MAT1041 · Week 11 · - The Kernel and Range of a Linear Transformation

Kernel and range of a linear transformation

You learn to find the kernel and the range of a linear transformation, to compute their bases and dimensions (nullity, rank), the dimension theorem, and to test whether a transformation is one-to-one, onto or an isomorphism.

kernelrange (image)rank of a transformationnullity of a transformationdimension theorem (rank-nullity)one-to-one (injective)onto (surjective)isomorphismisomorphic spaces

Last week we looked at what a linear transformation does. This week we ask two questions: What does it lose? and Where can it reach? The answer to the first is the kernel (everything that goes to zero), the answer to the second is the range (all outputs). For matrix transformations these are the null space and the column space of week 9; so the computation is completely familiar. The new language: one-to-one, onto, isomorphism, and the dimension theorem that ties them together with a single equation of numbers.

Golden rules:

The kernel lives in the domain, the range lives in the codomain.
For a matrix transformation: kernel = null space, range = column space. A single RREF gives both.
Dimension theorem: rank + nullity = the dimension of the domain.
One-to-one $\iff$ the kernel is $\{\mathbf{0}\}$ . Onto $\iff$ rank = the dimension of the codomain.

Definitions

Let $T : V \to W$ be linear.

Kernel: $\ker T = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\}$ .
Range: $\operatorname{range}T = \{T(\mathbf{v}) : \mathbf{v} \in V\}$ .

Both are subspaces. For the kernel: $T(\mathbf{0}) = \mathbf{0}$ ; if $T(\mathbf{u}) = T(\mathbf{v}) = \mathbf{0}$ then $T(\mathbf{u} + \mathbf{v}) = \mathbf{0}$ and $T(c\mathbf{u}) = \mathbf{0}$ . For the range: the sum of two images $T(\mathbf{u}) + T(\mathbf{v}) = T(\mathbf{u} + \mathbf{v})$ is again an image.

Their dimensions have special names: $\operatorname{nullity}T = \dim\ker T$ , $\operatorname{rank}T = \dim\operatorname{range}T$ .

Computing for a matrix transformation

Example. $T : \mathbb{R}^4 \to \mathbb{R}^3$ , $T(\mathbf{x}) = A\mathbf{x}$ , $A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 2 & 4 & 1 & 1 \\ 0 & 0 & 1 & 3 \end{bmatrix}$ .

$A \xrightarrow{R_2 \leftarrow R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 1 & 3 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 - R_2} \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

This is already the RREF (pivots in columns 1 and 3, zeros above the pivots).

Kernel. $x_1 + 2x_2 - x_4 = 0$ , $x_3 + 3x_4 = 0$ . With $x_2 = s$ , $x_4 = t$ :

$\mathbf{x} = s(-2, 1, 0, 0) + t(1, 0, -3, 1), \qquad \ker T \text{ basis: } \{(-2, 1, 0, 0),\ (1, 0, -3, 1)\}, \quad \operatorname{nullity}T = 2.$

Range. The originals of the pivot columns:

$\operatorname{range}T \text{ basis: } \{(1, 2, 0),\ (0, 1, 1)\}, \qquad \operatorname{rank}T = 2.$

The range is a plane in $\mathbb{R}^3$ . Its equation: when $[A \mid \mathbf{b}]$ is reduced with the same operations, $b_3 - b_2 + 2b_1$ is left on the right of the last row, so the range is $b_3 = b_2 - 2b_1$ . For example $(1, 3, 1)$ is an image of some vector but $(1, 0, 0)$ is not in the range.

Check: $2 + 2 = 4 = \dim\mathbb{R}^4$ .

The dimension theorem

If $V$ is finite-dimensional and $T : V \to W$ is linear, then

$\operatorname{rank}T + \operatorname{nullity}T = \dim V.$

Intuition: every dimension of the domain is either “crushed” in the kernel or “lives” in the range. For matrices this is the same as the rank-nullity theorem of week 9 ( $n$ = number of columns = $\dim V$ ).

On the exam, often only numbers are given (Q4, Q5): if one dimension is given, the other follows from the theorem.

One-to-one, onto, isomorphism

Concept	Definition	Test
One-to-one	$T(\mathbf{u}) = T(\mathbf{v}) \Rightarrow \mathbf{u} = \mathbf{v}$	$\ker T = \{\mathbf{0}\}$ , nullity 0
Onto	every $\mathbf{w} \in W$ is an image	$\operatorname{rank}T = \dim W$
Isomorphism	one-to-one and onto	both

Why one-to-one $\iff$ zero kernel? $T(\mathbf{u}) = T(\mathbf{v}) \iff T(\mathbf{u} - \mathbf{v}) = \mathbf{0} \iff \mathbf{u} - \mathbf{v} \in \ker T$ . If the kernel contains only $\mathbf{0}$ , then $\mathbf{u} = \mathbf{v}$ .

