Question

Can you find a $\mathbf{3}\mathbf{\times }\mathbf{3}$ matrix $\mathit{A}$such that $\mathit{i}\mathit{m}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$? Explain.

Short answer

We cannot find the$3\times 3$ matrix$A$ such that $im\left(A\right)=\ker \left(A\right)$.

Step-by-step solution

To mention given data and recall theorem 3.3.7

Let $A$be a $3\times 3$matrix.

We have given that $\mathrm{im}\left(\mathrm{A}\right)=\ker \left(\mathrm{A}\right)$.

Theorem 3.3.7, is stated as follows:

For any$n\times m$ matrix $A$, the equation,

$\dim \left(\ker \left(\mathrm{A}\right)\right)+\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)=\mathrm{m}$.

Here, in this case$n=m=3$ .

To find the dimension of the space

By Theorem 3.3.7, we have,

$\begin{array}{c}\mathrm{m}=\dim \left(\ker \left(\mathrm{A}\right)\right)+\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)\\ \Rightarrow 3=\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)+\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)\\ \Rightarrow 3=2\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)\\ \Rightarrow \dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)=\frac{3}{2}\end{array}$

This is impossible.

Hence, we cannot find the$3\times 3$ matrix $A$such that $im\left(A\right)=\ker \left(A\right)$.