Let $p(t), q(t) ∈ \mathbb C[t]$ be relatively prime, $A ∈ M_n(\mathbb{C})$. Show that $\operatorname{rank}(p(A))+\operatorname{rank}(q(A)) ≥ n$.

Question

Let $p(t), q(t) ∈ \mathbb C[t]$ be relatively prime, $A ∈ M_n(\mathbb{C})$. Show that $\operatorname{rank}(p(A))+\operatorname{rank}(q(A)) ≥ n$.

I have been stumped on this question for quite awhile. Could someone please enlighten me in regards to a fitting theorem? I'm assuming this is related to Bilinear and Quadratic forms but I couldn't find anything in regards to relatively prime functions in Friedberg's textbook.

Answer 1

Using $1=pr+qs$ one can easily deduce $\operatorname{ker}(p(A)) \subset \operatorname{Im}(q(A))$, thus we get

$$n = \operatorname{rank}(p(A)) + \dim \operatorname{ker}(p(A)) \leq \operatorname{rank}(p(A))+\operatorname{rank}(q(A)).$$

This proof gives you for free: Equality holds iff $\operatorname{ker}(p(A)) \supset \operatorname{Im}(q(A))$ iff $(pq)(A)=0$.

Answer 2

We are going to show that $Im(p(A))+Im(q(A))=C^n$.

Since $p,q$ are relatively prime, there exists $u(t),v(t)\in C[t]$ such that $up+vq=1$. This implies that for every $x\in C^n$, $p(A)(u(A(x)))+q(A)(v(A(x)))=x$ since $p(A)(u(A(x)))\in Im(p(A))$ and $q(A)(v(A(x)))\in Im(q(A))$, we deduce that $Im(p(A))+Im(q(A))=C^n$ this implies that $dim(Im(p(A))+dim(Im(q(A))-dim(Im(p(A))\cap Im(q(A)))=n$ since $dim(Im(p(A))=rank(p(A))$ and $dim(Im(q(A))=rank(q(A))$, we deduce that $rank(p(A))+rank(q(A))\geq n$ done.