Let $A, B \in \mathcal{M}_n (\mathbb{R})$ such that $A^2 = A$ and $B^2 = B$. If $\det(2A + B)=0$, prove that $$\det(A + 2B) = 0$$
My attempt:
I know that both matrices are diagonalizable and their all of their eignvalues are either $0$ or $1$. I can assume that one of them, lets say $A$ is in its diagonalized form but this doesn't help very much as long as I don't have any information about $B$'s form. I was also thinking about multiplying the matrix $2A+B$ with another matrix to use that the matrices are idempotent and simplify things, but I haven't managed to get anything too useful.
Another good idea of mine is to take a nontrivial vector $v$ such that
$(2A+B)\cdot v=0$ and then $2A\cdot v+B\cdot v=0$ and now my idea was to multiply the left hand side by $ \overline{{v^t}}$ and use that $A^2=A$ and $B^2=B$ but I don't know that $A=A^t$ so I can't form a scalar product of 2 vectors.
Best Answer
Here's a straightforward solution.
$A$ and $B$ are projections.
Let $\text{Im}(A)=S,\text{Ker}(A)=T,\text{Im}(B)=U,\text{Ker}(B)=V$.
So you can interpret the conditions in the problem as follows:
$\mathbb{R}^n=S\oplus T=U\oplus V$
There exist $s\in S, t\in T, u\in U, v\in V$ such that $s+t=u+v\neq0$ and $2s+u=0$.
$\\$
It follows that $u=-2s$ and $v=3s+t$.
Now, $as+bt=(a-3b)s+(3bs+bt)$, where $(a-3b)s\in U$ and $3bs+bt\in V$.
So look for $a,b\in\mathbb{R}$ such that $A(as+bt)+2B(as+bt)=as+2(a-3b)s=(3a-3b)s=0$.
We can set $a,b=1$. We have already stated that $s+t\neq0$.