Is containment of images of linear maps Zariski closed?

Question

Consider the set $$\\\{ (A,B) \in \mathbb{P}^{n\times n-1} \times \mathbb{P}^{n\times n -1} : \text{im}(A) \subseteq \text{im}(B)\}.$$ That is, this is the set of pairs of square matrices $(A,B)$ so that the image of $A$ is contained in the image of $B$. Is this Zariski closed? I would be happy if this were at least true over an algebraically closed field.

I tried to write down explicit equations for this, but my first attempt would involved using determinants of submatrices of $A$, and would fail if $A$ was singular. I had thought that this would follow from the projection theorem for for projective varieties, but am unsure.

Answer 1

I am making my comment an answer. The specified set is not Zariski closed. If it were, then its intersection with every Zariski closed subset $C$ would be relatively closed in $C$. But now let $C$ be the curve, a copy of the affine line, where the first component $A$ is held fixed as the identity $n\times n$ matrix, and the second component is a varying diagonal matrix whose first $n-1$ diagonal entries all equal $1$, yet whose last entry $t$ varies in a copy of the affine line. The intersection of $C$ with the specified set is a non-closed, dense Zariski open in $C$, namely the open subset where $t$ is invertible.

Answer 2

Another way to see your set is not closed: it is evidently not the whole space of pairs, but yet it contains the open dense set $U\times U$ where $U \subset \mathbb{P}^{n^2-1}$ consists of (classes of) invertible matrices. So your set is different from its closure.