Kernel and range of a linear transformation – Transformation of Matrices (Senior 6 Mathematics) | Ganzaa

Transformation of Matrices

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Senior 6 · Mathematics

Transformation of Matrices

5 lessons

Lesson 3 of 5

Kernel and range of a linear transformation

35 min

What a transformation loses, and what it reaches

A shadow on the ground is a picture of a three-dimensional object squashed onto a plane. Two things can be asked of any such squashing. Which vectors are flattened to nothing at all? And which vectors can appear as a result? For a linear transformation the first set is the kernel and the second is the range. In S4 you found them for transformations of the plane by inspection; with row reduction you can now find them in any dimension.

Recall that $f : \mathbb{R}^n \to \mathbb{R}^m$ is linear when $f(\alpha\mathbf{u} + \beta\mathbf{v}) = \alpha f(\mathbf{u}) + \beta f(\mathbf{v})$ for all vectors $\mathbf{u}, \mathbf{v}$ and scalars $\alpha, \beta$ . Every linear $f$ is $f (X) = A X$ for one $m \times n$ matrix $A$ , the matrix of $f$ , whose columns are the images of the standard basis vectors $\mathbf{e}_1, \dots, \mathbf{e}_n$ .

Definition

Let $f : \mathbb{R}^n \to \mathbb{R}^m$ be linear with matrix $A$ .

The kernel (or null space) of $f$ is $\operatorname{Ker} f = \{\mathbf{v} \in \mathbb{R}^n : f(\mathbf{v}) = \mathbf{0}\}$ : the solutions of the homogeneous system $A X = 0$ . Its dimension is the nullity of $f$ .
The range (or image) of $f$ is $\operatorname{Im} f = \{f(\mathbf{v}) : \mathbf{v} \in \mathbb{R}^n\}$ : the span of the columns of $A$ . Its dimension is the rank of $f$ .

Example: for the projection $p (x, y, z) = (x, y, 0)$ , the kernel is the $z$ -axis and the range is the $x y$ -plane.

The kernel lives in the domain and is crushed to the zero vector; the range lives in the codomain.

Important

Theorem. $\operatorname{Ker} f$ is a subspace of $\mathbb{R}^n$ , $\operatorname{Im} f$ is a subspace of $\mathbb{R}^m$ , $\dim \operatorname{Im} f = \operatorname{rank} A$ , and

\dim \operatorname{Ker} f + \dim \operatorname{Im} f = n \quad \text{(rank–nullity theorem).}

Reasons. If $f(\mathbf{u}) = f(\mathbf{v}) = \mathbf{0}$ , then $f(\alpha\mathbf{u} + \beta\mathbf{v}) = \alpha\mathbf{0} + \beta\mathbf{0} = \mathbf{0}$ , so the kernel is closed under linear combinations, and it contains $\mathbf{0}$ . In an echelon form of $A$ , each of the $r = \operatorname{rank} A$ pivot columns gives one independent column of $A$ in the range, and each of the $n - r$ columns without a pivot gives one free parameter of the kernel. So $\dim \operatorname{Im} f = r$ , $\dim \operatorname{Ker} f = n - r$ , and the two add up to $n$ .

The method

To find	Do this	Result
$\operatorname{Ker} f$	Reduce $A$ and solve $A X = 0$ ; write the solution with its free parameters.	The vectors multiplying the parameters form a basis of the kernel.
$\operatorname{Im} f$	Use the same echelon form: note which columns hold pivots.	The original columns of $A$ in those positions form a basis of the range.
a check	Add the two dimensions.	They must add up to $n$ , the dimension of the domain.

Worked example 1: a transformation of $\mathbb{R}^3$

Let $f(x, y, z) = (x + y - z,\ 2x + 3y + z,\ 3x + 4y)$ . Find a basis of $\operatorname{Ker} f$ and of $\operatorname{Im} f$ , and an equation of $\operatorname{Im} f$ .

