# If $A$ and $B$ are orthogonal projection matrices, how can I show that trace$(AB) \le$ rank$(AB)$?

If $$A$$ and $$B$$ are orthogonal projection matrices, how can I show that trace$$(AB) \le$$ rank$$(AB)$$?

I was using C-S inequality to get tr$$(AB) \le \sqrt{tr(A^2)tr(B^2)}$$ and I know that $$tr(A^2)=$$rank$$(A)$$. But I can't get the rank of $$AB$$.

• It might help to note that $\operatorname{rank}(AB) = \operatorname{rank}(B) - \dim(\operatorname{im}(B) \cap \ker(A))$ Oct 14 at 1:08

## 1 Answer

If either $$A$$ or $$B$$ is zero, this holds trivially.

Suppose that both $$A$$ and $$B$$ are non-zero. It suffices to show that all eigenvalues of $$AB$$ have magnitude at most equal to $$1$$. To that end, note that if $$\|\cdot\|$$ denotes the spectral norm, then we have $$\|A\| = \|B\| = 1,$$ so that $$\|AB\| \leq \|A\| \cdot \|B\| = 1$$. It follows that all eigenvalues $$\lambda$$ of $$AB$$ satisfy $$|\lambda| \leq \|AB\| \leq 1$$. Thus, if $$AB$$ has rank $$r$$ and $$\lambda_1,\dots,\lambda_k$$ (with $$k \leq r$$) are the non-zero eigenvalues of $$AB$$, then we have $$\operatorname{tr}(AB) \leq |\operatorname{tr}(AB)| = \left|\sum_{i=1}^k\lambda_i\right| \leq \sum_{i=1}^k|\lambda_i| \leq \sum_{i=1}^k 1 = k \leq r,$$ which is what we wanted.

• @Lawrence $A$ is an orthogonal projection matrix, this is not the same as saying that $A$ is "orthogonal" matrix. An orthogonal projection is a matrix $A$ satisfying $A^2 = A$ and $A = A^T$. Oct 14 at 2:54
• Oh! Sorry. I didn’t read the word projection at all! Oct 14 at 14:57
• Please tell me how rank AB=r implies number of non zero eigenvalues =r Oct 14 at 23:40
• @Lawrence Actually, for an arbitrary matrix $M$ we only know that the number of non-zero eigenvalues is at most $r$. It turns out that in this case, $AB$ is necessarily diagonalizable which means that the number of non-zero eigenvalues is exactly $r$, but this is beyond the scope of a reasonable proof. Oct 14 at 23:47
• @LawrenceMano The fact that $AB$ has at most $r$ non-zero eigenvalues is a consequence of the rank nullity theorem. Note that the dimension of the nullspace is the equal to $n-r$, which is also the geometric multiplicity of the eigenvalue $0$. The algebraic multiplicity of $0$ is therefore at least $n-r$, and the number of non-zero eigenvalues is correspondingly at most $r$. Oct 14 at 23:50