Note to Student: In this section we will often use \(U\), \(V\) and \(W\) to denote subspaces of \(\RR ^n\), or any other finite-dimensional vector space, such as those we study in Vector Spaces.

Image and Kernel of a Linear Transformation

The Image of a Linear Transformation

In Example 2 we observed that the image of the linear transformation was equal to the column space of its standard matrix. In general, it is easy to see that if \(T:\RR ^n\rightarrow \RR ^m\) is a linear transformation with standard matrix \(A\) then the following relationship holds:

\[\mbox {im}(T)=\mbox {col}(A)\]

In addition, by Theorem ??, we know that

\[\mbox {dim}(\mbox {im}(T))=\mbox {dim}(\mbox {col}(A))=\mbox {rank}(A)\]

By Theorem ?? and Definition ??, we know that for an \(m\times n\) matrix \(A\), \(\mbox {col}(A)\) is a subspace of \(\RR ^m\). However, when vector spaces other than \(\RR ^m\) are involved, it is not yet clear that \(\mbox {im}(T)\) is a subspace of the codomain. The following theorem resolves this issue.

To show that \(\mbox {im}(T)\) is a subspace, we need to show that \(\mbox {im}(T)\) is closed under addition and scalar multiplication.

Suppose \(\vec {w}_1\) and \(\vec {w}_2\) are in \(\mbox {im}(T)\). Then there are vectors \(\vec {v}_1\) and \(\vec {v}_2\) in \(V\) such that \(T(\vec {v}_1)=\vec {w}_1\) and \(T(\vec {v}_2)=\vec {w}_2\). Then

\[\vec {w}_1+\vec {w}_2=T(\vec {v}_1)+T(\vec {v}_2)=T(\vec {v}_1+\vec {v}_2)\]
This shows that \(\vec {w}_1+\vec {w}_2\) is in \(\mbox {im}(T)\).

For any scalar \(a\), we have:

\[a\vec {w}_1=aT(\vec {v}_1)=T(a\vec {v}_1)\]
This shows that \(a\vec {w}_1\) is in \(\mbox {im}(T)\).

We can now define the rank of a linear transformation.

This definition gives us the following relationship between the rank of a linear transformation \(T:\RR ^n\rightarrow \RR ^m\) and the rank of the standard matrix \(A\) associated with it.

The Kernel of a Linear Transformation

It is important to pay attention to the locations of the kernel and the image. We already proved that \(\mbox {im}(T)\) is a subspace of the codomain. In contrast, \(\mbox {ker}(T)\) is located in the domain. (We will prove shortly that it is a subspace of the domain.)

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Recall that the null space of a matrix \(A\) is defined to be set of all solutions to the homogeneous equation \(A\vec {x}=\vec {0}\). This means that if \(T:\RR ^n\rightarrow \RR ^m\) is a linear transformation with standard matrix \(A\) then

\[\mbox {ker}(T)=\mbox {null}(A)\]

We know that \(\mbox {null}(A)\) of an \(m\times n\) matrix is a subspace of \(\RR ^n\). (See Theorem ??.) We conclude this section by showing that even when vector spaces other than \(\RR ^n\) are involved, the kernel of a linear transformation is a subspace of the domain of the transformation.

To show that \(\mbox {ker}(T)\) is a subspace, we need to show that \(\mbox {ker}(T)\) is closed under addition and scalar multiplication.

Suppose that \(\vec {v}_1\) and \(\vec {v}_2\) are in \(\mbox {ker}(T)\). Then,

\[T(\vec {v}_1+\vec {v}_2)=T(\vec {v}_1)+T(\vec {v}_2)=\vec {0}+\vec {0}=\vec {0}\]
This shows that \(\vec {v}_1+\vec {v}_2\) is in \(\mbox {ker}(T)\).

For any scalar \(a\) we have:

\[T(a\vec {v}_1)=aT(\vec {v}_1)=a\vec {0}=\vec {0}\]
This shows that \(a\vec {v}_1\) is in \(\mbox {ker}(T)\).

This definition gives us the following relationship between nullity of a linear transformation \(T:\RR ^n\rightarrow \RR ^m\) and the nullity of the standard matrix \(A\) associated with it.

Rank-Nullity Theorem for Linear Transformations

In Examples 4 and 11, we found the image and the kernel of the linear transformation \(T:\RR ^5\rightarrow \RR ^4\) with standard matrix

\[A=\begin{bmatrix}1 & 2 & 2 &-1 & 0\\-1 & 3 & 1 & 0 & -1\\3 & 0 & 0 & 3 & 6\\ 1 & -1 & 1 & -2 & -1\end{bmatrix}\]

We also found that

\[\mbox {rank}(T)=\mbox {dim}(\mbox {im}(T))=\mbox {dim}(\mbox {col}(A))=\mbox {rank}(A)=3\]

and

\[\mbox {nullity}(T)=\mbox {dim}(\mbox {ker}(T))=\mbox {dim}(\mbox {null}(A))=\mbox {nullity}(A)=2\]

Because of Rank-Nullity Theorem for matrices (Theorem ??), it is not surprising that

\[\mbox {rank}(T)+\mbox {nullity}(T)=3+2=5=\mbox {dim}(\RR ^5)\]

The following theorem is a generalization of this result.

