: with $\dim V = \dim W$ , look for a map that is injective but not surjective; then make $\dim V < \dim W$ and find one at once.

IIIExistence Without Solving First

On $\mathcal{P}_2(\mathbf{R})$ , $Tp = ((x^2 + 5x + 7)p)''$ is injective, so every $q \in \mathcal{P}_2(\mathbf{R})$ equals $Tp$ for some $p$ .

c₂ (x²) 12 c₁ (x) 6 c₀ (1) 2 Rows solved 0

IVOne-Sided Inverses: Finite Versus Infinite

With $\dim V = \dim W < \infty$ , $ST = I$ forces $TS = I$ ; on $\mathbf{F}^\infty$ the shifts give $ST = I$ but $TS \neq I$ .

Space Product Stage x

VQuiz Questions

Problem 1 · The Null Space Decides

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ defined by $T(x, y, z) = (x - y,\ y - z,\ z - x)$ — decide which properties $T$ has.

✅ Correct!

(1, 1, 1) \in \text{null}\, T

, so

\dim \text{null}\, T = 1

and

\dim \text{range}\, T = 3 - 1 = 2 < 3

❌ Test the null space.

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

, so

\text{null}\, T \neq \{0\}

and

T

is not injective. With

W = V = \mathbf{R}^3

, it is then not surjective either.

❌ Impossible here.

T

maps the finite-dimensional

\mathbf{R}^3

to itself, so injective and surjective are the same condition:

T

has both or neither.

Show solution

Null space. $T(x, y, z) = 0$ means $x - y = 0$ , $y - z = 0$ , $z - x = 0$ , so $x = y = z$ . Hence $\text{null}\, T = \operatorname{span}\big((1, 1, 1)\big)$ , which has dimension $1$ , and $T$ is not injective.

Range. By the fundamental theorem,

\dim \text{range}\, T = \dim \mathbf{R}^3 - \dim \text{null}\, T = 3 - 1 = 2 < 3,

so $T$ is not surjective. This is what the equal-dimensions theorem predicts: on $\mathbf{R}^3 \to \mathbf{R}^3$ , failing injectivity forces failing surjectivity. Directly, the coordinates of every output add to $(x - y) + (y - z) + (z - x) = 0$ , so for example $(1, 0, 0) \notin \text{range}\, T$ .

Problem 2 · Count the Dimensions

Given: $T \in \mathcal{L}(V, W)$ is injective — decide in which case $T$ must also be surjective.

✅ Correct!

\dim \mathcal{P}_4(\mathbf{R}) = 5 = \dim \mathbf{R}^5

, both finite, so injective implies surjective.

❌ The dimensions differ. An injective

T

has

\dim \text{range}\, T = 3 < 4

; for example

(x, y, z) \mapsto (x, y, z, 0)

is injective and misses

(0, 0, 0, 1)

❌ The theorem needs finite dimension. On

\mathcal{P}(\mathbf{R})

p \mapsto xp

is injective, but the constant

1

is not

x

times any polynomial.

❌ Count again.

\mathcal{P}_4(\mathbf{R})

has basis

1, x, x^2, x^3, x^4

, so its dimension is

5

, not

4

. An injective

T

from

\mathbf{R}^4

has

\dim \text{range}\, T = 4 < 5

Show solution

The equal-dimensions theorem needs $V$ and $W$ finite-dimensional with $\dim V = \dim W$ . Check each case:

(a) $\dim V = 3 \neq 4 = \dim W$ : an injective $T$ has $\dim \text{range}\, T = 3 - 0 = 3 < 4$ , so it is never surjective.

(b) $\dim \mathcal{P}_4(\mathbf{R}) = 5 = \dim \mathbf{R}^5$ : the theorem applies, and $T$ is surjective (indeed invertible).

(d) $\dim \mathbf{R}^4 = 4 \neq 5 = \dim \mathcal{P}_4(\mathbf{R})$ , so as in (a) $\dim \text{range}\, T = 4 < 5$ .

Only (b) forces surjectivity.

Problem 3 · A One-Sided Inverse Between Different Dimensions

Given: $T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^3)$ , $T(x, y) = (x, y, x + y)$ , and $S \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $S(x, y, z) = (x, y)$ , so $ST = I$ on $\mathbf{R}^2$ — decide whether $TS = I$ on $\mathbf{R}^3$ .

✅ Correct!

\text{range}\, TS \subseteq \text{range}\, T

, which has dimension at most

2

, so

TS

can never be the identity on

\mathbf{R}^3

❌ The dimensions are not equal.