Solution. The matrix and its reduced form are

A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & 4 & 0 \end{pmatrix} \xrightarrow[R_3 \to R_3 - 3R_1]{R_2 \to R_2 - 2R_1} \begin{pmatrix} 1 & 1 & -1 \\ 0 & 1 & 3 \\ 0 & 1 & 3 \end{pmatrix} \xrightarrow[R_1 \to R_1 - R_2]{R_3 \to R_3 - R_2} \begin{pmatrix} 1 & 0 & -4 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{pmatrix} .

Kernel. $z = t$ is free; $x = 4 t$ , $y = - 3 t$ . So $\operatorname{Ker} f = \operatorname{span}\{(4, -3, 1)\}$ , a line through the origin. Check: $f(4, -3, 1) = (4 - 3 - 1,\ 8 - 9 + 1,\ 12 - 12) = (0, 0, 0)$ .

Range. The pivots are in columns 1 and 2, so the first two columns of $A$ form a basis: $\operatorname{Im} f = \operatorname{span}\{(1, 2, 3), (1, 3, 4)\}$ , a plane through the origin. A normal vector is $(1, 2, 3) \times (1, 3, 4) = (-1, -1, 1)$ , so $\operatorname{Im} f$ has equation $Z = X + Y$ .

Rank–nullity: $1 + 2 = 3$ . Meaning: $f$ is neither one-to-one (the whole line $\operatorname{Ker} f$ goes to $\mathbf{0}$ ) nor onto (the output $(0, 0, 1)$ is never reached, since $1 \ne 0 + 0$ ).

Important

One-to-one and onto. A linear $f : \mathbb{R}^n \to \mathbb{R}^m$ with matrix $A$ is

injective (one-to-one) if and only if $\operatorname{Ker} f = \{\mathbf{0}\}$ , that is $\operatorname{rank} A = n$ ;
surjective (onto) if and only if $\operatorname{Im} f = \mathbb{R}^m$ , that is $\operatorname{rank} A = m$ ;
bijective if both hold, which needs $m = n$ and $\det A \ne 0$ ; then $f^{- 1}$ has matrix $A^{- 1}$ .

Reason for the first: $f(\mathbf{u}) = f(\mathbf{v}) \iff f(\mathbf{u} - \mathbf{v}) = \mathbf{0} \iff \mathbf{u} - \mathbf{v} \in \operatorname{Ker} f$ . If the kernel holds only $\mathbf{0}$ , then $\mathbf{u} = \mathbf{v}$ .

Worked example 2: from $\mathbb{R}^4$ to $\mathbb{R}^3$

Let $g(x, y, z, t) = (x + 2y + t,\ 2x + 4y + z + 4t,\ x + 2y + z + 3t)$ . Find bases of $\operatorname{Ker} g$ and $\operatorname{Im} g$ , and decide whether $g$ is injective or surjective.

Solution. This is the matrix of practice question 3 in lesson 1, now read as a transformation:

\begin{pmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 4 \\ 1 & 2 & 1 & 3 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} .

Kernel. Pivots in columns 1 and 3, so $y = s$ and $t = u$ are free. Row 2: $z = - 2 u$ . Row 1: $x = - 2 s - u$ . So

(x, y, z, t) = s(-2, 1, 0, 0) + u(-1, 0, -2, 1), \qquad \operatorname{Ker} g = \operatorname{span}\{(-2, 1, 0, 0), (-1, 0, -2, 1)\} .

Range. Columns 1 and 3 of the original matrix: $\operatorname{Im} g = \operatorname{span}\{(1, 2, 1), (0, 1, 1)\}$ , the plane $X - Y + Z = 0$ .

Check: $2 + 2 = 4 = n$ . Since $\dim \operatorname{Ker} g = 2 \ne 0$ , $g$ is not injective; since $\operatorname{rank} = 2 \lt 3$ , it is not surjective.

