Kernel and range of a linear transformation – Operations on Linear Transformations in 2D (Senior 4 Mathematics) | Ganzaa

Operations on Linear Transformations in 2D

Lessons

Senior 4 · Mathematics

Operations on Linear Transformations in 2D

5 lessons

Lesson 5 of 5

Kernel and range of a linear transformation

25 min

What a shadow forgets

On a sunny afternoon a tree throws a flat shadow on the ground. Every point of the tree lands somewhere on the ground, but many different points land on the same spot: all the points on one sun ray share a single shadow point. From the shadow alone you cannot tell how high a branch was. Two questions describe any such "flattening" exactly: what lands on the origin? and where can the images lie?

For a linear transformation of the plane, the answers are two subspaces, the kernel and the range. Together they tell you whether the transformation can be undone.

The shadow of a tree on a field in the late-winter sun, near South Acre in Norfolk, England. Sunlight projects the tree onto the ground along parallel rays; all the points on one ray fall on the same point of the shadow. Photo: Keith Evans, CC BY-SA 2.0, via Wikimedia Commons.

Definition

Kernel and range: for a linear transformation $f : \mathbb{R}^2 \to \mathbb{R}^2$ ,

\ker f = \{ \vec{v} \in \mathbb{R}^2 : f(\vec{v}) = \vec{0} \}, \qquad \operatorname{Im} f = \{ f(\vec{v}) : \vec{v} \in \mathbb{R}^2 \}.

The kernel (null space) is the set of vectors sent to $\vec{0}$ . The range (image) is the set of all outputs. If $f$ has matrix $A$ , then $\ker f$ is the solution set of the homogeneous system $AX = \vec{0}$ , and since $A\begin{pmatrix} x \\ y \end{pmatrix} = x(\text{column 1}) + y(\text{column 2})$ , $\operatorname{Im} f$ is the span of the columns of $A$ .

Example: the projection $(x, y) \mapsto (x, 0)$ has kernel the $y$ -axis and range the $x$ -axis.

Important

Theorem 1. $\ker f$ and $\operatorname{Im} f$ are subspaces of $\mathbb{R}^2$ .

Proof. Kernel: $f(\vec{0}) = \vec{0}$ , so $\vec{0} \in \ker f$ . If $f(\vec{u}) = f(\vec{v}) = \vec{0}$ , then $f(\vec{u} + \vec{v}) = \vec{0} + \vec{0} = \vec{0}$ and $f(k\vec{u}) = k\vec{0} = \vec{0}$ . Range: $\vec{0} = f(\vec{0}) \in \operatorname{Im} f$ ; $f(\vec{u}) + f(\vec{v}) = f(\vec{u} + \vec{v})$ and $kf(\vec{u}) = f(k\vec{u})$ are again outputs of $f$ . $\blacksquare$

Theorem 2 (dimension theorem in the plane). $\dim \ker f + \dim \operatorname{Im} f = 2$ . For a matrix $A$ there are exactly three cases:

Matrix $A$	$\ker f$	$\operatorname{Im} f$
$\det A \ne 0$	$\{ \vec{0} \}$ , dimension 0	$\mathbb{R}^2$ , dimension 2
$\det A = 0$ , $A \ne O$	a line through $O$ , dimension 1	a line through $O$ , dimension 1
$A = O$	$\mathbb{R}^2$ , dimension 2	$\{ \vec{0} \}$ , dimension 0

Why. If $\det A \ne 0$ , $AX = \vec{0}$ has only $X = A^{-1}\vec{0} = \vec{0}$ , and the two columns are independent, so they span $\mathbb{R}^2$ . If $\det A = 0$ but $A \ne O$ , the columns are dependent and not both zero, so they span a line; and $AX = \vec{0}$ has non-zero solutions (the columns are dependent) but not every $X$ is a solution (as $A \ne O$ ), so the kernel is a line.

