Does there exists $n\in \Bbb{N}$ such that $A^n=I_2,$ where $A=\begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix}?$

Question

Here's the matrix

$$A= \begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix},$$ where $I_2$ represents the $2\times 2$ identity matrix.

My trial

I'm claiming that there do not exists $n\in \Bbb{N}$ such that $A^n=I_2.$ Suppose it exists, then

$$ \begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix}^n=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}.$$

I know that $$ \begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix}^2=\begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix}$$ but I don't see a contradiction coming. Any help, please?

Answer 1

No, because your matrix $A$ is singular.

Particularly, if you multiply the matrix with all zeroes in its first row, the resulting matrix has again all zeroes in its first row:

$$A\cdot \begin{bmatrix} x & y\\ v & w \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 3 & 1 \end{bmatrix}\cdot \begin{bmatrix} x & y\\ v & w \end{bmatrix}= \begin{bmatrix} 0 & 0\\ ? & ? \end{bmatrix}$$

so

$$A.A = \begin{bmatrix} 0 & 0\\ ? & ? \end{bmatrix}$$

and, similarly

$$A.A.A = \begin{bmatrix} 0 & 0\\ ? & ? \end{bmatrix}$$

and so on, and you want to obtain $1$ in the upper left corner.