Warning

Take the range basis from the ORIGINAL matrix. Row operations keep the pivot positions but change the columns themselves. In worked example 2 the reduced columns 1 and 3 are $(1, 0, 0)$ and $(0, 1, 0)$ , which span the plane $Z = 0$ , not the range $X - Y + Z = 0$ .

Do not confuse where they live. $\operatorname{Ker} g \subseteq \mathbb{R}^4$ (vectors with 4 components) while $\operatorname{Im} g \subseteq \mathbb{R}^3$ (3 components).

Tip

Rank–nullity gives impossibility results at once. A linear map $\mathbb{R}^4 \to \mathbb{R}^3$ has rank at most 3, so its kernel has dimension at least $4 - 3 = 1$ : it can never be one-to-one. A linear map $\mathbb{R}^2 \to \mathbb{R}^3$ has rank at most 2, so it can never be onto.

3Blue1Brown: Inverse matrices, column space and null space, Chapter 7 of Essence of Linear Algebra (about 12 min). Watch how a transformation that squashes space onto a plane has a column space (the range) and a null space (the kernel), and why it then has no inverse.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the kernel and the range of $f(x, y) = (x - 2y,\ -3x + 6y)$ , and draw them.
Let $f(x, y, z) = (x + y,\ y + z,\ x - z)$ . Find a basis of $\operatorname{Ker} f$ , a basis of $\operatorname{Im} f$ and an equation of $\operatorname{Im} f$ .
Is $h(x, y, z) = (x + 2y,\ y - z,\ x + z)$ bijective? Give a reason.
A linear map $f : \mathbb{R}^5 \to \mathbb{R}^3$ is onto. What is the dimension of its kernel?
Let $f(x, y, z, t) = (x - y + 2z,\ 2x - 2y + 4z + t)$ . Show that $f$ is onto $\mathbb{R}^2$ and find a basis of its kernel.
For which $k$ is $f(x, y, z) = (x + y + z,\ x + 2y + 3z,\ x + 3y + kz)$ not bijective? For that value, find $\operatorname{Ker} f$ .
Let $f : \mathbb{R}^n \to \mathbb{R}^m$ be linear. Prove that $\operatorname{Im} f$ is a subspace of $\mathbb{R}^m$ .

Show answersHide answers

$\operatorname{Ker} f = \operatorname{span}\{(2, 1)\}$ , the line $x = 2 y$ ; $\operatorname{Im} f = \operatorname{span}\{(1, -3)\}$ , the line $Y = - 3 X$ .
Reduced form $\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ : $\operatorname{Ker} f = \operatorname{span}\{(1, -1, 1)\}$ ; $\operatorname{Im} f = \operatorname{span}\{(1, 0, 1), (1, 1, 0)\}$ , the plane $X = Y + Z$ .
Yes: $\det \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 1 \end{pmatrix} = -1 \ne 0$ .
Onto means rank 3, so $\dim \operatorname{Ker} f = 5 - 3 = 2$ .
Reduced form $\begin{pmatrix} 1 & -1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ has rank $2 = m$ , so $f$ is onto; $\operatorname{Ker} f = \operatorname{span}\{(1, 1, 0, 0), (-2, 0, 1, 0)\}$ .
$\det = k - 5$ , so $k = 5$ ; then $\operatorname{Ker} f = \operatorname{span}\{(1, -2, 1)\}$ .
$\mathbf{0} = f(\mathbf{0}) \in \operatorname{Im} f$ . If $\mathbf{a} = f(\mathbf{u})$ and $\mathbf{b} = f(\mathbf{v})$ , then $\alpha\mathbf{a} + \beta\mathbf{b} = f(\alpha\mathbf{u} + \beta\mathbf{v}) \in \operatorname{Im} f$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagram, the worked examples, the practice items and the quiz questions. Every kernel vector was checked by substituting it, and every range equation against all the columns.
Video: Inverse matrices, column space and null space, Chapter 7 of Essence of Linear Algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1What a transformation loses, and what it reaches

2The method

What a transformation loses, and what it reaches

The method