I am trying to understand the relation between the controllability matrix and the PHB test. More precisely, I want to know the relation between the image space of the controllability matrix (also known as the reachable space) and the image space of the PHB matrix.
Let $C = [B, AB , A^2B , \ldots , A^{n-1}B]$ be the controllability matrix and $[A-\lambda I , B]$ be the PHB matrix. We know that the system is controllable iff $\operatorname{rank}(C) = n$ or $\operatorname{rank}[A-\lambda_i I , B] = n$ for every eigenvalue $\lambda_i$ of $A$ ($i=1,\ldots,n$).
Question is: Is it true that $\operatorname{Im}(C) = \operatorname{Im}([A-\lambda_1 I , B]) \cap \operatorname{Im}([A-\lambda_2 I , B]) \cap \ldots \cap \operatorname{Im}([A-\lambda_n I , B])$?
Edit: I've just seen a (deleted) answer to my question and I want to use this opportunity to clarify my question. It is clear to me that my statement holds if the system is controllable (since it implies that $\operatorname{rank}(C) = \operatorname{rank}[A-\lambda_i I, B] = n$, $\forall i$). My question is more general in the sense that I also want to consider the case where $\operatorname{rank}(C)<n$. Thank you!
I believe the claim to be false. Please check this counter-example. Consider the linear control system with,
$$ A = \begin{pmatrix} 1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 0\end{pmatrix}\qquad B = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. $$
Notice that $B \in \mathrm{null}(A)$ so the system cannot be controllable. Check that the controllability matrix is,
$$ [B, AB,A^2B] = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{pmatrix} $$
so it follows that,
$$ \mathrm{im}(C) = \mathrm{span}\left\{B\right\} $$
Now let us compute the PBH matrices. The eigenvalues are $1$ (repeated twice) and $0.$ So,
$$ \begin{aligned}{} [A - 1 I, B] &= \left(\begin{array}{ccc|c} 0 & -1 & 0 & 1\\ 0 & 0 & -1 & 1\\ 0 & 0 & -1 & 1 \end{array}\right)\\ [A - 0 I, B] &= \left(\begin{array}{ccc|c} 1 & -1 & 0 & 1\\ 0 & 1 & -1 & 1\\ 0 & 0 & 0 & 1 \end{array}\right) \end{aligned} $$
Clearly $\mathrm{im}([A - 0 I, B]) = \mathbb{R}^3$ so it suffices to compute the image of the former matrix. Observe,
$$ \mathrm{im}([A - 1 I, B]) = \mathrm{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}\right\} $$
It follows that the intersection of images of the PBH matrices do not agree with the image of the controllability matrix.