$A\in \Bbb R^{n\times n}$ and $B \in \Bbb R ^{n\times m}$. Show that $\exists p\in\Bbb R ^m$ s.t. $(A,Bp)$ is controllable iff $(A,B)$ is controllable Let $A\in \Bbb R^{n\times n}$ and $B \in \Bbb R ^{n\times m}$. Show that there exists a vector $p\in \Bbb R ^m$ such that $(A,Bp)$ is controllable iff $(A,B)$ is controllable.
Here when I say $(A,B)$ is controllable i mean that the system $\dot{x}=Ax +Bu$ is controllable.
First I show that if $(A,B)$ is controllable then $(A,Bp)$ is controllable also. One characterization of $(A,B)$ being controllable is that there exists a row vector $\eta \in \Bbb R^{n}$ such that $\eta e^{At}B=0$ for all $t\in [0,T]$ where $T$ is some time greater than zero. Now using this same $\eta$ we find that
$$\eta e^{At} (Bp) = (\eta e^{At} B)p =0$$ 
and thus $(A,Bp)$ is also controllable for any $p \in \Bbb R^m$ actually (is this right, it doesn't seem right, for instance if $p$ is all zeros the system is obviously not controllable). So this result is probably already not right, however I cannot find the mistake. 
For the reverse implication I am completely stuck. I am sorry if I made any silly mistakes here, the whole topic of controllability is still new to me. If anyone could help me out I would be very thankful!
EDIT: Reading the wording of the original question again, I think we can just take $p$ to be a column vector of all ones. Then $(A,B)$ is controllable iff $(A,Bp)$ is controllable, so we have shown the existence of the vector $p$. Is this too simple?
 A: Here's a more general approach based on the Hautus test:
Assume $w^T_k, \: k = 1,...,n$ to be a right eigenvector of $A$. Further, let $B = [b_1, ..., b_m]$. If there is an eigenvector $w^T_k$ with $w^T_k b_i = 0$ for all $i=1,...,m$, then for all $p \in \mathbb{R}^m$ we have that $w_k^TBp = 0$. The Hautus test states that $\text{rk}[\lambda I -A, B] = n \leftrightarrow (A,B)$ controllable, where $[\lambda I - A, B]$ has full column rank iff $w^T[\lambda I - A, B] \neq 0^T$ for all $w^T$. 
Hence, we have that
$$ 
\begin{align}
& (A,Bp) \text{ controllable}\leftrightarrow \text{rk}[\lambda I - A, Bp] = n  \leftrightarrow (\exists p) \: w_k^TBp \neq 0 \\ \rightarrow & (\forall w_k^T)(\exists i) \: w_k^Tb_i\neq 0 \leftrightarrow \text{rk}[\lambda I - A, B] = n \leftrightarrow (A,B) \text{ controllable}.
\end{align}
$$
The other direction can be proven by assuming that there is an eigenvector $w^T_k$ such that for all $p$ we have $w_k^TBp=0$ which requires that $w^T_kB = 0^T$. The contrapositive will then yield
$$ (A,B) \text{ controllable} \rightarrow (A,Bp) \text{ controllable}. $$