T

is injective, but "injective implies invertible" needs

\dim V = \dim W

, and here

2 \neq 3

T

is not surjective:

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

❌ Compute the composition.

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

: the third coordinate is rebuilt from

x

and

y

, and

z

is lost.

❌ No choice of $S$ helps. For any

S

TS(w) = T(S w) \in \text{range}\, T

, and

\dim \text{range}\, T \le \dim \mathbf{R}^2 = 2

, so

TS

is not surjective onto

\mathbf{R}^3

Show solution

First, $ST(x, y) = S(x, y, x + y) = (x, y)$ , so $ST = I$ on $\mathbf{R}^2$ . In the other order,

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

so $TS(0, 0, 1) = (0, 0, 0) \neq (0, 0, 1)$ and $TS \neq I$ .

No other $S$ can fix this. For every $w \in \mathbf{R}^3$ , $TS(w) = T(Sw) \in \text{range}\, T$ , and by the fundamental theorem $\dim \text{range}\, T = 2 - \dim \text{null}\, T \le 2 < 3$ . So $TS$ is never surjective, and in particular never $I$ . The theorem $ST = I \iff TS = I$ fails here only because $\dim \mathbf{R}^2 \neq \dim \mathbf{R}^3$ .

Problem 4 · Existence First, Then a Formula

Given: $T \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}))$ defined by $Tp = p + p'$ — justify that every $q \in \mathcal{P}_3(\mathbf{R})$ equals $p + p'$ for some $p \in \mathcal{P}_3(\mathbf{R})$ , then find $p$ for $q = x^3$ .

Why does such a $p$ exist?

For $q = x^3$ , $p =$

✅ Correct! Injectivity on the

4

-dimensional

\mathcal{P}_3(\mathbf{R})

guarantees

p

; matching coefficients from

x^3

down finds

p = x^3 - 3x^2 + 6x - 6

❌ Linearity alone is not enough. The zero map on

\mathcal{P}_3(\mathbf{R})

is linear and not surjective. What makes

T

surjective is

\text{null}\, T = \{0\}

on a finite-dimensional space mapped to itself.

❌ That does not transfer. Surjectivity of

p \mapsto p'

says nothing directly about

p \mapsto p + p'

; and on

\mathcal{P}_3(\mathbf{R})

differentiation is not even surjective (

x^3

is not a derivative of a polynomial of degree

\le 3

❌ Wrong space. "Injective implies surjective" needs finite dimension, and

\mathcal{P}(\mathbf{R})

is infinite-dimensional (

p \mapsto xp

is injective, not surjective). Restrict to

\mathcal{P}_3(\mathbf{R})

, which contains

q

❌ Check the signs. For this

p

p + p' = x^3 + 6x^2 + 12x + 12

: the

3x^2

from

p'

must be cancelled, so the

x^2

coefficient of

p

-3

❌ Keep going down. Each new term of

p

creates a lower-degree term in

p'

that must also be cancelled: this

p

gives

p + p' = x^3 - 6x

x^3 + 6

, not

x^3

❌ Not quite. Write

p = x^3 + ax^2 + bx + c

and match the coefficients of

p + p'

with those of

x^3

Show solution

Existence. $T$ is linear, and for $p \in \mathcal{P}_3(\mathbf{R})$ we have $\deg p' < \deg p$ when $p$ is a nonzero non-constant polynomial ( $p' = 0$ when $p$ is constant), so $T$ maps $\mathcal{P}_3(\mathbf{R})$ into itself. If $p \neq 0$ , the leading term of $p + p'$ is the leading term of $p$ , so $p + p' \neq 0$ . Hence $\text{null}\, T = \{0\}$ and $T$ is injective. Since $\dim \mathcal{P}_3(\mathbf{R}) = 4 < \infty$ and $T$ maps the space to itself, $T$ is surjective: every $q \in \mathcal{P}_3(\mathbf{R})$ is $p + p'$ for some $p$ .

The formula for $q = x^3$ . The leading coefficient of $p + p'$ equals that of $p$ , so write $p = x^3 + ax^2 + bx + c$ . Then $p' = 3x^2 + 2ax + b$ and

p + p' = x^3 + (a + 3)x^2 + (b + 2a)x + (c + b).

Matching with $x^3$ : $a = -3$ , then $b = -2a = 6$ , then $c = -b = -6$ . So $p = x^3 - 3x^2 + 6x - 6$ .

Check: $p' = 3x^2 - 6x + 6$ , and $p + p' = x^3 + (-3 + 3)x^2 + (6 - 6)x + (-6 + 6) = x^3$ .

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