Q:

Show that the system is controllable (i.e. prove P has full rank)

Given the matrix: 
$$A = \begin{pmatrix}m&1&0&0&0\\  
0&m&1&0&0&...\\
0&0&m&1&0&...\\
0&0&0&m&1&...&\\
...&&&&&1\\
0&0&0&0&...&m\end{pmatrix}$$
and vector $ B = [b_1, b_2, b_3,...,b_n]^T $
and matrix $ P = [B, AB,  A^2B, ... ,A^{n-1}B]$
show that P has full rank if and only if $b_n$ does not equal $0$
What I have done:
I have shown that if $b_n = 0$ that one row of P is all zeroes and hence the matrix does not have all linearly independent rows. 
I now that to prove the other side of the statement, that as long as $b_n$ is not equal to zero, that P is necessarily full rank. 
I have tried to solve this by proving that each row/column is necessarily independent of the others, but this hasn't brought me any success. 
Suggestions please and thank you!

A:

Denote by $J$ the nilpotent Jordan block of size $n$, then you have $A=mI+J$. Therefore, $P=[B,mB+JB,m^2B+2JB+J^2B,\ldots]$. Applying the Gaussian elementary transformations to the cloumns of $P$, we see that the matrix $[B,JB,J^2B,\ldots,J^{n-1}B]$ has the same rank as $P$. In other words, it is now sufficient to solve your problem in special case $m=0$, which is trivial.

