Show that if A = B + C, then BC = 0 A,B,C are n $\times$ n symmetric and idempotent matrices.
The question is:
If
$$
\textbf{A = B + C }
$$
then show
$$
\textbf{BC = 0 }
$$
I'm not sure where to start?
 A: Here's my expansion of Brian's argument:
$A^{2}=A$, $B^{2}=B$, $C^{2}=C$, hence
$A^{2}=(B+C)^{2}\rightarrow BC+CB=0 \implies BC+(BC)^{T}=0$ Hence $BC$ is an anti-symmetric matrix, thus it has only imaginary eigenvalues.
$ BC+CB=0 \implies B(BC+CB)C=0 \implies BC + (BC)^{2}=0$,
If $BC\neq0$ then $BC + (BC)^{2}=0$ is the minimal-polynomial for $BC$, hence $-1$
is an eigenvalue of $BC$, which is a contradiction since $BC$ can have only imaginary eigenvalues
A: Recall that a matrix $M$ is idempotent if $M^2=M$. 
Thus, since $A=B+C$ where $A$, $B$, and $C$ are idempotent implies that
$$
B+C=(B+C)^2=B^2+C^2+BC+CB=B+C+BC+CB
$$
Simplifying this equation gives
$$
0=BC+CB=BC+(B^\top C^\top)^\top
$$
How might we proceed from here?
A: We can also proceed from the step $0=BC+CB$ this way:   
Multiply both sides by $B$ we have 
$0=B^2C+BCB=BC+BCB=BC(I+B)$, 
but $I+B$ is full rank matrix  (it doesn't have zero eigenvalue because for $B+I$ all eigenvalues: - $0,1$ for idempotent matrix $B$,  were shifted by $+1$ with reference to eig. of $B$) so $BC$ annihilates full rank matrix,
hence  $BC$ has to be zero matrix.
A: Note that the claim is not true over a field of characteristic $2$: consider the matrices $B=C=\mathbf 1$ and $A=\mathbf0$.
This counterexample shows that you're not going to be able to prove the claim simply by performing the sort of algebraic manipulations that work over all fields, such as $(XY)^T=Y^TX^T$ and so on. You're going to need some argument that involves the base field. For example, see zimbra314's answer in terms of complex eigenvalues.
