Opening the book…
- Describe the kernel and image of a linear transformation.
- Use the kernel and image to determine if a linear transformation is one to one or onto.
Here we consider the case where the linear map is not necessarily an isomorphism. First here is a definition of what is meant by the image and kernel of a linear transformation.
Let \(V\) and \(W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. Then the image of \(T\) denoted as \(\mathrm{im}\left( T\right)\) is defined to be the set \[\left\{ T(\vec{v}):\vec{v}\in V\right\}\nonumber \] In words, it consists of all vectors in \(W\) that equal \(T(\vec{v})\) for some \(\vec{v}\in V\). The kernel, \(\ker \left( T\right)\), consists of all \(\vec{v}\in V\) such that \(T(\vec{v})=\vec{0}\). That is, \[\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\nonumber \]
Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively.
Let \(V,W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). In fact, they are both subspaces.
- 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 \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]
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.
Let \(T:\mathbb{P}_1\to\mathbb{R}\) be the linear transformation defined by \[T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.\nonumber \] Find the kernel and image of \(T\).
Solution
We will first find the kernel of \(T\). It consists of all polynomials in \(\mathbb{P}_1\) that have \(1\) for a root. \[\begin{aligned} \mathrm{ker}(T) & = \{ p(x)\in \mathbb{P}_1 ~|~ p(1)=0\} \\ & = \{ ax+b ~|~ a,b\in\mathbb{R} \mbox{ and }a+b=0\} \\ & = \{ ax-a ~|~ a\in\mathbb{R} \}\end{aligned}\] Therefore a basis for \(\mathrm{ker}(T)\) is \[\left\{ x-1 \right\}\nonumber \] Notice that this is a subspace of \(\mathbb{P}_1\).
Now consider the image. It consists of all numbers which can be obtained by evaluating all polynomials in \(\mathbb{P}_1\) at \(1\). \[\begin{aligned} \mathrm{im}(T) & = \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = \mathbb{R}\end{aligned}\] Therefore a basis for \(\mathrm{im}(T)\) is \[\left\{ 1 \right\}\nonumber \] Notice that this is a subspace of \(\mathbb{R}\), and in fact is the space \(\mathbb{R}\) itself.
Let \(T: \mathbb{M}_{22} \mapsto \mathbb{R}^2\) be defined by \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]\nonumber \] Then \(T\) is a linear transformation. Find a basis for \(\mathrm{ker} (T)\) and \(\mathrm{im}(T)\).
Solution
You can verify that \(T\) represents a linear transformation.
Now we want to find a way to describe all matrices \(A\) such that \(T(A) = \vec{0}\), that is the matrices in \(\mathrm{ker}(T)\). Suppose \(A = \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ]\) is such a matrix. Then \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \] 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 is \(a = s, b = s, c = t, d = -t\) where \(s, t\) are scalars. We can describe \(\mathrm{ker}(T)\) as follows. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\).
We now wish to find a basis for \(\mathrm{im}(T)\). We can write the image of \(T\) as \[\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}\nonumber \] Notice that this can be written as \[\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]
However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of \(\mathrm{im}(T)\). \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]
Notice that these vectors have the same span as the set above but are now linearly independent.
A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. A special case was done earlier in the context of matrices. Recall that for an \(m\times n\) matrix \(% A,\) it was the case that the dimension of the kernel of \(A\) added to the rank of \(A\) equals \(n\).
Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. Suppose the dimension of \(V\) is \(n\). Then \(n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\).
- Proof
-
From Proposition \(\PageIndex{1}\), \(\mathrm{im}\left( T\right)\) is a subspace of \(W.\) By Theorem 7.6.6, 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 \[\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u} _{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}\nonumber \] 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. It follows that \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots ,\vec{v} _{r}\right\}\) is a basis for \(V\) and so \[n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber \]
Let \(T:V\rightarrow W\) be a linear transformation and suppose \(V,W\) are finite dimensional vector spaces. Then the rank of \(T\) denoted as \(\mathrm{rank}\left( T\right)\) is defined as the dimension of \(\mathrm{im}\left( T\right) .\) The nullity of \(T\) is the dimension of \(\ker \left( T\right) .\) Thus the above theorem says that \(\mathrm{rank}\left( T\right) + \left( nullity \left( T\right) \right) =\dim \left( V\right) .\)
Let \(V\) be a vector space of dimension \(n\) and let \(W\) be a subspace. Then \(W=V\) if and only if the dimension of \(W\) is also \(n\).
Let \(T:V\rightarrow W\) be a linear map where the dimension of \(V\) is \(n\) and the dimension of \(W\) is \(m\). Then \(T\) is one to one if and only if \(\ker \left( T\right) =\left\{ \vec{0}\right\}\) and \(T\) is onto if and only if \(\mathrm{rank}\left( T\right) =m\).
- Proof
-
The statement \(\ker \left( T \right) =\left\{ \vec{0}\right\}\) is equivalent to saying if \(T \left( \vec{v} \right)=\vec{0},\) it follows that \(\vec{v}=\vec{0}\). Thus by Proposition 5.5.1 \(T\) is one to one. If \(T\) is onto, then \(\mathrm{im}\left( T\right) =W\) and so \(\mathrm{rank}\left( T\right)\) which is defined as the dimension of \(\mathrm{im}\left( T\right)\) is \(m\). If \(\mathrm{ rank}\left( T\right) =m,\) then by Theorem 7.6.3, since \(\mathrm{im} \left( T\right)\) is a subspace of \(W,\) it follows that \(\mathrm{im}\left( T\right) =W\).
Let \(S:\mathbb{P}_2\to\mathbb{M}_{22}\) be a linear transformation defined by \[S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.\nonumber \] Prove that \(S\) is one to one but not onto.
Solution
Here we will determine that \(S\) is one to one, but not onto, using the method provided in Corollary \(\PageIndex{1}\).
By definition, \[\ker(S)=\{ax^2+bx+c\in \mathbb{P}_2 ~|~ a+b=0, a+c=0, b-c=0, b+c=0\}.\nonumber \]
Suppose \(p(x)=ax^2+bx+c\in\ker(S)\). This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in reduced row-echelon form:
\[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]
Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\).
Similarly, by Corollary \(\PageIndex{1}\), if \(S\) is onto it will have \(\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4\). The image of \(S\) is given by
\[\mathrm{im}(S) = \left\{ \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \right\} = \mathrm{span} \left\{ \left [\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ], \left [\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right ] \right\}\nonumber \]
These matrices are linearly independent which means this set forms a basis for \(\mathrm{im}(S)\). Therefore the dimension of \(\mathrm{im}(S)\), also called \(\mathrm{rank}(S)\), is equal to \(3\). It follows that \(S\) is not onto.
This page titled 7.10: The Kernel and Image of a Linear Map was last modified on Sat, 06 Sep 2025 05:17:43 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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