Rosen brock matrix -nullspaces-subspaces

Question

Let \begin{align}P_{\Sigma}= \left[ \begin{array}{cc} \lambda_iI-A & -B \\ C & 0 \\ \end{array} \right] \end{align} where $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$ and $C\in\mathbb{R}^{p\times n}$. When $\lambda_i$, for $i=1,\dotsc, n-p$ are distinct, then the nullspaces of $P_{\Sigma}(\lambda_i)$ for $i=1,\dotsc, n-p$ are one dimensional subspaces of $\mathbb{R}^{n+p}$. This is a statement from the paper I am reading. I do not see Why when $\lambda_i$ are distinct the associated subspace is one dimensional? Please help me relate. Thank you.

Answer 1

Note that $(9)$ follows after Assumptions 3.1 and 3.2 in Schmid-Ntogramatzidis (which is the paper OP is reading, see here). So by assumption we have $p = m$ and hence the Rosenbrock matrix $P_\Sigma(s)$ is $(p + n) \times (p + n)$. Note also that we use $B$ (rather than $-B$) and also we allow for a nonzero $D$ matrix.

The invariant zeros of the system $(A, B, C, D)$ are defined to be those $z \in \mathbb{C}$ such that the rank of $P_\Sigma(z)$ is less than its normal rank. The normal rank of $P(s)$ is $n + p$, as $\Sigma$ is assumed to be invertible, and hence the kernel is trivial (contains only the zero vector) for all $s \in \mathbb{C}$ that are not invariant zeros. For those invariant zeros, the rank of $P_\Sigma(z)$ is less than $n + p$ and hence the kernel is nontrivial. The assumption that "the zeros are distinct" means that these kernels are $1$-dimensional.