Given a polynomial matrix $A(t)$ of rank $r$, I would like to know at what complex evaluations of $t$ the rank decreases. Some research with google told me these values are sometimes called the zeros or roots of $A(s)$. I also read that these zeroes can be computed using the Smith normal form of the matrix (taking the zeroes of the product of the values on the diagonal).
1) If I am not mistaken this in particular tells us that there are only finitely many roots of $A(s)$. Can this be seen without using the Smith normal form?
2) What is known in the multivariable case (i.e. for a matrix $A(t_1,...,t_n)$)? Is there a way to compute the zeros? Are there still finitely many?
Regarding (1), this can be seen without using Smith normal form as follows. We use the fact that a matrix over a field has rank at least r iff there is an r by r submatrix with nonzero determinant.
So let's consider our polynomial matrix $A \in \mathbb C[t]^{m \times n}$, say. First, we determine $q := \max\{rank(A(z)) | z \in \mathbb C \}$. Note that this step is non-constructive, but $rank(A(z))$ is an integer between $0$ and $\min(m, n)$ inclusive, so $q$ exists.
Now, we define the set of polynomials $G := \{ \det(A'(t)) | A'$ is a q by q submatrix of $A \} \subseteq \mathbb C[t]$. Pick any $z \in \mathbb C$. By construction of $q$, and using the fact from the first paragrah, $rank(A(t_0)) < q$ iff $z$ is a zero of every element of $G$. Or equivalently, $rank(A(t_0)) = q$ iff there is an $f \in G$ such that $f(z) \neq 0$. If, for $f(t) \in \mathbb C[t]$, we denote its zero set by $V(f) := \{ z \in \mathbb C | f(z) = 0 \}$, then we have $$W := \left\{z \in \mathbb C | rank(A(z)) < q \right\} = \bigcap_{f \in G}V(f).$$ Next we show that the r.h.s is a finite set. Some of the elements of $G$ might be identically $0$, and we have $V(0) = \mathbb C$. But, by definition of $q$, we know that there is at least one $f_0 \in G$ such that $f_0$ is not identically zero. For this $f_0$,we have $|V(f_0)| \lt \infty$. More specifically, if $f_0$ is constant nonzero, then $V(f_0) = \emptyset$, and if $f_0$ is nonconstant and nonzero, then $V(f_0)$ is finite and nonempty. Using this $f_0$, we find that $W$ is finite. So we have shown that $rank(A(z))$ drops below maximum $q$ only at a finite number of complex values $z$.
In fact, $W$ is the set of common zeroes of the elements of $G$, and even of the ideal generated by $G$ in $\mathbb C[t]$. If you know a little bit about algebraic geometry, you will recognize this as an algebraic variety, albeit a one-dimensional one.
Regarding (2), this can be handled completely analogous to (1). The difference is that now we have multivariate polynomials,and $W$ will be an algebraic variety of dimension greater than $1$. Such a set can be infinite. If you want to find out more about that, read up on elementary algebraic geometry.