Seeing the kernel and the range

The transformation $f(x, y) = \left(x - \frac{1}{2}y,\ 0\right)$ , with matrix $\begin{pmatrix} 1 & -\frac{1}{2} \\ 0 & 0 \end{pmatrix}$ , slides each point parallel to the direction $(1, 2)$ until it reaches the $x$ -axis, like a shadow cast by slanting sunlight. Every output is on the $x$ -axis, so that is the range. The points sent to the origin are those with $x - \frac{1}{2}y = 0$ : the kernel is the line $y = 2 x$ , the "sun ray" through $O$ .

Points on any line parallel to the kernel share one image: $f (4, 2) = f (5, 4) = (3, 0)$ , because $(5, 4) - (4, 2) = (1, 2)$ is in the kernel. The dimensions add up as Theorem 2 says: $1 + 1 = 2$ .

Worked example 1: kernel and range from a matrix

Find the kernel and the range of the transformation with matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ .

Solution. $\det A = 4 - 4 = 0$ and $A \ne O$ , so both are lines.

Kernel: $x + 2 y = 0$ and $2 x + 4 y = 0$ are the same equation; put $y = t$ , $x = - 2 t$ : $\ker f = \{ t(-2, 1) : t \in \mathbb{R} \}$ , the line $x + 2 y = 0$ .

Range: the columns $(1, 2)$ and $(2, 4)$ are parallel, so $\operatorname{Im} f = \operatorname{span}\{ (1, 2) \}$ , the line $y = 2 x$ .

Check: $A\begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 2 \\ -4 + 4 \end{pmatrix} = \vec{0}$ ✓, and $\dim \ker f + \dim \operatorname{Im} f = 1 + 1 = 2$ .

Worked example 2: is a vector in the range?

Let $B = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix}$ . Is $(2, - 4)$ in the range? Is $(1, 1)$ ?

Solution. $\det B = -6 + 6 = 0$ ; the columns $(3, - 6)$ and $(1, - 2)$ are parallel, so $\operatorname{Im} f$ is the line through $O$ in the direction $(1, - 2)$ : the line $y = - 2 x$ . $(2, - 4)$ satisfies $y = - 2 x$ : it is in the range (for example $B\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}$ ). $(1, 1)$ does not: no vector is sent to it, and the system $BX = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ has no solution.

Important

Theorem 3. For a linear transformation $f$ of $\mathbb{R}^2$ with matrix $A$ , these statements are equivalent:

$f$ is one-to-one (different vectors have different images);
$\ker f = \{ \vec{0} \}$ ;
$\operatorname{Im} f = \mathbb{R}^2$ (every vector is an image);
$\det A \ne 0$ , so $f$ has an inverse.

Proof of the first equivalence. $f(\vec{u}) = f(\vec{v}) \iff f(\vec{u} - \vec{v}) = \vec{0} \iff \vec{u} - \vec{v} \in \ker f$ . So equal images force $\vec{u} = \vec{v}$ exactly when the kernel contains only $\vec{0}$ . The rest follows from the table in Theorem 2. $\blacksquare$

Worked example 3: a parameter

For which $k$ is the kernel of $\begin{pmatrix} k & 2 \\ 3 & k - 1 \end{pmatrix}$ a line? Find the kernel in each case.

Solution. The kernel is a line exactly when $\det = 0$ (the matrix is never $O$ ): $k (k - 1) - 6 = k^{2} - k - 6 = (k - 3) (k + 2) = 0$ .

$k = 3$ : rows $(3, 2)$ and $(3, 2)$ ; kernel $3 x + 2 y = 0$ , that is $\operatorname{span}\{ (2, -3) \}$ .
$k = - 2$ : rows $(- 2, 2)$ and $(3, - 3)$ ; kernel $x - y = 0$ , that is $\operatorname{span}\{ (1, 1) \}$ .

For every other $k$ , $\ker f = \{ \vec{0} \}$ and the transformation is one-to-one.

Warning

Kernel and range are different lines. In worked example 1 the kernel is $x + 2 y = 0$ but the range is $y = 2 x$ . The kernel lives among the inputs (solve $AX = \vec{0}$ ); the range lives among the outputs (span the columns). Do not read the range off the rows.

