If $A$, $B$ idempotent and $AB=0$, then $A+B$ idempotent.

Question

We know that if $A,B$ idempotent then we have (see edit) $$(A+B)^2=A+B\implies AB=0,$$ and I'm wondering whether the converse, i.e. that if $A,B$ idempotent and $AB=0$ then $A+B$ idempotent is true. Expanding we get $$(A+B)^2=A^2+AB+BA+B^2=A+B+BA$$ so we just need to show that $BA=0$. It certainly isn't true in general that $AB=0\implies BA=0$, but I couldn't think of an example where $A$ and $B$ are idempotent. For orthogonal projections at least I think that the statement is true (by thinking geometrically), but I need to consider general projections.

My intuition is that this statement probably is true. Is this correct, and if so how can I show this?

Edit: $(A+B)^2=A+B\implies AB=-BA$, but then $BAB=-BA$ and $AB=-BAB$, so $AB=BA$ and therefore $AB=BA=0$.

Answer 1

What if $A =\left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right)$ and $B =\left(\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right)$?

$AB = 0$, but $BA = \left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}\right)$.

And $A+B = \left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right)$ is not idempotent since $(A+B)^2 = \left(\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix}\right)$.

Answer 2

Counterexample: Let $$ A = \begin{bmatrix} 0 & -1 \\ 0 & 1 \end{bmatrix} \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ Both matrices are idempotent. Then $AB = 0$ but $BA \neq 0$.

Thinking about this in terms of general projections: $A$ projects a vector along a horizontal line to the line $x = -y$, while $B$ projects along a vertical line to the line $y = 0$ (the $x$-axis). Applying $B$ and then $A$ always yields the zero vector, since $Bv$ always yields a vector with no $y$-component. But applying $A$ and then $B$ does not necessarily yield the zero vector, since $Av$ will usually have a non-zero $x$-component.