5.7: The Kernel and Image of A Linear Map

Outcomes

1. Describe the kernel and image of a linear transformation, and find a basis for each.

In this section we will consider the case where the linear transformation is not necessarily an isomorphism. First consider the following important definition.

Definition $\PageIndex{1}$: Kernel and Image

Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and let $T:V\mapsto W$ be a linear transformation. Then the image of $T$ denoted as $\mathrm{im}\left( T\right)$ is defined to be the set $$\mathrm{im}\left( T\right) = \left\{T (\vec{v}):\vec{v}\in V\right\}\nonumber$$ In words, it consists of all vectors in $W$ which equal $T(\vec{v})$ for some $\vec{v}\in V$.

The kernel of $T$, written $\ker \left( T\right)$, consists of all $\vec{v}\in V$ such that $T(\vec{v})=\vec{0}$. That is, $$\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\nonumber$$

Below is a video on the kernel and image of a linear transformation.

Below is a video on which vectors are in the range and kernel of a linear transformation.

 

Below is a video on which vectors are in the kernel of a 2x3 matrix.

Below is a video on finding a nonzero vector in the kernel of a linear transformation.

Below is another video on finding a nonzero vector in the kernel of a linear transformation.

It follows that $\mathrm{im}\left( T\right)$ and $\ker \left( T\right)$ are subspaces of $W$ and $V$ respectively.

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

Let $V, W$ be subspaces of $\mathbb{R}^n$ and let $T:V\rightarrow W$ be a linear transformation. Then $\ker \left( T\right)$ is a subspace of $V$ and $\mathrm{im}\left( T\right)$ is a subspace of $W$.

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

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.

We will now examine how to find the kernel and image of a linear transformation and describe the basis of each.

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

Let $T: \mathbb{R}^4 \mapsto \mathbb{R}^2$ be defined by

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

Then $T$ is a linear transformation. Find a basis for $\mathrm{ker}(T)$ and $\mathrm{im}(T)$.

Solution

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

First we will find a basis for $\mathrm{ker}(T)$. To do so, we want to find a way to describe all vectors $\vec{x} \in \mathbb{R}^4$ such that $T(\vec{x}) = \vec{0}$. Let $\vec{x} = \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right]$ be such a vector. Then

$$T \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right] = \left[ \begin{array}{c} a - b \\ c + d \end{array} \right] = \left ( \begin{array}{c} 0 \\ 0 \end{array} \right ) \nonumber$$

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 to this system is $a = s, b = s, c = t, d = -t$ where $s, t$ are scalars. We can describe $\mathrm{ker}(T)$ as follows.

$$\mathrm{ker}(T) = \left\{ \left[ \begin{array}{r} s \\ s \\ t \\ -t \end{array} \right] \right\} = \mathrm{span} \left\{ \left[ \begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \end{array} \right], \left[ \begin{array}{r} 0 \\ 0 \\ 1 \\ -1 \end{array} \right] \right\} \nonumber$$

Notice that this set is linearly independent and therefore forms a basis for $\mathrm{ker}(T)$.

We move on to finding a basis for $\mathrm{im}(T)$. We can write the image of $T$ as $$\mathrm{im}(T) = \left\{ \left[ \begin{array}{c} a - b \\ c + d \end{array} \right] \right\} \nonumber$$

We can write this in the form $$\mathrm{span} = \left\{ \left[ \begin{array}{r} 1 \\ 0 \end{array} \right], \left[ \begin{array}{r} -1 \\ 0 \end{array} \right], \left[ \begin{array}{r} 0 \\ 1 \end{array} \right], \left[ \begin{array}{r} 0 \\ 1 \end{array} \right] \right\}\nonumber$$

This set is clearly not linearly independent. By removing unnecessary vectors from the set we can create a linearly independent set with the same span. This gives a basis for $\mathrm{im}(T)$ as $$\mathrm{im}(T) = \mathrm{span} \left\{ \left[ \begin{array}{r} 1 \\ 0 \end{array} \right], \left[ \begin{array}{r} 0 \\ 1 \end{array} \right] \right\}\nonumber$$

Below is a video on finding the kernel of a transformation matrix.

