Question

Using your work in Exercise 42 as a guide, explain how you can write the image of any matrix A as the kernel of some matrix B.

Short answer

$\mathrm{im}\left(\mathrm{A}\right)=\ker \left(\mathrm{B}\right)$

Step-by-step solution

Expressing

Considering A to be the matrix and B to be the kernel of the matrix. The objective is to express the image of matrix A as the kernel of matrix B.

Definition of image and kernel.

Recall the definition of image and kernel of T. consider the T to be the linear transformation from the ${\mathrm{ℝ}}^{\mathrm{m}}\mathrm{to}{\mathrm{ℝ}}^{\mathrm{n}}$. The image and the kernel of T are denoted by$\mathrm{im}\left(\mathrm{T}\right)$ and ker(T) respectively and defined as follows:

$$\begin{gathered} \mathrm{im}\left(\mathrm{T}\right)=\left\{\mathrm{T}\left(\overrightarrow{\mathrm{x}}\right):\overrightarrow{\mathrm{x}}\mathrm{in}{\mathrm{ℝ}}^{\mathrm{m}}\right\}\subseteq {\mathrm{ℝ}}^{\mathrm{n}} \\ \ker \left(\mathrm{T}\right)=\left\{\overrightarrow{\mathrm{x}}\mathrm{in}{\mathrm{ℝ}}^{\mathrm{m}}:\mathrm{T}\left(\overrightarrow{\mathrm{x}}\right)=0\right\}\subseteq {\mathrm{ℝ}}^{\mathrm{m}} \end{gathered}$$

Proving im(A)=ker(B)

Consider A to be a nxm matrix with rank r.

Assume, $\mathrm{r}=\mathrm{n}$.

Recall the results. Consider A to be a nxn matrix. If the rank of A is n, then

$\mathrm{im}\left(\mathrm{A}\right)={\mathrm{ℝ}}^{\mathrm{n}}$and $\ker \left(\mathrm{A}\right)=\left\{0\right\}$.

This implies, B will be a$\mathrm{n}\times \mathrm{n}$ zero matrix. Then, $\mathrm{im}\left(\mathrm{A}\right)=\ker \left(\mathrm{B}\right)$.

Writing image of matrix A for the kernel of matrix B

Assume the rank of the matrix A to be $\mathrm{r}<\mathrm{n}$and B represents the matrix after removing the first r rows and first m columns of data-custom-editor="chemistry" $\mathrm{rref}\left[\mathrm{A}:{\mathrm{I}}_{\mathrm{n}}\right]$. This implies, B is the data-custom-editor="chemistry" $\left(\mathrm{n}-\mathrm{r}\right)\times \mathrm{n}$matrix. This implies, im(A)=ker(B). Hence, in both cases, it is concluded that

$\mathrm{im}\left(\mathrm{A}\right)=\ker \left(\mathrm{B}\right)$.