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Suppose we have two matrices $A, B$ where $A \in \mathbb R^{n \times n}$ and $B \in \mathbb R^{n \times m}$. Further assume the rank of $[\lambda I - A, B]$ is $n$ for any $\lambda \in \mathbb C$ where $[\lambda I -A, B]$ denotes concatenating the columns.

I am wondering whether we can choose a continuous function $v\colon \mathbb C \to \mathbb C^{n+m}$ by $\lambda \mapsto v(\lambda)$ where $v(\lambda) \in \ker([\lambda I -A, B])$. We need $v$ not to be identically $0$.

user1101010
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  • What do you mean by "choose a continuous function"? The kernel is non-trivial, so clearly such a (non-zero) exists. – Ben Grossmann Jun 09 '18 at 01:44
  • @Omnomnomnom: For each $\lambda$, yes, the kernel $\ker([\lambda I - A, B]$ is nontrivial and we can choose a vector $v$. This $v$ is dependent on $\lambda$. So for each $\lambda$, we can choose a $v(\lambda)$. But can we get a mapping $\lambda \mapsto v(\lambda)$ that is continuous? – user1101010 Jun 09 '18 at 02:18
  • This answer seems like a step in the right direction. Hope that helps – Ben Grossmann Jun 09 '18 at 02:51

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