\(\mbox {im}(T)=\RR ^2\) \(\mbox {im}(T)\) is a line in \(\RR ^2\) \(\mbox {im}(T)=\{\vec {0}\}\) \(\mbox {im}(T)=\RR ^5\) \(\mbox {im}(T)\) is a plane in \(\RR ^5\)
\(\mbox {rank}(T)=\answer {2}\)
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
Let \(V\) and \(W\) be vector spaces, and let \(T:V\rightarrow W\) be a linear transformation. The image of \(T\),
denoted by \(\mbox {im}(T)\), is the set
\[\mbox {im}(T)=\{T(\vec {v}):\vec {v}\in V\}\]
In other words, the image of \(T\) consists of individual images of all vectors of \(V\).
Consider the linear transformation \(T:\RR ^3\rightarrow \RR ^2\) with standard matrix
\[A=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}\]
(a) Let \(\vec {v}=\begin{bmatrix}a\\b\\c\end{bmatrix}\) then
\[T(\vec {v})=A\vec {v}=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=a\begin{bmatrix}1\\2\end{bmatrix}+b\begin{bmatrix}2\\4\end{bmatrix}+c\begin{bmatrix}3\\6\end{bmatrix}\]
Thus, every element of the image can be written as a linear combination of the columns of \(A\). We conclude that
\[\mbox {im}(T)=\mbox {span}\left (\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\4\end{bmatrix}, \begin{bmatrix}3\\6\end{bmatrix}\right )=\mbox {col}(A)\]
Every column of \(A\) is a scalar multiple of \(\begin{bmatrix}1\\2\end{bmatrix}\). Thus,
\[\mbox {im}(T)=\mbox {span}\left (\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\4\end{bmatrix}, \begin{bmatrix}3\\6\end{bmatrix}\right )=\mbox {span}\left (\begin{bmatrix}1\\2\end{bmatrix}\right )\]
The image of \(T\) is a line in \(\RR ^2\) determined by the vector \(\begin{bmatrix}1\\2\end{bmatrix}\).
(b) The action of \(T\) can be illustrated with a sketch.
![[Picture]](main.online-6fd8556b41145fba91c3809bdd25d9dc.svg)
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
\[\mbox {dim}(\mbox {im}(T))=\mbox {dim}(\mbox {col}(A))=\mbox {rank}(A)\]
Let \(T:\RR ^5\rightarrow \RR ^4\) be a linear transformation with standard matrix
\[A=\begin{bmatrix}1 & 2 & 2 &-1 & 0\\-1 & 3 & 1 & 0 & -1\\3 & 0 & 0 & 3 & 6\\ 1 & -1 & 1 & -2 & -1\end{bmatrix}\]
Find \(\mbox {im}(T)\) and \(\mbox {dim}(\mbox {im}(T))\). As in Example 2, the image of \(T\) is given by
\[\mbox {im}(T)=\mbox {span}\left (\begin{bmatrix}1\\-1\\3\\1\end{bmatrix}, \begin{bmatrix}2\\3\\0\\-1\end{bmatrix}, \begin{bmatrix}2\\1\\0\\1\end{bmatrix}, \begin{bmatrix}-1\\0\\3\\-2\end{bmatrix}, \begin{bmatrix}0\\-1\\6\\-1\end{bmatrix}\right )=\mbox {col}(A)\]
This time it is harder to detect the vectors that can be eliminated from the spanning
set without affecting the span. We have to rely on the reduced row-echelon form of
\(A\).
\[\begin{bmatrix}1 & 2 & 2 &-1 & 0\\-1 & 3 & 1 & 0 & -1\\3 & 0 & 0 & 3 & 6\\ 1 & -1 & 1 & -2 & -1\end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & 0 & 0 & 1 & 2\\0 & 1 & 0 & 1 & 1\\0 & 0 & 1 & -2 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\]
We can see that \(\mbox {rank}(A)=3\), so \(\mbox {dim}(\mbox {im}(T))=3\).
