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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?

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    $\begingroup$ Have you tried squaring both sides of the equality ? $\endgroup$ – Joel Cohen Feb 6 at 4:44
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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

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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?

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  • $\begingroup$ At this point, since we have, BC = -(B'C')', can we not argue that BC has to equal zero for this to be true? $\endgroup$ – user1992460 Feb 6 at 5:15
  • $\begingroup$ Note that for symmetric matrices $S$, we have $S = S^T$. $\endgroup$ – Trevor Kafka Feb 6 at 5:17
  • $\begingroup$ Just from the symmetric argument, neither $BC=CB$ must be true, nor $BC$ is necessarily symmetric as answered here math.stackexchange.com/questions/874469/… $\endgroup$ – cr001 Feb 6 at 5:21
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    $\begingroup$ I have tried to extend your argument to get to a contradiction; take a look. Thanks $\endgroup$ – piyush_sao Feb 6 at 5:44
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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.

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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.

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    $\begingroup$ How about using the spectrum decomposition in some way? $\endgroup$ – user1992460 Feb 6 at 6:25

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