Tip

Start every kernel-and-range question with $\det A$ . Non-zero: kernel $\{ \vec{0} \}$ , range $\mathbb{R}^2$ , done. Zero: one row gives the kernel's equation, one non-zero column gives the range's direction.

Professor Dave Explains: Image and Kernel (about 6 min). Watch for how the kernel is found by solving a homogeneous system and the image by looking at the columns.

✦

Practice

Work each one, then open the answers.

Find the kernel and the range of the transformation with matrix (a) $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ (b) $\begin{pmatrix} 2 & -1 \\ -4 & 2 \end{pmatrix}$ (c) $\begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix}$ .
For $B = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix}$ , find the kernel and check the dimension theorem.
Is $(4, 6)$ in the range of $\begin{pmatrix} 2 & -1 \\ -4 & 2 \end{pmatrix}$ ? Is $(- 3, 6)$ ?
The orthogonal projection onto the line $y = x$ has matrix $\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ . Find its kernel and range.
A linear transformation has $\ker f = \operatorname{span}\{ (1, -1) \}$ and $f (1, 0) = (2, 3)$ . Find its matrix and its range.
Prove that $\operatorname{Im} f$ is a subspace for every linear transformation $f$ .
Explain why no linear transformation of $\mathbb{R}^2$ can have kernel $\{ \vec{0} \}$ and a line as its range.
Show that a rotation about $O$ has kernel $\{ \vec{0} \}$ , using (a) its determinant (b) lengths.

Show answersHide answers

(a) kernel: the $y$ -axis; range: the $x$ -axis. (b) $\det = 4 - 4 = 0$ ; kernel $2 x - y = 0$ , $\operatorname{span}\{ (1, 2) \}$ ; range $\operatorname{span}\{ (2, -4) \}$ , the line $y = - 2 x$ . (c) $\det = -1 \ne 0$ : kernel $\{ \vec{0} \}$ , range $\mathbb{R}^2$ .
$3 x + y = 0$ : $\ker f = \operatorname{span}\{ (1, -3) \}$ ; range the line $y = - 2 x$ ; $1 + 1 = 2$ . ✓
The range is the line $y = - 2 x$ : $(4, 6)$ is not on it ( $6 \ne -8$ ); $(- 3, 6)$ is ( $6 = - 2 (- 3)$ ).
Kernel: $x + y = 0$ , the line $y = - x$ ; range: $\operatorname{span}\{ (1, 1) \}$ , the line $y = x$ .
$f(1, -1) = \vec{0}$ gives $f(\vec{e}_2) = f(\vec{e}_1) = (2, 3)$ : matrix $\begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}$ ; range $\operatorname{span}\{ (2, 3) \}$ .
$\vec{0} = f(\vec{0})$ is in it; $f(\vec{u}) + f(\vec{v}) = f(\vec{u} + \vec{v})$ and $kf(\vec{u}) = f(k\vec{u})$ are images too.
$\dim \ker f + \dim \operatorname{Im} f = 2$ , but $0 + 1 = 1 \ne 2$ .
(a) $\det R_\theta = 1 \ne 0$ . (b) $\|R_\theta\vec{v}\| = \|\vec{v}\|$ , so $R_\theta\vec{v} = \vec{0}$ forces $\|\vec{v}\| = 0$ , that is $\vec{v} = \vec{0}$ .

Quick check

Key points

$\ker f = \{ \vec{v} : f(\vec{v}) = \vec{0} \}$ (solve $AX = \vec{0}$ ); $\operatorname{Im} f$ = all outputs = span of the columns of $A$ .
Both are subspaces, and $\dim \ker f + \dim \operatorname{Im} f = 2$ .
$\det A \ne 0 \iff \ker f = \{ \vec{0} \} \iff \operatorname{Im} f = \mathbb{R}^2 \iff f$ is one-to-one and invertible.
$\det A = 0$ , $A \ne O$ : kernel and range are (usually different) lines through $O$ .

1What a shadow forgets

2Seeing the kernel and the range

What a shadow forgets

Seeing the kernel and the range