Let A $\in$ $\mathbb{R}^{n\times n}$ and B $\in$ $\mathbb{R}^{n\times m}$. Show that (A , B) is controllable if and only if (A + $\alpha$I, B) is controllable for all $\alpha \in \mathbb{R}$.
C = $[B \ \ \ \ \ AB \ \ \ \ \ A^{2}B \ \ .....\ \ A^{n-1}B]$
C' = $[B \ \ \ \ \ [A + \alpha I ]B \ \ \ \ \ [A + \alpha I ]^2 B \ \ .....\ \ [A + \alpha I ]^{n-1}B]$
I understand that if the rank of C = n; then (A,B) is controllable. How do I show that rank of C' is also 'n'?
Here is an outline of a way to solve this. First, notice that $rank(C) = n$ is equivalent to saying that for any $x$ there exists a vector $v$ such that $x = Cv$. So for $n=3$ this means that there exist a $v=(v_1, v_2, v_3)$ such that, \begin{align} x = Cv = \begin{bmatrix}B & AB & A^2B\end{bmatrix}\begin{bmatrix}v_1 \\ v_2 \\ v_3 \end{bmatrix} \end{align}
Using this fact, you want to show that for any $x$ there exists a $w = (w_1, w_2, w_3)$ such that, \begin{align} x = C'w = \begin{bmatrix}B & (A+\alpha I)B & (A+\alpha I)^2B\end{bmatrix}\begin{bmatrix}w_1 \\ w_2 \\ w_3 \end{bmatrix} \end{align} Notice that this can be written as, \begin{align} x = Cw' = \begin{bmatrix}B & AB & A^2B\end{bmatrix} \underbrace{\begin{bmatrix} w_1 + \alpha w_2 + \alpha^{2}w_3 \\ w_2 + 2\alpha w_3\\ w_3 \end{bmatrix}}_{:= w'} \end{align} The question now is if we can find $w$ such that $w' = v$ since we know that $x=Cv$. We can write \begin{align} \begin{bmatrix}v_1 \\ v_2 \\ v_3 \end{bmatrix} &= \begin{bmatrix} w_1 + \alpha w_2 + \alpha^{2}w_3 \\ w_2 + 2\alpha w_3\\ w_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 1 & \alpha & \alpha^{2} \\ 0 & 1 & 2\alpha \\ 0 & 0 & 1 \end{bmatrix}}_{:= M}\begin{bmatrix} w_1\\ w_2\\ w_3 \end{bmatrix} \end{align} Notice, that the matrix $M$ is invertible, and therefore we can always find a vector $w$ such that $w' = v$. Therefore, we also know that for any $x$ there exists a $w$ such that $x = C'w$, which means that $C'$ has rank $n$. Once you have shown that $rank(C) = n \implies rank(C') = n$, showing that $rank(C') = n \implies rank(C) = n$ is easy.