Question

Consider a square matrix A with $\mathit{k}\mathit{e}\mathit{r}\mathit{\left(}{\mathit{A}}^{\mathit{2}}\mathit{\right)}\mathit{=}\mathit{k}\mathit{e}\mathit{r}\mathit{\left(}{\mathit{A}}^{\mathit{3}}\mathit{\right)}$? Is $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}{\mathit{A}}^{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}{\mathit{A}}^{\mathbf{4}}\mathbf{\right)}$? Justify your answer.

Short answer

Hence, it is proved that;

$ker\left(A^3\right)=ker\left(A^4\right)$

Step-by-step solution

Step 1: Using property ker(A3)⊆ker(A4)

The kernel of a linear transformation $T\left(\overrightarrow{x}\right)=A\overrightarrow{x}$from

$$\begin{gathered} {\mathrm{ℝ}}^{\mathrm{m}}\mathrm{to}{\mathrm{ℝ}}^{\mathrm{n}}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{set}\mathrm{of}\mathrm{linear}\mathrm{system}\mathrm{A}\overrightarrow{\mathrm{x}}=0. \\ \mathrm{Also},\mathrm{the}\mathrm{property}\mathrm{below}\mathrm{is}\mathrm{true}\mathrm{for}\ker \left(\mathrm{A}\right); \\ ker\mathit{\left(}{A}^{\mathit{3}}\mathit{\right)}\mathit{\subseteq }ker\mathit{\left(}{A}^{\mathit{4}}\mathit{\right)} \end{gathered}$$

Proving ker (A2)=ker(A3)

$$\begin{gathered} \mathrm{Consider}\mathrm{a}\mathrm{matrix}\mathrm{A}\mathrm{with}\ker \left(A^{\mathit{2}}\mathit{\right)}=ker\left(A^3\right). \\ \mathrm{prove}\mathrm{the}\mathrm{equality}\ker \mathit{\left(}{A}^{\mathit{3}}\right)=\ker \mathit{\left(}{A}^{\mathit{4}}\right)\mathrm{as}\mathrm{shown}\mathrm{below}. \\ \mathrm{The}\mathrm{first}\mathrm{step}\mathrm{is}\mathrm{to}\mathrm{prove}\mathrm{thatker}\left({\mathrm{A}}^3\right)\subseteq \mathrm{ke}r\mathit{\left(}{A}^{\mathit{4}}\mathit{\right)}.\mathrm{The}\mathrm{same}\mathrm{can}\mathrm{be}\mathrm{bserved}\mathrm{from}\mathrm{the} \\ \mathrm{property}\left(1\right)\mathrm{of}\mathrm{kernel}\left(\mathrm{A}\right).\mathrm{Therefore},\ker \left({\mathrm{A}}^3\right) \\ \mathit{\subseteq }ker\mathit{}\mathit{\left(}{A}^{\mathit{4}}\right).\mathrm{This}\mathrm{concludes}\mathrm{the}\mathrm{first}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{proof}. \end{gathered}$$

Proving ker (A3)=ker(A4)

$$\begin{gathered} \mathrm{The}\mathrm{next}\mathrm{step}\mathrm{is}\mathrm{to}\mathrm{prove}\mathrm{that}\ker \left({\mathrm{A}}^4\right)\subseteq \ker \left({\mathrm{A}}^3\right).\mathrm{Let}\overrightarrow{\mathrm{x}}\in \ker \left({\mathrm{A}}^3\right). \\ \mathrm{Then}\mathrm{use}\mathrm{the}\mathrm{defination}\mathrm{of}\mathrm{kernal}\mathrm{of}\mathrm{a}\mathrm{linear}\mathrm{transformation}\mathrm{to}\mathrm{obtain}; \\ {\mathrm{A}}^4\overrightarrow{\mathrm{x}}=0 \\ \mathrm{Implies}; \\ {\mathrm{A}}^3\left(\mathrm{A}\overrightarrow{\mathrm{x}}\right)=0 \\ \mathrm{Implies}; \\ \mathrm{A}\overrightarrow{\mathrm{x}}\in \ker \left({\mathrm{A}}^3\right) \\ \mathrm{Since},\ker \left({\mathrm{A}}^2\right)=\ker \left({\mathrm{A}}^3\right) \\ \mathrm{Therefore}; \\ \mathrm{A}\overrightarrow{\mathrm{x}}\in \ker \left({\mathrm{A}}^2\right) \\ \mathrm{Again}\mathrm{use}\mathrm{it}\mathrm{and}\mathrm{the}\mathrm{defination}\mathrm{of}\mathrm{the}\mathrm{kernel}\mathrm{of}\mathrm{the}\mathrm{linear}\mathrm{transformation}\mathrm{to}\mathrm{obtain}; \\ {\mathrm{A}}^2\left(\mathrm{A}\overrightarrow{\mathrm{x}}\right)=0 \\ \mathrm{Implies}; \\ {\mathrm{A}}^3\overrightarrow{\mathrm{x}}=0 \\ \mathrm{Implies}; \\ \overrightarrow{\mathrm{x}}\in \ker \left({\mathrm{A}}^3\right) \\ \mathrm{Therefore}; \\ \ker \left({\mathrm{A}}^4\right)\subseteq \ker \left({\mathrm{A}}^3\right) \\ \mathrm{Hence}\mathrm{we}\mathrm{can}\mathrm{conclude}\mathrm{that}\boxed{\ker \left({\mathrm{A}}^2\right)\subseteq \ker \left({\mathrm{A}}^3\right)} \end{gathered}$$