Let $M_1(\mathbf{x})$ and $M_2(\mathbf{x})$ be $m$ by $m$ matrices with each entry a homogeneous form in $\mathbb{C}[x_1, \ldots, x_n]$.
I would like to show that
$$
\{ \mathbf{x} \in \mathbb{A}^n_{\mathbb{C}}:\dim (\ker M_1(\mathbf{x}) \cap \ker M_2(\mathbf{x})) \geq C \},$$
for any $C > 0$,
is 1) an affine variety (zero set of some polynomials), 2) it is defined by homogenous forms.
Hence, an affine cone over some projective variety.
This is easy to see if $M_1 = M_2$, but I was not sure how to proceed for this more general case. Any comments appreciated!
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1 Answer
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Evidently, $\mathrm{Ker} M_1\cap \mathrm{Ker} M_2=\mathrm{Ker}(M_1,M_2)$, where $(M_1,M_2)=:M$ is the $2m\times m$ matrix obtained by putting $M_1,M_2$ together ($k$-th column of $M$ consists the of the $k$-th column of $M_1$ followed by the $k$-th column of $M_2$).
Then $\mathrm{dim}\,\mathrm{Ker} M=2m-r$, where $r$ is the rank of $M$. So $\mathrm{dim}\,\mathrm{Ker} M\geq C$ is equivalent to $r\leq 2m-C$, and this means that determinants of all submartices of size $(2m-C+1)\times(2m-C+1)$ are zero. These determinants are polynomials defining your affine vriety.
answered Aug 14, 2021 at 13:16
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$\begingroup$ Do you mean $M$ is a matrix with $m$ columns and $2m$ rows? $\endgroup$ Commented Aug 14, 2021 at 16:58
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1$\begingroup$ @Takeshi Gouda: yes. I corrected. $\endgroup$ Commented Aug 14, 2021 at 17:05