# 5.7: The Kernel and Image of A Linear Map

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## 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.

This page titled 5.7: The Kernel and Image of A Linear Map is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .