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!



How can I show $\{\mathbf{x}: \dim (\ker M_1(\mathbf{x}) \cap \ker M_2(\mathbf{x})) \geq C \}$ is an affine variety?
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.


