Linear-Algebra-Done-Right · Unit 5 · Video 3 · Interactive Practice

Null Space, Range, Injectivity and Surjectivity

IKey Formulas

Statement	Name	Role
$\text{null}\, T = \{v \in V : Tv = 0\}$	Null space (kernel)	A subspace of the domain $V$
$\text{range}\, T = \{Tv : v \in V\}$	Range	A subspace of the target space $W$
$T \text{ injective} \iff \text{null}\, T = \{0\}$	Injectivity test	Needs $T \in \mathcal{L}(V, W)$ : it uses linearity
$T \text{ surjective} \iff \text{range}\, T = W$	Surjective (onto)	Depends on which space is taken as the target $W$

Key Insight: For a linear map, $Tu = Tv \iff T(u - v) = 0 \iff u - v \in \text{null}\, T$ . Two inputs share an output exactly when their difference is in the null space, so the only obstruction to injectivity is a nonzero vector sent to $0$ .

IIShared Outputs and the Null Space

For a linear map, $Tu = Tv$ exactly when $u - v$ lies in $\text{null}\, T$ .

Map

💡 Challenge: for each map, place $v \neq u$ with $Tv = Tu$ , or name the null space that makes it impossible.

IIIThe Theorem Needs Linearity

$f(x) = x^2$ sends only $0$ to $0$ , yet it is not injective: it is not linear.

Function Output c 1

IVSurjectivity Depends on the Target

The same rule $p \mapsto p'$ is surjective onto $\mathcal{P}_4(\mathbf{R})$ but not onto $\mathcal{P}_5(\mathbf{R})$ .

Map

c₀ 1 c₁ 0 c₂ 3 c₃ 0 c₄ 0 c₅ 0

VQuiz Questions

Problem 1 · Compute a Null Space

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ defined by $T(x, y, z) = (x - y,\ y - 2z)$ . Find $\text{null}\, T$ .

✅ Correct!

x = y

and

y = 2z

give

(x, y, z) = z(2, 2, 1)

; for instance

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

. Since the null space contains a nonzero vector,

T

is not injective.

❌ Not quite. Two equations in three unknowns leave a free variable:

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

, so

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

❌ Check the second equation.

y - 2z = 0

means

y = 2z

, not

z = 2y

: indeed

T(1, 1, 2) = (0,\ 1 - 4) = (0, -3) \neq 0

❌ Only one condition used.

Tv = 0

needs both coordinates of

Tv

to vanish. The vector

(1, 1, 0)

has

x = y

, yet

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

❌ Not quite. Solve

x - y = 0

and

y - 2z = 0

together.

Show solution

A vector $(x, y, z)$ is in $\text{null}\, T$ exactly when $T(x, y, z) = (0, 0)$ , that is,

x - y = 0 \quad\text{and}\quad y - 2z = 0.

So $y = 2z$ and $x = y = 2z$ , with $z \in \mathbf{R}$ free:

\text{null}\, T = \{(2z, 2z, z) : z \in \mathbf{R}\} = \operatorname{span}\big((2, 2, 1)\big).

Check: $T(2, 2, 1) = (2 - 2,\ 2 - 2 \cdot 1) = (0, 0)$ . This null space is a line through the origin, a subspace of the domain $\mathbf{R}^3$ , and because it is not $\{0\}$ , $T$ is not injective: for example $T(2, 2, 1) = T(0, 0, 0)$ .

Problem 2 · Which One Is Injective?

Given: four functions. Find the one that is injective.

✅ Correct!

x - y = 0

and

x + y = 0

force

x = y = 0

, so

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

and the linear map

T

is injective.

❌ Not injective.

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

, so

(1, -1, 0) \in \text{null}\, T

; for instance

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

❌ That is the linearity trap. Only

(0, 0)

maps to

(0, 0)

, yet

f(1, 0) = f(-1, 0) = (1, 0)

. The null-space test does not apply:

f

is not linear, since

f\big(2(1, 0)\big) = (4, 0) \neq 2f(1, 0) = (2, 0)

❌ Not injective. Every constant polynomial has derivative

0

, so

\text{null}\, D

is the set of constants; for instance

D(x^3 + 7) = D(x^3) = 3x^2

❌ Not quite. For each linear map, look for a nonzero vector sent to

0

Show solution

For a linear map, injective $\iff$ null space $= \{0\}$ . Check each option:

(a) $T(1, -1, 0) = (1 - 1,\ 0) = (0, 0)$ : the null space contains a nonzero vector, so $T$ is not injective.
(b) $f$ is not linear, so the theorem does not apply. Directly: $f(1, 0) = (1, 0) = f(-1, 0)$ with $(1, 0) \neq (-1, 0)$ , so $f$ is not injective, although $f(x, y) = (0, 0)$ only when $(x, y) = (0, 0)$ .
(c) $D1 = 0$ , so the constant $1 \neq 0$ lies in $\text{null}\, D$ : not injective.
(d) $T(x, y) = (0, 0, 0)$ means $x - y = 0$ , $x + y = 0$ and $2y = 0$ , so $x = y = 0$ . Thus $\text{null}\, T = \{0\}$ and $T$ is injective.

Problem 3 · Two Inputs, One Output

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (x + y,\ 2x + 2y)$ , and $u = (3, 1)$ . Find $\text{null}\, T$ , then a vector $v \neq u$ with $Tv = Tu$ .