For a matrix transformation $T(\mathbf{x}) = A\mathbf{x}$ , $A$ being $m \times n$ :

	Condition
One-to-one	$\operatorname{rank}A = n$ (a pivot in every column, columns independent)
Onto	$\operatorname{rank}A = m$ (a pivot in every row, columns span $\mathbb{R}^m$ )
Isomorphism	$m = n$ and $\det A \neq 0$

In the example the rank is 2: $n = 4 \neq 2$ , not one-to-one; $m = 3 \neq 2$ , not onto.

TIP · if the dimensions are equal one test is enough. If $\dim V = \dim W = n$ , the dimension theorem says: nullity 0 $\iff$ rank $n$ . That is, one-to-one $\iff$ onto. For a square matrix a single determinant calculation answers both (Q7).

Dimension obstructions

$\dim V < \dim W$ : rank $\le \dim V < \dim W$ , cannot be onto ( $\mathbb{R}^3 \to \mathbb{R}^4$ ).
$\dim V > \dim W$ : nullity $\ge \dim V - \dim W > 0$ , cannot be one-to-one ( $\mathbb{R}^4 \to \mathbb{R}^3$ ).

Isomorphic spaces

If there is an isomorphism from $V$ to $W$ , then $V$ and $W$ are isomorphic: the same structure in different clothes. Theorem: two finite-dimensional spaces are isomorphic $\iff$ their dimensions are equal. This is why $P_3$ , $M_{2,2}$ and $\mathbb{R}^4$ are “the same” space; it is why converting polynomial and matrix questions to coefficient vectors is legitimate.

Kernel and range in abstract spaces

Method: write the condition in terms of the coefficients, solve the equation.

$D : P_3 \to P_2$ derivative: the kernel is the constants (nullity 1), rank $4 - 1 = 3$ , onto.
$T : M_{2,2} \to \mathbb{R}$ , $T(A) = \operatorname{tr}A$ : the kernel is the matrices with trace zero (dimension 3), rank 1, onto.
$T(A) = A - A^T$ : the kernel is the symmetric matrices (dimension 3), the range is the skew-symmetric matrices (dimension 1). Q8.

Common mistakes

Wrong	Right
Looking for the kernel in the codomain	The kernel is in $V$ : the $\mathbf{v}$ ‘s with $T(\mathbf{v}) = \mathbf{0}$
Writing the RREF columns as the range basis	The original pivot columns
Using $\dim W$ in the dimension theorem	$\dim V$ (the domain)
Looking at the nullity for “onto”	Compare the rank with $\dim W$
Looking for injectivity for $\mathbb{R}^4 \to \mathbb{R}^3$	Dimension obstruction: impossible
Writing the basis of $\ker T = \{\mathbf{0}\}$ as $\{\mathbf{0}\}$	The basis is the empty set, dimension 0
Saying one-to-one because $\mathbf{0}$ is in the kernel	The kernel must contain only $\mathbf{0}$

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

Find the kernel of $T : \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x, y) = (x - y,\ 2x - 2y)$ .

Show answer

\ker T = \operatorname{span}\{(1, 1)\}

$T(x, y) = (0, 0)$ : $x - y = 0$ and $2x - 2y = 0$ , both are the same equation. $y = t$ , $x = t$ .
The kernel is the line $y = x$ . Nullity 1.

Question 2medium

Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & -3 & 4 \end{bmatrix}$ . Find a basis for $\ker T$ .

Show answer

\{(-5, -2, 1)\}

$R_2 \leftarrow R_2 - 2R_1$ : $(0, 1, 2)$ . $R_1 \leftarrow R_1 + 2R_2$ : $(1, 0, 5)$ . $x_1 = -5t$ , $x_2 = -2t$ , $x_3 = t$ . Check: $A(-5, -2, 1)^T = (-5 + 4 + 1,\ -10 + 6 + 4) = (0, 0)$ .

Question 3medium

For the same $T$ , find the range of $T$ . Is $T$ onto?

Show answer

\operatorname{range}T = \mathbb{R}^2

; yes, onto

Rank = number of pivots = 2. The range is a 2-dimensional subspace of $\mathbb{R}^2$ , that is, $\mathbb{R}^2$ itself. Onto.
Dimension theorem: $\operatorname{rank} + \operatorname{nullity} = 2 + 1 = 3 = \dim\mathbb{R}^3$ .

Question 4medium

$T : \mathbb{R}^4 \to \mathbb{R}^3$ is linear and $\operatorname{rank}T = 2$ . Find $\operatorname{nullity}T$ . Is $T$ one-to-one?

Show answer

Nullity 2; not one-to-one

Dimension theorem: $\operatorname{nullity} = \dim(\text{domain}) - \operatorname{rank} = 4 - 2 = 2$ .
The kernel is larger than zero, so it is not one-to-one.

Question 5medium

$T : \mathbb{R}^5 \to \mathbb{R}^3$ is linear with $\operatorname{nullity}T = 2$ . Is $T$ onto?

Show answer

Yes

$\operatorname{rank} = 5 - 2 = 3 = \dim\mathbb{R}^3$ . The range is a 3-dimensional subspace of $\mathbb{R}^3$ , that is, all of it. Onto.

