User:IssaRice/Linear algebra/Rank of polynomial matrix is constant everywhere except possibly at finitely many points

This is Corollary 6.2 in Linear Algebra Done Wrong.

I find the proof in the book pretty unclear, so I want to write up a clearer proof.

Corollary statement: Let $A(x)$ be an $m\times n$ polynomial matrix (i.e. a matrix whose entries are polynomials of $x$). Then $x\mapsto \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)$ is constant everywhere, except possibly at finitely many points, where the rank is smaller.

Proof:

We first show that $x\mapsto \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)$ takes on a maximum value, which we will call $r$. To show that $r$ exists, we start at $r:=min\{m,n\}$ (this is the largest rank that an $m\times n$ matrix can have, so it is safe to start here). If there exists some $x$ such that $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)=r$, then we have found our $r$. If not, we replace $r$ by $r-1$ and continue. After finitely many steps, we either return a value or hit $0$ (because the rank of a matrix cannot be negative). So $r$ exists.

Now we have two cases:

• $r=0$: For every $x$, we have $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)\le r=0$ from what we showed above. Since the rank of a matrix is a non-negative integer, we also know that $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)\ge 0$. Combining these two, we must have $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)=0$ for every $x$, so in this case $x\mapsto \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)$ is identically zero.
• $r>0$: Since $x\mapsto \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)$ takes on the maximum value $r$, we can find some point $x_0$ such that $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x_0)=r$.