5.7: The Kernel and Image of A Linear Map
- 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.
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.
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.
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.
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.
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.
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.
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.