
# 9.8: The Kernel and Image of a Linear Map


Learning Objectives
1. Describe the kernel and image of a linear transformation.
2. Use the kernel and image to determine if a linear transformation is one to one or onto.

Here we consider the case where the linear map is not necessarily an isomorphism. First here is a definition of what is meant by the image and kernel of a linear transformation.

Definition Kernel and Image

Let $$V$$ and $$W$$ be vector spaces and let $$T:V\rightarrow W$$ be a linear transformation. Then the image of $$T$$ denoted as $$\mathrm{im}\left( T\right)$$ is defined to be the set $\left\{ T(\vec{v}):\vec{v}\in V\right\}$ In words, it consists of all vectors in $$W$$ which equal $$T(\vec{v})$$ for some $$\vec{v}\in V$$. The kernel, $$\ker \left( T\right)$$, consists of all $$\vec{v}\in V$$ such that $$T(\vec{v})=\vec{0}$$. That is, $\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}$

Then in fact, both $$\mathrm{im}\left( T\right)$$ and $$\ker \left( T\right)$$ are subspaces of $$W$$ and $$V$$ respectively.

Proposition $$\PageIndex{1}$$: Kernel and Image as Subspaces

Let $$V,W$$ be vector spaces and let $$T:V\rightarrow W$$ be a linear transformation. Then $$\ker \left( T\right) \subseteq V$$ and $$\mathrm{im}\left( T\right) \subseteq W$$. In fact, they are both subspaces.

Proof

First consider $$\ker \left( T\right) .$$ It is necessary to show that if $$\vec{v}_{1},\vec{v}_{2}$$ are vectors in $$\ker \left( T\right)$$ and if $$a,b$$ are scalars, then $$a\vec{v}_{1}+b\vec{v}_{2}$$ is also in $$\ker \left( T\right) .$$ But

$T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}$

Thus $$\ker \left( T\right)$$ is a subspace of $$V$$.

Next suppose $$T(\vec{v}_{1}),T(\vec{v}_{2})$$ are two vectors in $$\mathrm{im}\left( T\right) .$$ Then if $$a,b$$ are scalars,

$aT(\vec{v}_{2})+bT(\vec{v}_{2})=T\left( a\vec{v}_{1}+b\vec{v}_{2}\right)$

and this last vector is in $$\mathrm{im}\left( T\right)$$ by definition.

Consider the following example.

Example $$\PageIndex{1}$$: Kernel and Image of a Transformation

Let $$T:\mathbb{P}_1\to\mathbb{R}$$ be the linear transformation defined by $T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.$ Find the kernel and image of $$T$$.

Solution

We will first find the kernel of $$T$$. It consists of all polynomials in $$\mathbb{P}_1$$ that have $$1$$ for a root. \begin{aligned} \mathrm{ker}(T) & = & \{ p(x)\in \mathbb{P}_1 ~|~ p(1)=0\} \\ & = & \{ ax+b ~|~ a,b\in\mathbb{R} \mbox{ and }a+b=0\} \\ & = & \{ ax-a ~|~ a\in\mathbb{R} \}\end{aligned} Therefore a basis for $$\mathrm{ker}(T)$$ is $\left\{ x-1 \right\}$ Notice that this is a subspace of $$\mathbb{P}_1$$.

Now consider the image. It consists of all numbers which can be obtained by evaluating all polynomials in $$\mathbb{P}_1$$ at $$1$$. \begin{aligned} \mathrm{im}(T) & = & \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = & \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = & \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = & \mathbb{R}\end{aligned} Therefore a basis for $$\mathrm{im}(T)$$ is $\left\{ 1 \right\}$ Notice that this is a subspace of $$\mathbb{R}$$, and in fact is the space $$\mathbb{R}$$ itself.

Example $$\PageIndex{1}$$: Kernel and Image of a Linear Transformation

Let $$T: \mathbb{M}_{22} \mapsto \mathbb{R}^2$$ be defined by $T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]$ Then $$T$$ is a linear transformation. Find a basis for $$\mathrm{ker} (T)$$ and $$\mathrm{im}(T)$$.

Solution

You can verify that $$T$$ represents a linear transformation.

Now we want to find a way to describe all matrices $$A$$ such that $$T(A) = \vec{0}$$, that is the matrices in $$\mathrm{ker}(T)$$. Suppose $$A = \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ]$$ is such a matrix. Then $T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]$ The values of $$a, b, c, d$$ that make this true are given by solutions to the system \begin{aligned} a - b &=& 0 \\ c + d &=& 0 \end{aligned} The solution is $$a = s, b = s, c = t, d = -t$$ where $$s, t$$ are scalars. We can describe $$\mathrm{ker}(T)$$ as follows. $\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}$ It is clear that this set is linearly independent and therefore forms a basis for $$\mathrm{ker}(T)$$.

We now wish to find a basis for $$\mathrm{im}(T)$$. We can write the image of $$T$$ as $\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}$ Notice that this can be written as $\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}$

However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of $$\mathrm{im}(T)$$. $\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}$

Notice that these vectors have the same span as the set above but are now linearly independent.

A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. A special case was done earlier in the context of matrices. Recall that for an $$m\times n$$ matrix $$% A,$$ it was the case that the dimension of the kernel of $$A$$ added to the rank of $$A$$ equals $$n$$.

