# Block diagonal matrix and difference equation

Compute $A^j~\text{for} ~~j=1,2,....,n.$ For the block diagonal matrix $A=\begin{bmatrix} J_2(1)& \\ &J_3(0) \end{bmatrix}$,

And show that the difference equation $x_{j+1}=Ax_{j}$ has a solution satisfying $|x_{j}|\rightarrow\infty~\text{as}~j\rightarrow\infty$

Attempt

So $A^1=\begin{bmatrix} 1&1&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \end{bmatrix} ,~~A^2=\begin{bmatrix} 1&2&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix} ,~~A^j=\begin{bmatrix} 1&j&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix}\forall ~~j\geq3$

I am certain that this isn't a difficult question, but I am not sure how to apply this to the difference equation. Which is probably due to my lack of experience with them.

Note that any solution to the difference equation is given by $$x_j = A^j x_0$$ where $x_0 \in \mathbf R^n$ is arbitrary. Choosing $x_0 = e_2$ (as the second column of $A^j$ is unbounded in $j$, we have $$|Ax_j| = \sqrt{j^2 + 1} \to \infty, \quad j \to \infty.$$