By Theorem 6, \(\mbox {im}(T)\) is a subspace of \(W\). There exists a basis for \(\mbox {im}(T)\) of the form \(\{T(\vec {v}_1), \ldots ,T(\vec {v}_r)\}\). By Theorem 13, \(\mbox {ker}(T)\) is a subspace of \(V\). Let \(\{\vec {u}_1,\ldots ,\vec {u}_s\}\) be a basis for \(\mbox {ker}(T)\).

We will show that \(\{\vec {u}_1,\ldots ,\vec {u}_s, \vec {v}_1,\ldots ,\vec {v}_r\}\) is a basis for \(V\).

For any vector \(\vec {v}\) in \(V\), we have:

\[T(\vec {v})=c_1T(\vec {v}_1)+\ldots +c_rT(\vec {v}_r)\]
for some scalars \(c_i\) \((1\leq i\leq r)\). Thus,
\[T(\vec {v})-\big (c_1T(\vec {v}_1)+\ldots +c_rT(\vec {v}_r)\big )=\vec {0}\]
By linearity,
\[T((\vec {v}-(c_1\vec {v}_1+\ldots +c_r\vec {v}_r))=\vec {0}\]
Therefore \(\vec {v}-(c_1\vec {v}_1+\ldots +c_r\vec {v}_r)\) is in \(\mbox {ker}(T)\).

Hence there are scalars \(a_i\) \((1\leq i\leq s)\) such that

\[\vec {v}-(c_1\vec {v}_1+\ldots +c_r\vec {v}_r)=a_1\vec {u}_1+\ldots +a_s\vec {u}_s\]
Thus,
\[\vec {v}=(c_1\vec {v}_1+\ldots +c_r\vec {v}_r)+(a_1\vec {u}_1+\ldots +a_s\vec {u}_s)\]

We conclude that

\[V=\mbox {span}(\vec {u}_1,\ldots ,\vec {u}_s, \vec {v}_1,\ldots ,\vec {v}_r)\]

Now we need to show that \(\{\vec {u}_1,\ldots ,\vec {u}_s, \vec {v}_1,\ldots ,\vec {v}_r\}\) is linearly independent.

Suppose

\begin{align}\label {eq:kerplusimproof} c_1\vec {v}_1+\ldots +c_r\vec {v}_r+a_1\vec {u}_1+\ldots +a_s\vec {u}_s=\vec {0}\end{align}

Applying \(T\) to both sides, we get

\[T(c_1\vec {v}_1+\ldots +c_r\vec {v}_r+a_1\vec {u}_1+\ldots +a_s\vec {u}_s)=T(\vec {0})\]
\[c_1T(\vec {v}_1)+\ldots +c_rT(\vec {v}_r)+a_1T(\vec {u}_1)+\ldots +a_sT(\vec {u}_s)=\vec {0}\]

But \(T(\vec {u}_i)=\vec {0}\) for \(1\leq i\leq s\), thus

\[c_1T(\vec {v}_1)+\ldots +c_rT(\vec {v}_r)=\vec {0}\]
Since \(\{T(\vec {v}_1),\ldots ,T(\vec {v}_r)\}\) is linearly independent, it follows that each \(c_i=0\).

But then Equation (??) implies that \(a_1\vec {u}_1+\ldots +a_s\vec {u}_s=\vec {0}\). Because \(\{\vec {u}_1, \ldots ,\vec {u}_s\}\) is linearly independent, it follows that each \(a_i=0\).

We conclude that \(\{\vec {u}_1,\ldots ,\vec {u}_s,\vec {v}_1,\ldots ,\vec {v}_r\}\) is a basis for \(V\). Thus,

\[\mbox {dim}(\mbox {ker}(T))+\mbox {dim}(\mbox {im}(T))=s+r=n\]

Practice Problems

Describe the image and find the rank of the linear transformation \(T:\RR ^5\rightarrow \RR ^2\) induced by \(A=\begin{bmatrix}3&2&4&7&1\\-1&-9&7&6&8\end{bmatrix}\).
\(\mbox {im}(T)=\RR ^2\) \(\mbox {im}(T)\) is a line in \(\RR ^2\) \(\mbox {im}(T)=\{\vec {0}\}\) \(\mbox {im}(T)=\RR ^5\) \(\mbox {im}(T)\) is a plane in \(\RR ^5\)

