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.

1 Answer

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 $$|Ax_j| = \sqrt{j^2 + 1} \to \infty, \quad j \to \infty.$$