Suppose we have $A, B \in M_{n,n}\left(\mathbb{C}\right)$ such that $\det A = \det B = 0$. Prove that there exists $X \neq 0 \in M_{n,n}\left(\mathbb{C}\right)$ s.t. $AX = XB = 0$
My idea on how to prove this was the following: Because $\det A = 0 = \det B = \det B^T$, all those matrix have $0$ as an eigenvalue and as such let's denote $V_A, V_T$ the set eigenvectors of $A,B^T$ relative to $0$.
Now, if we show that we can construct a matrix $X$ with elements of $V_A$ as columns and elements of $V_T$ as rows we are set.
This is because $AX = \left(AX^1, \dots, AX^n\right) = \left(0, \dots, 0\right) = 0$ and, with $X_* := X^T, XB = \left(B^TX_*\right)^T = \left(B^TX_*^1, \dots, B^TX_*^n\right)^T = \left(0, \dots, 0\right)^T = 0$
What I'm missing is how to formalize that such a matrix can always be constructed: my idea was that, in the worst case, i have $0$ "intersections" between the basis of $V_A$ and of $V_T$: for example let's suppose that $$V_A = Span\left(\left(\begin{matrix} 1 \\ 1 \\ 0\end{matrix}\right)\right), V_T = Span\left(\left(\begin{matrix}0 \\ 1 \\ 1\end{matrix} \right)\right)$$ then I can simply define a function $f$ which resolves the problem of mismatches.
What i mean is $X$ relative to the case explained before would be $$X = f\left(V_A,V_B\right) = \left(\begin{matrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}\right)$$
More in general, one could think of such a procedure in this way. Let $v_T$ be a vector in $V_T$, and define $I_T$ the set of the index s.t. $v_T^i \neq 0$. Do the same with $v_A \in V_A$ and the set $I_A$
Now, consider $i \in I_T^C$: set all columns $X^i = 0$. Do the same with the rows of $X$ considering $j \in I_A^C$ Now fill the remaining positions in a way that respects your vectors: we have thus created an $X$ s.t. it's rows are eigenvectors of $B^T$ and its columns are eigenvectors of $A$.
You probably have to ensure $rank{X} \geq 1$, do to so you just take $y \in \ker A$ and $z \in \ker B^T$, you have your matrix: $$X = y z^T$$
Then $AX = Ay z^T = 0\cdot z^T = 0 = y\cdot 0 = y (B^T z)^T = yz^T B$