Question 6medium

Let $D : P_3 \to P_2$ , $D(p) = p'$ . Find $\ker D$ , $\operatorname{nullity}D$ and $\operatorname{rank}D$ .

Show answer

\ker D

= constant polynomials; nullity 1, rank 3

$p' = 0 \iff p$ is constant. The kernel has basis $\{1\}$ , nullity 1. $\dim P_3 = 4$ , by the dimension theorem: rank $= 4 - 1 = 3 = \dim P_2$ . So $D$ is onto: every polynomial of degree at most 2 is the derivative of some polynomial.

Question 7medium

Is $T : \mathbb{R}^3 \to \mathbb{R}^3$ , $T(x, y, z) = (x + y,\ y + z,\ x + z)$ an isomorphism?

Show answer

Yes

The standard matrix is $\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$ , $\det = 1(1 - 0) - 1(0 - 1) + 0 = 2 \neq 0$ .
Square and invertible: the kernel is $\{\mathbf{0}\}$ (one-to-one), the range is $\mathbb{R}^3$ (onto). An isomorphism.

Question 8hard

Let $T : M_{2,2} \to M_{2,2}$ , $T(A) = A - A^T$ . Find $\dim\ker T$ and $\dim\operatorname{range}T$ .

Show answer

\dim\ker T = 3

\dim\operatorname{range}T = 1

$T(A) = O \iff A = A^T$ : the kernel is the symmetric matrices, with basis $\left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right\}$ , dimension 3.
Dimension theorem: rank $= 4 - 3 = 1$ . Indeed $T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = (b - c)\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ .

Question 9hard

Let $T : P_2 \to \mathbb{R}$ , $T(p) = p(1)$ . Find a basis for $\ker T$ and determine whether $T$ is onto.

Show answer

Basis

\{x - 1, x^2 - 1\}

; onto

$p = a + bx + cx^2$ , $p(1) = a + b + c = 0$ . $a = -b - c$ : $p = b(x - 1) + c(x^2 - 1)$ . Nullity 2.
Rank $= 3 - 2 = 1 = \dim\mathbb{R}$ : onto. (The constant polynomial $p = r$ gives every $r$ .)

Question 10medium

True or false: There is a linear transformation from $\mathbb{R}^3$ onto $\mathbb{R}^4$ .

Show answer

False

$\operatorname{rank}T \le \dim(\text{domain}) = 3 < 4$ . The range is at most 3-dimensional, it can't fill $\mathbb{R}^4$ .
In general: if $\dim V < \dim W$ it can't be onto; if $\dim V > \dim W$ it can't be one-to-one.

Question 11hard

For which $k$ is $T(\mathbf{x}) = A\mathbf{x}$ one-to-one, where $A = \begin{bmatrix} 1 & 2 \\ 3 & k \\ 2 & 4 \end{bmatrix}$ ?

Show answer

k \neq 6

One-to-one $\iff$ the kernel is $\{\mathbf{0}\}$ $\iff$ the columns are independent. Two columns are dependent if and only if one is a multiple of the other. $(2, k, 4) = 2(1, 3, 2)$ only when $k = 6$ . For $k \neq 6$ : rank 2, one-to-one.

Question 12medium

Let $T(x, y) = (x,\ y,\ x + y)$ . Is $\mathbf{w} = (1, 2, 3)$ in the range of $T$ ?

Show answer

Yes,

\mathbf{w} = T(1, 2)

$T(x, y) = (1, 2, 3)$ : $x = 1$ , $y = 2$ , and the third component $x + y = 3$ holds.
For $(1, 2, 4)$ it would not hold: the range is the plane $z = x + y$ .

Verification: The kernel bases were found in node as null spaces from the exact RREF with BigInt fractions, and the range bases from the pivot columns; the condition for membership in the range was checked by exact solution for many w vectors, and rank/nullity and injectivity in the parametric question at many values of k; the M₂₂ and P₂ transformations were computed by representing them with a matrix on coefficient vectors (verify/MAT1041/11.mjs).

Review card

ker T = {v : T(v) = 0} ⊆ V (in the DOMAIN), a subspace. range T = {T(v)} ⊆ W (in the CODOMAIN), a subspace. If T(x) = Ax: ker T = null(A), range T = col(A). ker basis: solve Ax = 0. range basis: the ORIGINAL PIVOT COLUMNS. nullity T = dim ker, rank T = dim range. DIMENSION THEOREM: rank T + nullity T = dim V (the domain). One-to-one ⇔ ker T = {0} ⇔ nullity 0. Onto ⇔ range T = W ⇔ rank T = dim W. Isomorphism = one-to-one + onto. dim V = dim W ⇔ isomorphic. If dim V = dim W: one-to-one ⇔ onto (one is enough). Square A: det ≠ 0 ⇔ isomorphism. dim V < dim W → cannot be onto; dim V > dim W → cannot be one-to-one. w ∈ range? → solve the system T(v) = w. Derivative P_n → P_{n-1}: ker = constants, onto.