\(\mbox {rank}(T)=\answer {2}\)

Describe the image and find the rank of the linear transformation \(T:\RR ^2\rightarrow \RR ^3\) induced by \(A=\begin{bmatrix}1&1\\1&1\\1&1\end{bmatrix}\)
\(\mbox {im}(T)=\RR ^3\) \(\mbox {im}(T)\) is a line in \(\RR ^2\) \(\mbox {im}(T)\) is a line in \(\RR ^3\) \(\mbox {im}(T)=\{\vec {0}\}\) \(\mbox {im}(T)\) is a plane in \(\RR ^3\)

\(\mbox {rank}(T)=\answer {1}\)

Suppose linear transformations \(T:\RR ^2\rightarrow \RR ^2\) and \(S:\RR ^2\rightarrow \RR ^2\) are such that \(\mbox {im}(T)=\mbox {im}(S)=\mbox {span}\left (\begin{bmatrix}1\\-3\end{bmatrix}\right )\). Does this mean that \(T\) and \(S\) are the same transformation? Justify your claim.
Describe the kernel and find the nullity of the linear transformation \(T:\RR ^3\rightarrow \RR ^2\) induced by \(A=\begin{bmatrix}2&1&0\\-1&1&-3\end{bmatrix}\).
\(\mbox {ker}(T)=\RR ^3\) \(\mbox {ker}(T)=\{\vec {0}\}\) \(\mbox {ker}(T)=\RR ^2\) \(\mbox {ker}(T)\) is a plane in \(\RR ^3\) \(\mbox {ker}(T)\) is a line in \(\RR ^3\)

\(\mbox {nullity}(T)=\answer {1}\)

Describe the kernel and find the nullity of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) induced by \(A=\begin{bmatrix}2&-1\\3&0\end{bmatrix}\).
\(\mbox {ker}(T)=\RR ^2\) \(\mbox {ker}(T)=\{\vec {0}\}\) \(\mbox {ker}(T)\) is a line in \(\RR ^2\)

\(\mbox {nullity}(T)=\answer {0}\)

Describe the kernel and find the nullity of the linear transformation \(T:\RR ^3\rightarrow \RR ^5\) induced by \(A=\begin{bmatrix}1&2&-1\\1&2&-1\\1&2&-1\\1&2&-1\\1&2&-1\end{bmatrix}\)
\(\mbox {ker}(T)\) is a plane in \(\RR ^3\) \(\mbox {ker}(T)\) is a line in \(\RR ^3\) \(\mbox {ker}(T)\) is a line in \(\RR ^5\) \(\mbox {ker}(T)=\RR ^3\) \(\mbox {ker}(T)=\{\vec {0}\}\)

\(\mbox {nullity}(T)=\answer {2}\)

Suppose a linear transformation \(T:\RR ^3\rightarrow \RR ^3\) is such that \(\mbox {im}(T)\) is a plane in \(\RR ^3\). Then
\[\mbox {rank}(T)=\answer {2}\]
\[\mbox {nullity}(T)=\answer {1}\]
Suppose a linear transformation \(T:\RR ^5\rightarrow \RR ^5\) is such that \(T(\vec {v})=\vec {0}\) for all \(\vec {v}\) in \(\RR ^5\). Then
\[\mbox {rank}(T)=\answer {0}\]
\[\mbox {nullity}(T)=\answer {5}\]
Let \(T:\RR ^6\rightarrow \RR ^4\) be a linear transformation with standard matrix
\[A=\begin{bmatrix}2&-1&1&-2&1&1\\1&2&3&6&-4&1\\0&2&2&4&-2&-1\\1&3&2&6&-3&2\end{bmatrix}\]
Find \(\mbox {im}(T)\) and \(\mbox {ker}(T)\) if the reduced row-echelon form of \(A\) is
\[\text {rref}(A)=\begin{bmatrix}1&0&0&1&-1&0\\0&1&0&1&0&0\\0&0&1&1&-1&0\\0&0&0&0&0&1\end{bmatrix}\]
Let \(V=\mbox {span}\left (\begin{bmatrix}1\\1\end{bmatrix}\right )\), and let \(T:V\rightarrow \RR ^2\) be a linear transformation defined by \(T(\vec {v})=2\vec {v}\). Find \(\mbox {im}(T)\) and \(\mbox {ker}(T)\).
Suppose a linear transformation \(T\) is induced by a \(4\times 6\) matrix \(A\). Let \(S\) be a linear transformation induced by \(A^T\). Find \(\mbox {nullity}(S)\), if \(\mbox {nullity}(T)=3\). Prove your claim.