Find the determinant of a sum of idempotent matrices

Question

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 $$\det(A + 2B) = 0$$

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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.

Answer 1

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 $s+t\neq0$.

Answer 2

Here is a solution which I think is more natural, although it is somewhat longer than those on AoPS.

Suppose $Ax=kBx$ for some scalar $k\notin\{0,1\}$ and some nonzero vector $x$ (in your case $k=-\frac12$). Then $Ax$ and $Bx$ are either both zero or both nonzero. In the former case, clearly any linear combination of $A$ and $B$ are singular.

Suppose $Ax$ and $Bx$ are nonzero. If $x$ and $Ax$ are linearly dependent, we must have $Ax=x$. But then $Bx=\frac1kAx=\frac1kx$, which is a contradiction to the assumption that $B$ is idempotent. Therefore $x$ and $Ax$ are linearly independent. It follows that $V=\operatorname{span}\{x,Ax\}$ is a two-dimensional invariant subspace of both $A$ and $B$ (and this is the key observation in my proof). The matrix representations of the restrictions of $A$ and $B$ on $V$ with respect to the ordered basis $\{x,Ax\}$ are respectively $$ \pmatrix{0&0\\ 1&1}\text{ and }\pmatrix{0&0\\ \frac1k&1}. $$ It is now evident that every linear combination of $A$ and $B$ is singular. Using these matrix representations, one can also easily find a non-trivial solution to $(\lambda A+\mu B)v=0$. For instance, we have $$ (B-kA)\big[(k+1)Ax-kx\big]=0. $$