Question

Consider a square matrix.

1. What is the relationship among ker($\mathit{A}$) and ker (${\mathit{A}}^{\mathbf{2}}$)? Are they necessarily equal? Is one of them necessarily contained in the other? More generally, what can you say about ker($\mathit{A}$),ker (${\mathit{A}}^{\mathbf{2}}$), ker (${\mathit{A}}^{\mathbf{3}}$),…?
2. What can you say about im($\mathit{A}$),im (${\mathit{A}}^{\mathbf{2}}$), im(${\mathit{A}}^{\mathbf{3}}$),…?

Hint: Exercise 37 is helpful.

Short answer

1. $\ker \left({\mathrm{A}}^{\mathrm{k}}\right)\subseteq \ker \left({\mathrm{A}}^{\mathrm{k}+1}\right)$
   
2. $\mathrm{im}\left({\mathrm{A}}^{\mathrm{k}+1}\right)\subseteq \mathrm{im}\left({\mathrm{A}}^{\mathrm{k}}\right)$
   

Step-by-step solution

Define kernel of linear transformation

Thekernel of linear transformation is defined as follows:

The kernel of a linear transformation $T\left(\overrightarrow{x}\right)=A\overrightarrow{x}$from ${\mathrm{ℝ}}^m$to ${\mathrm{ℝ}}^n$consists of all zeros of the transformation, i.e., the solutions of the equations $T\left(\overrightarrow{x}\right)=A\overrightarrow{x}=0$.

It is denoted by ker$\left(T\right)$or ker$\left(A\right)$.

(a) Find relationship among kernels

Consider $\overrightarrow{x}\in \ker \left(A^k\right), k\in \mathrm{\mathbb{N}}$.

Therefore, $A^k\overrightarrow{x}=0 \cdots \cdots \left(1\right)$.

Now,

$$\begin{gathered} A^{k+1}\overrightarrow{x}=\left(A^k·A\right)\overrightarrow{x} \\ =A\left(A^k\overrightarrow{x}\right) \\ =A·0 \cdots \left(\because by \left(1\right)\right) \\ =0 \end{gathered}$$

$\therefore A^{k+1}\overrightarrow{x}=0$.

Thus, $\overrightarrow{x}\in \ker \left(A^{k+1}\right), k\in \mathrm{\mathbb{N}}$.

Therefore,$\ker \left(A^k\right)\subseteq \ker \left(A^{k+1}\right)$

Hence, by Exercise 37, they do not need to be equal.

(b) Find relationship among images

Consider $\overrightarrow{z}\in im\left(A^{k+1}\right)$.

This implies that there exists a $\overrightarrow{y}$such that

$$\begin{gathered} A^{k+1}\overrightarrow{y}=\overrightarrow{z} \\ \Rightarrow \left(A^{k}A\right)\overrightarrow{y}=\overrightarrow{z} \\ \Rightarrow \overrightarrow{z}\in im\left(A^k\right) \end{gathered}$$

Thus, $\mathrm{im}\left({\mathrm{A}}^{\mathrm{k}+1}\right)\subseteq \mathrm{im}\left({\mathrm{A}}^{\mathrm{k}}\right)$.

Hence, by Exercise 37, they do not need to be equal.