What is null T?

Which v ≠ u has Tv = Tu?

✅ Correct!

u - v = (3, 1) - (6, -2) = (-3, 3) = -3(1, -1) \in \text{null}\, T

, so

Tv = Tu = (4, 8)

❌ Check the null space.

T(x, y) = 0

means

x + y = 0

(the second coordinate is twice the first), so

\text{null}\, T = \{(x, -x)\}

❌ That is the range. Every output

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

is a multiple of

(1, 2)

, so

\operatorname{span}\big((1, 2)\big) = \text{range}\, T

, a subspace of the target. The null space lives in the domain:

T(1, 2) = (3, 6) \neq 0

❌ Check the difference.

Tv = Tu

exactly when

u - v \in \text{null}\, T

, i.e. when the coordinates of

u - v

add to

0

❌ That is the output, not an input.

Tu = (4, 8)

, but

T(4, 8) = (12, 24) \neq (4, 8)

❌ Sum, not difference. Here

u + v = (2, -2) \in \text{null}\, T

, which gives

Tv = -Tu = (-4, -8)

. The condition for

Tv = Tu

u - v \in \text{null}\, T

Show solution

Null space. $T(x, y) = (x + y,\ 2(x + y)) = (0, 0)$ exactly when $x + y = 0$ :

\text{null}\, T = \{(x, -x) : x \in \mathbf{R}\} = \operatorname{span}\big((1, -1)\big).

Shared output. $Tu = T(3, 1) = (4, 8)$ . Since $Tv = Tu \iff u - v \in \text{null}\, T$ , we need the coordinates of $u - v$ to add to $0$ , i.e. $v_1 + v_2 = 3 + 1 = 4$ :

$(4, 8)$ : sum $12$ , $T(4, 8) = (12, 24)$ .
$(3, -1)$ : sum $2$ , $T(3, -1) = (2, 4)$ .
$(6, -2)$ : sum $4$ , $T(6, -2) = (4, 8)$ ✓, and $u - v = (-3, 3) \in \text{null}\, T$ .
$(-1, -3)$ : sum $-4$ , $T(-1, -3) = (-4, -8)$ .

So $v = (6, -2)$ . Every vector of the set $u + \text{null}\, T = \{(3 + t,\ 1 - t) : t \in \mathbf{R}\}$ has the output $(4, 8)$ .

Problem 4 · Second Derivative, Choice of Target

Given: the rule $Tp = p''$ on $\mathcal{P}_3(\mathbf{R})$ , regarded as a linear map $T \colon \mathcal{P}_3(\mathbf{R}) \to W$ . Find the target $W$ that makes $T$ surjective, and $\text{null}\, T$ .

For which W is T surjective?

What is null T?

✅ Correct! Every

c_0 + c_1 x

\big(\tfrac{c_0}{2} x^2 + \tfrac{c_1}{6} x^3\big)''

, so

\text{range}\, T = \mathcal{P}_1(\mathbf{R})

; and

p'' = 0

exactly when

\deg p \le 1

, so

\text{null}\, T = \mathcal{P}_1(\mathbf{R})

❌ Check the degrees. If

\deg p \le 3

then

\deg p'' \le 1

, so the range is at most

\mathcal{P}_1(\mathbf{R})

❌ Not surjective onto $\mathcal{P}_2(\mathbf{R})$ .

x^2

would need

p = \tfrac{1}{12} x^4 \notin \mathcal{P}_3(\mathbf{R})

: since

\deg p'' \le 1

x^2 \notin \text{range}\, T

❌ Not a valid target.

T(x^3) = 6x \notin \mathcal{P}_0(\mathbf{R})

, so

T

does not map

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

into

\mathcal{P}_0(\mathbf{R})

at all.

❌ Check the null space. Which polynomials of degree at most

3

have second derivative

0

? Try

p = x

❌ Too small. Constants are in the null space, but so is

x

(x)'' = (1)' = 0

. The null space of the second derivative is

\mathcal{P}_1(\mathbf{R})

Show solution

Write $p = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ . Then

p'' = 2a_2 + 6a_3 x.

Target. Every output has degree at most $1$ , so $T$ maps into $\mathcal{P}_1(\mathbf{R})$ ; it does not map into $\mathcal{P}_0(\mathbf{R})$ , since $T(x^3) = 6x$ . Given any $q = c_0 + c_1 x \in \mathcal{P}_1(\mathbf{R})$ , take $a_2 = \tfrac{c_0}{2}$ and $a_3 = \tfrac{c_1}{6}$ :

\Big(\tfrac{c_0}{2} x^2 + \tfrac{c_1}{6} x^3\Big)'' = c_0 + c_1 x = q.

So $\text{range}\, T = \mathcal{P}_1(\mathbf{R})$ : $T$ is surjective onto $W = \mathcal{P}_1(\mathbf{R})$ , and not onto $\mathcal{P}_2(\mathbf{R})$ or $\mathcal{P}_3(\mathbf{R})$ , which contain $x^2 \notin \text{range}\, T$ .

Null space. $p'' = 0 \iff a_2 = a_3 = 0 \iff p = a_0 + a_1 x$ , so $\text{null}\, T = \mathcal{P}_1(\mathbf{R})$ . In particular $T$ is not injective: $T(x) = T(0) = 0$ .

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