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Mathematics LibreTexts

9.8: The Kernel and Image of a Linear Map

  • Page ID
    29492
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    Learning Objectives
    1. Describe the kernel and image of a linear transformation.
    2. 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.

    Definition Kernel and Image

    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 \[\left\{ T(\vec{v}):\vec{v}\in V\right\}\] In words, it consists of all vectors in \(W\) which 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, \[\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\]

    Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively.

    Proposition \(\PageIndex{1}\): Kernel and Image as Subspaces

    Let \(V,W\) be 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

    \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\]

    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,

    \[aT(\vec{v}_{2})+bT(\vec{v}_{2})=T\left( a\vec{v}_{1}+b\vec{v}_{2}\right)\]

    and this last vector is in \(\mathrm{im}\left( T\right)\) by definition.

    Consider the following example.

    Example \(\PageIndex{1}\): Kernel and Image of a Transformation

    Let \(T:\mathbb{P}_1\to\mathbb{R}\) be the linear transformation defined by \[T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.\] 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 \[\left\{ x-1 \right\}\] 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 \[\left\{ 1 \right\}\] Notice that this is a subspace of \(\mathbb{R}\), and in fact is the space \(\mathbb{R}\) itself.

    Example \(\PageIndex{1}\): Kernel and Image of a Linear Transformation

    Let \(T: \mathbb{M}_{22} \mapsto \mathbb{R}^2\) be defined by \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]\] 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 \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\] 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. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\] 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 \[\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}\] Notice that this can be written as \[\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\]

    However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of \(\mathrm{im}(T)\). \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\]

    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 \(n\).

    Theorem \(\PageIndex{1}\): Dimension of Kernel + Image

    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 [prop:kernelimagevectorspaces], \(\mathrm{im}\left( T\right)\) is a subspace of \(W.\) By Theorem [thm:basisvectorspace], 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 \[T(\vec{v})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})\] 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 \[\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}=\sum_{j=1}^{s}a_{j}\vec{u}_{j}\] 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 \[V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}\] 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 \[\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u}_{j}=0\] Apply \(T\) to both sides to obtain \[\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u} _{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}\] 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 \[n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\]

    Consider the following definition.

    Definition \(\PageIndex{1}\): Rank of Linear Transformation

    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 \(\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .\)

    Recall the following important result.

    Theorem \(\PageIndex{1}\): Subspace of Same Dimension

    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 \(n\).

    From this theorem follows the next corollary.

    Corollary \(\PageIndex{1}\): One to One and Onto Characterization

    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 Lemma [lem:onetooneabstract] \(T\) 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 [thm:subspacevectorspace], since \(\mathrm{im} \left( T\right)\) is a subspace of \(W,\) it follows that \(\mathrm{im}\left( T\right) =W\).

    Example \(\PageIndex{1}\): One to One Transformation

    Let \(S:\mathbb{P}_2\to\mathbb{M}_{22}\) be a linear transformation defined by \[S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.\] Prove that \(S\) is one to one but not onto.

    Solution

    You may recall this example from earlier in Example [exa:onetoonegeneral]. Here we will determine that \(S\) is one to one, but not onto, using the method provided in Corollary [cor:oneoneontochar].

    By definition, \[\ker(S)=\{ax^2+bx+c\in \mathbb{P}_2 ~|~ a+b=0, a+c=0, b-c=0, b+c=0\}.\]

    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 :

    \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\]

    Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary [cor:oneoneontochar].

    Similarly, by Corollary [cor:oneoneontochar], if \(S\) is onto it will have \(\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4\). The image of \(S\) is given by

    \[\mathrm{im}(S) = \left\{ \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \right\} = \mathrm{span} \left\{ \left [\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ], \left [\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right ] \right\}\]

    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.