To identify vectors that span \(\mbox {im}(T)\), we turn to Procedure ??. We identify the first three columns as pivot columns. These columns are linearly independent and span \(\mbox {col}(A)\). Therefore,
\[\mbox {im}(T)=\mbox {col}(A)=\mbox {span}\left (\begin{bmatrix}1\\-1\\3\\1\end{bmatrix}, \begin{bmatrix}2\\3\\0\\-1\end{bmatrix}, \begin{bmatrix}2\\1\\0\\1\end{bmatrix}\right )\]
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.
The rank of a linear transformation \(T:V\rightarrow W\), is the dimension of the image of \(T\).
\[\mbox {rank}(T)=\mbox {dim}(\mbox {im}(T))\]
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
Let \(V\) and \(W\) be vector spaces, and let \(T:V\rightarrow W\) be a linear transformation. The kernel of \(T\),
denoted by \(\mbox {ker}(T)\), is the set
\[\mbox {ker}(T)=\{\vec {v}:T(\vec {v})=\vec {0}\}\]
In other words, the kernel of \(T\) consists of all vectors of \(V\) that map to \(\vec {0}\) in
\(W\). 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.)
![[Picture]](main.online-e7b9c82134a326def499c20abe8dfa1c.svg)
Let \(T:\RR ^5\rightarrow \RR ^4\) be a linear transformation with standard matrix
\[A=\begin{bmatrix}1 & 2 & 2 &-1 & 0\\-1 & 3 & 1 & 0 & -1\\3 & 0 & 0 & 3 & 6\\ 1 & -1 & 1 & -2 & -1\end{bmatrix}\]
- (a)
- Find \(\mbox {ker}(T)\)
- (b)
- Is \(\mbox {ker}(T)\) a subspace of \(\RR ^5\)? If so, find \(\mbox {dim}(\mbox {ker}(T))\).
(a) To find the kernel of \(T\), we need to find all vectors of \(\RR ^5\) that map to \(\vec {0}\) in \(\RR ^4\). This
amounts to solving the equation \(A\vec {x}=\vec {0}\).
Gauss-Jordan elimination yields:
\[\begin{bmatrix}1 & 2 & 2 &-1 & 0\\-1 & 3 & 1 & 0 & -1\\3 & 0 & 0 & 3 & 6\\ 1 & -1 & 1 & -2 & -1\end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & 0 & 0 & 1 & 2\\0 & 1 & 0 & 1 & 1\\0 & 0 & 1 & -2 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\]
Thus, the kernel of \(T\) consists of all elements of the form:
\[\begin{bmatrix}-1\\-1\\2\\1\\0\end{bmatrix}s+\begin{bmatrix}-2\\-1\\2\\0\\1\end{bmatrix}t\]
We conclude that
\[\mbox {ker}(T)=\mbox {span}\left (\begin{bmatrix}-1\\-1\\2\\1\\0\end{bmatrix}, \begin{bmatrix}-2\\-1\\2\\0\\1\end{bmatrix}\right )\]
(b) Since \(\mbox {ker}(T)\) is the span of two vectors of \(\RR ^5\), we know that \(\mbox {ker}(T)\) is a subspace of \(\RR ^5\). (See Theorem ??.) Observe that the two vectors in the spanning set are linearly independent. (How can we see this without performing computations?) Therefore \(\mbox {dim}(\mbox {ker}(T))=2\).
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.
Let \(T:V\rightarrow W\) be a linear transformation, then \(\mbox {ker}(T)\) is a subspace of \(V\).
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 \(\mbox {ker}(T)\).
The nullity of a linear transformation \(T:V\rightarrow W\), is the dimension of the kernel of
\(T\).
\[\mbox {nullity}(T)=\mbox {dim}(\mbox {ker}(T))\]
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.
\[\mbox {nullity}(A) = \mbox {dim}(\mbox {null}(A))=\mbox {dim}(\mbox {ker}(T))=\mbox {nullity}(T)\]
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.
Let \(T:V\rightarrow W\) be a linear transformation. Suppose \(\mbox {dim}(V)=n\), then
\[\mbox {rank}(T)+\mbox {nullity}(T)=n\]
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
\(A=\begin{bmatrix}3&2&4&7&1\\-1&-9&7&6&8\end{bmatrix}\).
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
\[\text {rref}(A)=\begin{bmatrix}1&0&0&1&-1&0\\0&1&0&1&0&0\\0&0&1&1&-1&0\\0&0&0&0&0&1\end{bmatrix}\]