4 Linear algebra 4.13 Linear maps 4.15 The rank-nullity theorem

4.14 Kernel and image

4.14.1 Definition of kernel and image

To every linear transformation we associate two important subspaces.

Definition 4.14.1.

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

1.

the kernel of $T$ , written $\ker T$ , is $\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}_{W}\}$
2.

the image of $T$ , written $\operatorname{im}T$ , is $\{T(\mathbf{v}):\mathbf{v}\in V\}$

In other words, the image is what we normally mean by the image of a function.

An important family of examples are the linear maps $T_{A}:\mathbb{F}^{n}\to\mathbb{F}^{m}$ defined by left-multiplication by an $m\times n$ matrix $A$ with entries from the field $\mathbb{F}$ . In that case the image $\operatorname{im}T_{A}$ is equal to the column space $C(A)$ by Proposition 3.2.1, and the kernel $\ker T_{A}$ is the nullspace $N(A)$ .

4.14.2 A property of all linear maps

Lemma 4.14.1.

Let $T:V\to W$ be a linear map. Then $T(\mathbf{0}_{V})=\mathbf{0}_{W}$ .

Proof.

	$\displaystyle T(\mathbf{0}_{V})$	$\displaystyle=T(\mathbf{0}_{V}+\mathbf{0}_{V})$
		$\displaystyle=T(\mathbf{0}_{V})+T(\mathbf{0}_{V})$

by the second part of the definition of linearity. Now add $-T(\mathbf{0}_{V})$ to both sides:

	$\displaystyle T(\mathbf{0}_{V})-T(\mathbf{0}_{V})$	$\displaystyle=T(\mathbf{0}_{V})+T(\mathbf{0}_{V})-T(\mathbf{0}_{V})$
	$\displaystyle\mathbf{0}_{W}$	$\displaystyle=T(\mathbf{0}_{V})\qed$

4.14.3 Kernels and images are subspaces

Lemma 4.14.2.

Let $T:V\to W$ be linear. Then $\ker T\leqslant V$ and $\operatorname{im}T\leqslant W$ .

Proof.

To show something is a subspace you must check the three conditions: it contains the zero vector, it is closed under addition, it is closed under scalar multiplication.

First, the kernel.

1.

To show that the kernel contains $\mathbf{0}_{V}$ , we must show that $T(\mathbf{0}_{V})=\mathbf{0}_{W}$ . That’s exactly Lemma 4.14.1.
2.

If $\mathbf{v},\mathbf{w}\in\ker T$ then $T(\mathbf{v}+\mathbf{w})=T(\mathbf{v})+T(\mathbf{w})=\mathbf{0}_{W}+\mathbf{0}% _{W}=\mathbf{0}_{W}$ , so $\mathbf{v}+\mathbf{w}\in\ker T$ .
3.

If $\mathbf{v}\in\ker T$ and $\lambda\in\mathbb{F}$ then $T(\lambda\mathbf{v})=\lambda T(\mathbf{v})$ by the second part of the definition of linearity, and this is $\lambda\mathbf{0}_{W}$ which equals $\mathbf{0}_{W}$ . Since $T(\lambda\mathbf{v})=\mathbf{0}_{W}$ , we have $\lambda\mathbf{v}\in\ker T$ .

Next, the image.

1.

We know from Lemma 4.14.1 that $T(\mathbf{0}_{V})=\mathbf{0}_{W}$ , so $\mathbf{0}_{W}\in\operatorname{im}T$ .
2.

Any two elements of $\operatorname{im}T$ have the form $T(\mathbf{u}),T(\mathbf{v})$ some $\mathbf{u},\mathbf{v}\in V$ . Then $T(\mathbf{u})+T(\mathbf{v})=T(\mathbf{u}+\mathbf{v})$ (second part of the linearity definition), which is an element if $\operatorname{im}T$ , so $\operatorname{im}T$ is closed under addition.
3.

If $T(\mathbf{u})\in\operatorname{im}T$ and $\lambda\in\mathbb{F}$ then $\lambda T(\mathbf{u})=T(\lambda\mathbf{u})$ by the second part of the definition of linearity, and this is an element of $\operatorname{im}T$ as it is $T$ applied to something, so $\operatorname{im}T$ is closed under scalar multiplication.

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Example 4.14.1.

