linear algebra - Invertible block matrix

Could I find $A_1,A_2,A_3,A_4 \in M_n(\mathbb C)$, such that, for all $z_1,z_2,z_3,z_4\in \mathbb C$, $\det(z_1A_1+z_2A_2+z_3A_3+z_4A_4)=0$ and $\det \begin{pmatrix} A_1&A_2\\A_3&A_4\end{pmatrix}\neq 0$?

If so, could this result be extended to $m^2$ matrices $A_1,\ldots,A_{m^2} \in M_n(\mathbb C)$?

Answer

No, such matrices do not exist. Already for $n=1$, where the matrices are replaced by complex numbers, the fact that $\operatorname{det}\left(\sum_i z_i A_i\right)=0$ should hold for any $z_{1\ldots 4}\in \mathbb{C}$ implies that all $A_i=0$ (set for example $z_1=1,z_2=z_3=z_4=0$).

Increasing $n$ does not help. Indeed, zero determinant of the matrix pencil of $A_{1\ldots 4}$ means that the null spaces of $A_i$ have a nontrivial intersection$^{\sharp}$. Tensor product of any vector from this common kernel with any non-zero vector from $\mathbb{C}^2$ gives a nontrivial element of the kernel of $\left(\begin{array}{cc} A_1 & A_2 \\ A_3 & A_4\end{array}\right)$.

$^{\sharp}$Edit: show this first for a pair of matrices $A,B$ such that $\operatorname{det}(A+\lambda B)=0$ for any $\lambda\in \mathbb{C}$ (using that vanishing determinant is equivalent to linear dependence of columns of $A+\lambda B$).