5.7: The Kernel and Image of A Linear Map
-
- Last updated
- Save as PDF
Outcomes
- 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.