9.8: The Kernel and Image of a Linear Map

Outcomes

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 $\PageIndex{1}$: 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

$$\left\{ T(\vec{v}):\vec{v}\in V\right\}\nonumber$$ 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, $$\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\nonumber$$

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

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

$$T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.\nonumber$$ 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 $$\left\{ x-1 \right\}\nonumber$$ 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 $$\left\{ 1 \right\}\nonumber$$ Notice that this is a subspace of $\mathbb{R}$, and in fact is the space $\mathbb{R}$ itself.

Example $\PageIndex{2}$: Kernel and Image of a Linear Transformation

Let $T: \mathbb{M}_{22} \mapsto \mathbb{R}^2$ be defined by

$$T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]\nonumber$$ 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

$$T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$$ 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. $$\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber$$ 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

$$\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}\nonumber$$ Notice that this can be written as $$\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber$$

However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of $\mathrm{im}(T)$.

$$\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber$$

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 $\PageIndex{1}$, $\mathrm{im}\left( T\right)$ is a subspace of $W.$ By Theorem 9.4.8, 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 $$T(\vec{v})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})\nonumber$$ 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 $$\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}=\sum_{j=1}^{s}a_{j}\vec{u}_{j}\nonumber$$ 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 $$V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}\nonumber$$ 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 $$\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u}_{j}=0\nonumber$$ Apply $T$ to both sides to obtain $$\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u} _{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}\nonumber$$ 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 $$n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber$$

Consider the following definition.

Definition $\PageIndex{2}$: 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 $\PageIndex{2}$: 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 9.7.1 $T$ 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 $\PageIndex{2}$, since $\mathrm{im} \left( T\right)$ is a subspace of $W,$ it follows that $\mathrm{im}\left( T\right) =W$.

Example $\PageIndex{3}$: One to One Transformation

Let $S:\mathbb{P}_2\to\mathbb{M}_{22}$ be a linear transformation defined by

$$S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.\nonumber$$ Prove that $S$ is one to one but not onto.

Solution

You may recall this example from earlier in Example 9.7.1. Here we will determine that $S$ is one to one, but not onto, using the method provided in Corollary $\PageIndex{1}$.

By definition,

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

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 reduced row-echelon form:

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

Since the unique solution is $a=b=c=0$, $\ker(S)=\{\vec{0}\}$, and thus $S$ is one-to-one by Corollary $\PageIndex{1}$.

Similarly, by Corollary $\PageIndex{1}$, if $S$ is onto it will have $\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4$. The image of $S$ is given by

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

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.