Theorem $$\PageIndex{1}$$: Dimension of Kernel + Image

Let $$T:V\rightarrow W$$ be a linear transformation where $$V,W$$ are vector spaces. Suppose the dimension of $$V$$ is $$n$$. Then $$n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)$$.

Proof

From Proposition [prop:kernelimagevectorspaces], $$\mathrm{im}\left( T\right)$$ is a subspace of $$W.$$ By Theorem [thm:basisvectorspace], there exists a basis for $$\mathrm{im}\left( T\right) ,\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{r})\right\} .$$ Similarly, there is a basis for $$\ker \left( T\right) ,\left\{ \vec{u} _{1},\cdots ,\vec{u}_{s}\right\}$$. Then if $$\vec{v}\in V,$$ there exist scalars $$c_{i}$$ such that $T(\vec{v})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})$ Hence $$T\left( \vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}\right) =0.$$ It follows that $$\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}$$ is in $$\ker \left( T\right)$$. Hence there are scalars $$a_{i}$$ such that $\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}=\sum_{j=1}^{s}a_{j}\vec{u}_{j}$ Hence $$\vec{v}=\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u} _{j}.$$ Since $$\vec{v}$$ is arbitrary, it follows that $V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}$ If the vectors $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}$$ are linearly independent, then it will follow that this set is a basis. Suppose then that $\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u}_{j}=0$ Apply $$T$$ to both sides to obtain $\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u} _{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}$ Since $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{r})\right\}$$ is linearly independent, it follows that each $$c_{i}=0.$$ Hence $$\sum_{j=1}^{s}a_{j}\vec{u }_{j}=0$$ and so, since the $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s}\right\}$$ are linearly independent, it follows that each $$a_{j}=0$$ also. It follows that $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots ,\vec{v} _{r}\right\}$$ is a basis for $$V$$ and so $n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)$

Consider the following definition.

Definition $$\PageIndex{1}$$: Rank of Linear Transformation

Let $$T:V\rightarrow W$$ be a linear transformation and suppose $$V,W$$ are finite dimensional vector spaces. Then the rank of $$T$$ denoted as $$\mathrm{rank}\left( T\right)$$ is defined as the dimension of $$\mathrm{im}\left( T\right) .$$ The nullity of $$T$$ is the dimension of $$\ker \left( T\right) .$$ Thus the above theorem says that $$\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .$$

Recall the following important result.

Theorem $$\PageIndex{1}$$: Subspace of Same Dimension

Let $$V$$ be a vector space of dimension $$n$$ and let $$W$$ be a subspace. Then $$W=V$$ if and only if the dimension of $$W$$ is also $$n$$.

From this theorem follows the next corollary.

Corollary $$\PageIndex{1}$$: One to One and Onto Characterization

Let $$T:V\rightarrow W$$ be a linear map where the dimension of $$V$$ is $$n$$ and the dimension of $$W$$ is $$m$$. Then $$T$$ is one to one if and only if $$\ker \left( T\right) =\left\{ \vec{0}\right\}$$ and $$T$$ is onto if and only if $$\mathrm{rank}\left( T\right) =m$$.

Proof

The statement $$\ker \left( T \right) =\left\{ \vec{0}\right\}$$ is equivalent to saying if $$T \left( \vec{v} \right)=\vec{0},$$ it follows that $$\vec{v}=\vec{0}$$ . Thus by Lemma [lem:onetooneabstract] $$T$$ is one to one. If $$T$$ is onto, then $$\mathrm{im}\left( T\right) =W$$ and so $$\mathrm{rank}\left( T\right)$$ which is defined as the dimension of $$\mathrm{im}\left( T\right)$$ is $$m$$. If $$\mathrm{ rank}\left( T\right) =m,$$ then by Theorem [thm:subspacevectorspace], since $$\mathrm{im} \left( T\right)$$ is a subspace of $$W,$$ it follows that $$\mathrm{im}\left( T\right) =W$$.

Example $$\PageIndex{1}$$: One to One Transformation

Let $$S:\mathbb{P}_2\to\mathbb{M}_{22}$$ be a linear transformation defined by $S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.$ Prove that $$S$$ is one to one but not onto.

Solution

You may recall this example from earlier in Example [exa:onetoonegeneral]. Here we will determine that $$S$$ is one to one, but not onto, using the method provided in Corollary [cor:oneoneontochar].

By definition, $\ker(S)=\{ax^2+bx+c\in \mathbb{P}_2 ~|~ a+b=0, a+c=0, b-c=0, b+c=0\}.$

Suppose $$p(x)=ax^2+bx+c\in\ker(S)$$. This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in :

$\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].$

Since the unique solution is $$a=b=c=0$$, $$\ker(S)=\{\vec{0}\}$$, and thus $$S$$ is one-to-one by Corollary [cor:oneoneontochar].

Similarly, by Corollary [cor:oneoneontochar], if $$S$$ is onto it will have $$\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4$$. The image of $$S$$ is given by

$\mathrm{im}(S) = \left\{ \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \right\} = \mathrm{span} \left\{ \left [\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ], \left [\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right ] \right\}$

These matrices are linearly independent which means this set forms a basis for $$\mathrm{im}(S)$$. Therefore the dimension of $$\mathrm{im}(S)$$, also called $$\mathrm{rank}(S)$$, is equal to $$3$$. It follows that $$S$$ is not onto.