Below is a video on finding the basis for the kernel of a linear transformation.

Below is a video on finding the kernel of a linear transformation given a matrix.

Recall that a linear transformation $T$ is called one to one if and only if $T(\vec{x}) = \vec{0}$ implies $\vec{x} = \vec{0}$. Using the concept of kernel, we can state this theorem in another way.

Theorem $\PageIndex{1}$: One to One and Kernel

Let $T$ be a linear transformation where $\mathrm{ker}(T)$ is the kernel of $T$. Then $T$ is one to one if and only if $\mathrm{ker}(T)$ consists of only the zero vector.

A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. In the previous example $\mathrm{ker}(T)$ had dimension $2$, and $\mathrm{im}(T)$ also had dimension of $2$. Is it a coincidence that the dimension of $\mathbb{M}_{22}$ is $4 = 2 + 2$? Consider the following theorem.

Theorem $\PageIndex{2}$: Dimension of Kernel and Image

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

Proof

From Proposition $\PageIndex{1}$, $\mathrm{im}\left( T\right)$ is a subspace of $W.$ We know that 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 $$\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u}) _{j}=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})=0\nonumber$$ 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. Therefore $\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots ,\vec{v} _{r}\right\}$ is a basis for $V$ and so $$n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber$$

The above theorem leads to the next corollary.

Corollary $\PageIndex{1}$

Let $T:V\rightarrow W$ be a linear transformation where $V,W$ are subspaces of $\mathbb{R}^n$. Suppose the dimension of $V$ is $m$. Then $$\dim \left( \ker \left( T\right) \right) \leq m\nonumber$$ $$\dim \left( \mathrm{im}\left( T \right) \right) \leq m\nonumber$$

This follows directly from the fact that $n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)$.

Consider the following example.

Example $\PageIndex{2}$

Let $T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{3}$ be defined by $$T(\vec{x})=\left[ \begin{array}{rr} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] \vec{x}\nonumber$$ Then $\mathrm{im}\left( T\right) =V$ is a subspace of $\mathbb{R}^{3}$ and $T$ is an isomorphism of $\mathbb{R}^{2}$ and $V$. Find a $2\times 3$ matrix $A$ such that the restriction of multiplication by $A$ to $V=\mathrm{im}\left( T\right)$ equals $T^{-1}$.

Solution

Since the two columns of the above matrix are linearly independent, we conclude that $\mathrm{dim}(\mathrm{im}(T)) = 2$ and therefore $\mathrm{dim}(\mathrm{ker}(T)) = 2 - \mathrm{dim}(\mathrm{im}(T)) = 2-2 = 0$ by Theorem $\PageIndex{2}$. Then by Theorem $\PageIndex{1}$ it follows that $T$ is one to one.

Thus $T$ is an isomorphism of $\mathbb{R }^{2}$ and the two dimensional subspace of $\mathbb{R}^{3}$ which is the span of the columns of the given matrix. Now in particular, $$T(\vec{e}_{1})=\left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] ,\ T(\vec{e}_{2})=\left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] \nonumber$$

Thus $$T^{-1}\left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] =\vec{e}_{1},\ T^{-1}\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] =\vec{e}_{2} \nonumber$$

Extend $T^{-1}$ to all of $\mathbb{R}^{3}$ by defining $$T^{-1}\left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] =\vec{e}_{1}\nonumber$$ Notice that the vectors $$\left\{ \left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] \right\} \nonumber$$ are linearly independent so $T^{-1}$ can be extended linearly to yield a linear transformation defined on $\mathbb{R}^{3}$. The matrix of $T^{-1}$ denoted as $A$ needs to satisfy $$A\left[ \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] =\left[ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] \nonumber$$ and so $$A=\left[ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]^{-1}=\left[ \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \nonumber$$

Note that $$\left[ \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] =\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\nonumber$$ $$\left[ \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] =\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \nonumber$$ so the restriction to $V$ of matrix multiplication by this matrix yields $T^{-1}.$

Below is a video on describing the kernel of a linear transformation:  projection onto y=x.

Below is a video on describing the kernel of a linear transformation:  reflection across the y-axis.