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 $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x_0)=r$. By Theorem 6.1, there exists an $r\times r$ submatrix $B(x_0)$ of $A(x_0)$ with non-zero determinant, i.e. a non-zero minor of order $r$. Call this minor $M(x_0)=det(B(x_0))\ne 0$, and consider the corresponding submatrix $B(x)$ and minor $M(x)$ of $A(x)$. Since $M(x)$ is the determinant of an $r\times r$ polynomial matrix, we see that $x\mapsto M(x)$ is a polynomial.

We note that if $M(x)=0$, then the submatrix $B(x)$ is not invertible, so the rank of $A(x)$ may be smaller than $r$. If $M(x)\ne 0$, then the submatrix $B(x)$ is invertible, so the rank of $A(x)$ must be at least $r$ by Theorem 6.1 (and so it must be equal to $r$, since $r$ was chosen as the maximum rank).

Since $M(x_0)\ne 0$, the polynomial $x\mapsto M(x)$ is not identically zero, so it can be zero only at finitely many points. Thus at finitely many points the rank of $A(x)$ may be smaller than $r$, but everywhere else we must have $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A(x)=r$.