# 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?

• Have you tried squaring both sides of the equality ? – Joel Cohen Feb 6 at 4:44

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

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?

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

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.

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.

• How about using the spectrum decomposition in some way? – user1992460 Feb 6 at 6:25