Showing a system is always controllable?
There's a very simple way to check controllability, indeed if you define the reachability matrix
$$
R = \begin{pmatrix}B &amp; AB &amp; \dots &amp; A^{n-1}B\end{pmatrix}
$$
then the reachable subspace is the image of R. Hence to check complete controllability you just have to check that $R$ is full rank. 
First, I think there's an error in the question, $B$ should be $nk\times k$ as you defined it.
Anyway, in your case $R$ will be a $nk\times nk$ matrix, the first $k$ columns are given by the columns of $B$, which are linearly independent by definition of $B$ (in fact you have the identity as the first block), thus $rankR\ge k$. The second $k$ columns are given by $AB$ which has the form
$$
AB = \begin{pmatrix}\star\\I_k\\0_{k\times k}\\\vdots\\0_{k\times k}\end{pmatrix}
$$
where $\star$ means a block you dont care about. Note that the columns of the matrix
$$
\begin{pmatrix}B &amp; AB\end{pmatrix} = \begin{pmatrix}I_k &amp; \star \\ 0_{k\times k} &amp; I_k \\ 0_{k\times k} &amp; 0_{k\times k} \\ \vdots &amp; \vdots \\ 0_{k\times k} &amp; 0_{k\times k}\end{pmatrix}
$$
are linearly independent reguardless of $\star$. Hence $rankR\ge 2k$. In the same way, for the term $A^k B$ you get the form
$$
A^k B = \begin{pmatrix}\star_1 \\ \star_2 \\ \vdots\\ \star_{k}\\ I_k\\0_{k\times k}\\ \vdots \\ 0_{k\times k}\end{pmatrix}
$$
where the Identity block is at position $k+1$. hence for any $0\ge k&lt;j\ge n-1$ $A^{k}B$ and $A^jB$ have linearly independent columns, hence you obtain thart $R$ is full rank an the system is completely controllable