We know that
$\text{rank}(\lambda I-E, F)= 2n$ (full row rank) for all $\{\lambda\in\mathbb{C} \mid \Re(\lambda) \ge 0\}$ where E=$\left(\begin{matrix} A_{n\times n} & 0_{n\times n} \\ -I_{n\times n} & 0_{n\times n} \\ \end{matrix}\right)$,F=$\left(\begin{matrix} B_{n\times p} \\ 0_{n\times p} \\ \end{matrix}\right)$, $A$ is a $n\times n$ matrix, $B$ is a $n \times p$ matrix,and $\lambda$ is a complex number with zero or positive real part.
With this assumption, I want to prove that the following equation
$$Ax + Bu = 0$$
for every possible $x$, we can always find a solution for $u$, where $x$ is a $n\times 1$ real vector and $u$ is a $p\times 1$ real vector.
I also guess that these two conditions are equivalent. However I cannot figure out the proof. At least, I want to prove that the first condition implies the second one.
(This is about the Hautus test/theorem in control theory)
(Edit: this question is only meaningful when n <= p)
Since $rank[\lambda I-E, F]=2n$ for all $\lambda\in \mathbb{C}$ with $Re(\lambda)\geq 0$ it also holds for $\lambda=0$. Hence $$rank\left[\matrix{-A & 0 & B\\\mathbb{I} & 0 & 0}\right]=2n$$ From standard rank properties we also have $$rank\left[\matrix{-A & 0 & B\\\mathbb{I} & 0 & 0}\right]=rank\left[\matrix{-A & B\\\mathbb{I} & 0}\right]$$ and it is easy to prove that this matrix has full row rank $2n$ iff $rank(B)=n$. This means that the $n\times n$ matrix $BB^T$ is nonsingular and a solution to $Ax+Bu=0$ is given by $$u^*=-B^T(BB^T)^{-1}Ax$$.