If a non-zero vector $\mathbf{x}$ satisfies $A\mathbf{x}=\mathbf{0}$ and $B\mathbf{x}=\mathbf{0}$ does it imply that $A-B$ is non-invertible

Question

Here is the question rephrased in a better way:

Let $A$ and $B$ be $n\times n$ matrices ($A$ can equal $B$). If an $n\times 1$ nonzero vector $\mathbf{x}$ satisfies both homogeneous equations $A\mathbf{x}=\mathbf{0}$ and $B\mathbf{x}=\mathbf{0}$, does this imply $A-B$ is non-invertible? State a counterexample if it is false.

Here is what I have tried (I am not very proficient in linear algebra so I might be missing something important here)

$A-B$ is non-invertible $\iff$ $det(A-B)=0$

Since $A \:adj(A)= det(A)\: I$, we see that $(A-B)\:adj(A-B)=\mathbf{0}$

However, I do not know how to proceed. Am I supposed to prove the statement from the other way around? Or is the statement false? I can't find any counterexamples though. Thanks for your help!

Answer 1

I searched for some information about homogeneous systems and got an easy solution: (very similar to the one @Bungo had pointed out in the comments) $$A\mathbf{x}=B\mathbf{x}=\mathbf{0}$$ $$\implies A\mathbf{x}-B\mathbf{x}=(A-B)\mathbf{x}=0$$ $\implies A-B$ is non-invertible (since it is given that $\mathbf{x}$ is non-zero)

This is because for a homogeneous system $A\mathbf{x}=\mathbf{0}$ there only exists one solution (where $\mathbf{x}$ is zero) if $A$ is invertible.