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4.C Images and kernels

We now define two subspaces which help us to understand a linear transformation (in much the same way that prime factors help us to understand an integer).

Definition 4.15:

The image of a linear transformation $T:V\to W$ is the set

\operatorname{im}\nolimits T:=\{T(\vec{x})\in W\;|\;\vec{x}\in V\}.

In other words, it is the set of elements $\vec{y}\in W$ such that there exists an $\vec{x}\in V$ with $\vec{y}=T(\vec{x})$ . This set is also sometimes written $\operatorname{im}\nolimits T=T(V)$ . One might also refer to the image of a subset $S\subset V$ , which will be denoted $T(S)$ .

The image of a matrix in $\operatorname{M}_{{n\times m}}({F})$ , will mean the image of the associated linear transformation $\vec{x}\mapsto A\vec{x}$ (using the standard basis, unless stated otherwise). Here $\vec{x}$ is viewed as a column vector in $F^{m}$ .

[Aside: The image of a function is also sometimes called its range.]

Theorem 4.16.

Let $A\in\operatorname{M}_{{n\times m}}({F})$ . Then $\operatorname{im}\nolimits A$ equals the span of its column vectors.

Proof.

Since $F^{m}=\{\alpha_{1}\vec{e_{1}}+\cdots+\alpha_{m}\vec{e_{m}}\;|\;\alpha_{i}\in F\}$ , we can rewrite the image as follows:

\operatorname{im}\nolimits A=\{A(\alpha_{1}\vec{e_{1}}+\cdots+\alpha_{m}\vec{e% _{m}})\;|\;\alpha_{i}\in F\}=\{\alpha_{1}A\vec{e_{1}}+\cdots+\alpha_{m}A\vec{e% _{m}}\;|\;\alpha_{i}\in F\}.

But since $A\vec{e_{i}}$ is the $i$ th column of the matrix $A$ , the right hand side is equal to the span of the column vectors. ∎

Example 4.17.

Let $A=\begin{bmatrix}3&-3\\ -1&1\end{bmatrix}\in\operatorname{M}_{{2}}({{\mathbb{R}}})$ . The image of $A$ is

\operatorname{im}\nolimits A=\operatorname{span}_{{\mathbb{R}}}\{\begin{% bmatrix}3\\ -1\end{bmatrix},\begin{bmatrix}-3\\ 1\end{bmatrix}\}=\operatorname{span}\{\begin{bmatrix}3\\ -1\end{bmatrix}\}.

So the image of $A$ is a 1-dimensional subspace of ${\mathbb{R}}^{2}$ with basis $(3,-1)$ .

Next we define the kernel of a linear transformation. I recommend thinking of the word “kernel” as the “core” of the transformation; because they are the elements which are lost (i.e. sent to zero) when mapped to $W$ .

Definition 4.18:

The kernel of a linear transformation $T:V\to W$ is the set

\ker T:=\{\vec{x}\in V\;|\;T(\vec{x})=\vec{0}\}.

The kernel of a matrix $A$ is the kernel of its linear transformation $\vec{x}\mapsto A\vec{x}$ . So it’s the set of vectors such that $A\vec{x}=\vec{0}$ .

[Aside: The kernel is also sometimes called the null space of a transformation.]

Example 4.19.

Let $A=\begin{bmatrix}3&-3\\ -1&1\end{bmatrix}\in\operatorname{M}_{{2}}({{\mathbb{R}}})$ . The kernel of $A$ is

\ker A=\left\{\begin{bmatrix}x\\ y\end{bmatrix}\;|\;3x-3y=0,\;x-y=0,\;x,y\in{\mathbb{R}}\right\}=\left\{\begin{% bmatrix}x\\ x\end{bmatrix}\;|\;x\in{\mathbb{R}}\right\}=\operatorname{span}\{\begin{% bmatrix}1\\ 1\end{bmatrix}\}.

So the kernel of $A$ is a 1-dimensional subspace of ${\mathbb{R}}^{2}$ with basis $(1,1)$ .

As in the above example, finding the kernel of a matrix is always equivalent to finding the solution set to a system of linear equations.

Exercise 4.20:

Check that $\operatorname{im}\nolimits T$ satisfies the three conditions in Theorem 2.8.

Exercise 4.21:

Check that $\ker T$ satisfies the three conditions in Theorem 2.8.

Exercise 4.22:

Define the linear transformation $T:{\mathbb{R}}^{3}\to{\mathbb{R}}^{3}$ by the formula $T(x,y,z):=(x+y,y+z,x-z).$

i.

Find the matrix ${}_{\mathcal{C}}[T]_{\mathcal{C}}$ with respect to the standard basis $\mathcal{C}$ .
ii.

Find a basis of $\operatorname{im}\nolimits T\subset{\mathbb{R}}^{3}$ .
iii.

Find a basis of $\ker T\subset{\mathbb{R}}^{3}$ .
iv.

Combine the bases from parts (ii) and (iii), and verify that the result is a (non-standard) basis $\mathcal{B}$ of ${\mathbb{R}}^{3}$ . Find the matrix ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ .

[End of Exercise]

Theorem 4.23.

Row operations on a matrix $A$ do not change the kernel (but they do change the image).

[Aside: Column operations don’t change the image of $A$ , but do change the kernel.]

Note that the kernel of a matrix could be thought of as the solution set to a system of linear equations, and those solutions are unchanged by row operations (a fact that was heavily used in MATH105), and indeed this is why row operations are what they are.