Let $A=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ so that we have a linear map $T_{A}:\mathbb{R}^{2}\to\mathbb{R}^{2}$ given by $T_{A}(\mathbf{x})=A\mathbf{x}$ . We will find $\operatorname{im}T_{A}$ and $\ker T_{A}$ .

	$\displaystyle\operatorname{im}T_{A}$	$\displaystyle=\left\{T_{A}\begin{pmatrix}x\\ y\end{pmatrix}:x,y\in\mathbb{R}\right\}$
		$\displaystyle=\left\{\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}:x,y\in\mathbb{R}\right\}$
		$\displaystyle=\left\{\begin{pmatrix}y\\ 0\end{pmatrix}:x,y\in\mathbb{R}\right\}$

Another way to write this is that $\operatorname{im}T_{A}=\operatorname{span}\begin{pmatrix}1\\ 0\end{pmatrix}$ , and so $\dim\operatorname{im}T_{A}=1$ .

Now we’ll do the kernel.

	$\displaystyle\ker T_{A}$	$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^{2}:T_{A}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\right\}$
		$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^{2}:\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\right\}$
		$\displaystyle=\left\{\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^{2}:\begin{pmatrix}y\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\right\}$
		$\displaystyle=\left\{\begin{pmatrix}x\\ 0\end{pmatrix}:x\in\mathbb{R}\right\}$

Again we could write this as $\ker T_{A}=\operatorname{span}\begin{pmatrix}1\\ 0\end{pmatrix}$ . The kernel and image are equal in this case.

Example 4.14.2.

Let $D:\mathbb{R}_{\leqslant n}[x]\to\mathbb{R}_{\leqslant n}[x]$ be $D(f)=\frac{df}{dx}$ . We will describe $\ker D$ and $\operatorname{im}D$ .

A polynomial has derivative zero if and only if it is constant, so $\ker D$ is the set of all constant polynomials. This is spanned by any (nonzero) constant polynomial, so it has dimension one.

Next consider $\operatorname{im}D$ . Let $S\leqslant\mathbb{R}_{\leqslant n}[x]$ be the subspace spanned by $1,x,\ldots,x^{n-1}$ , that is, the subspace consisting of all polynomials of degree at most $n-1$ . Certainly $\operatorname{im}D\leqslant S$ , since when you differentiate a polynomial of degree at most n you get a polynomial of degree at most $n-1$ . But if $s(x)\in S$ then $s(x)$ has an indefinite integral $t(x)$ in $\mathbb{R}_{\leqslant n}[x]$ and $D(t)=s$ , so every $s\in S$ is in $\operatorname{im}D$ , so $\operatorname{im}D=S$ .

A useful property of the kernel of a linear map is that it tells you whether or not the map is injective.

Proposition 4.14.3.

Let $T$ be a linear map $T:U\to V$ .

1.

$T$ is injective if and only if $\ker T=\{\mathbf{0}_{U}\}$ .
2.

$T$ is surjective if and only if $\dim\operatorname{im}T=\dim V$ .

Proof.

1.

Suppose $T$ is injective. By Lemma 4.14.1, $T(\mathbf{0}_{U})=\mathbf{0}_{V}$ , so $\mathbf{0}_{U}\in\ker T$ . Now if $\mathbf{u}\in\ker T$ then $T(\mathbf{u})=\mathbf{0}_{V}$ so $T(\mathbf{u})=T(\mathbf{0}_{U})$ , so by injectivity $\mathbf{u}=\mathbf{0}_{U}$ . It follows $\ker T=\{\mathbf{0}_{U}\}$ .

Conversely suppose $\ker T=\{\mathbf{0}_{U}\}$ and that $T(\mathbf{x})=T(\mathbf{y})$ . By linearity $T(\mathbf{x}-\mathbf{y})=\mathbf{0}_{V}$ , so $\mathbf{x}-\mathbf{y}\in\ker T$ , so $\mathbf{x}-\mathbf{y}=\mathbf{0}_{U}$ , so $\mathbf{x}=\mathbf{y}$ and $T$ is injective.
2.

If $T$ is surjective then $\operatorname{im}T=V$ so certainly $\dim\operatorname{im}T=\dim V$ .

Conversely, suppose $\dim\operatorname{im}T=\dim V$ . The image $\operatorname{im}T$ is a subspace of $V$ by Lemma 4.14.2. It has the same dimension as $V$ so it equals $V$ by Proposition 4.12.3. We have shown $\operatorname{im}T=V$ , so $T$ is surjective.

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