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If $A$ and $B$ are orthogonal projection matrices, show that

$$ \operatorname{tr} (AB) \le \operatorname{rank} (AB) $$

I was using the Cauchy-Schwarz inequality to get

$$\operatorname{tr} (AB) \le \sqrt{\operatorname{tr} \left( A^2 \right) \operatorname{tr} \left(B^2\right)}$$

and I know that $\operatorname{tr}\left(A^2\right)= \operatorname{rank} (A)$, but I can't get the rank of $AB$.

Rodrigo de Azevedo
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Iron Maiden 42
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  • It might help to note that $\operatorname{rank}(AB) = \operatorname{rank}(B) - \dim(\operatorname{im}(B) \cap \ker(A))$ – Ben Grossmann Oct 14 '21 at 01:08

1 Answers1

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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.

Ben Grossmann
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  • @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 $A = A^T$. – Ben Grossmann Oct 14 '21 at 02:54
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    Oh! Sorry. I didn’t read the word projection at all! – Lawrence Mano Oct 14 '21 at 14:57
  • Please tell me how rank AB=r implies number of non zero eigenvalues =r – Lawrence Mano Oct 14 '21 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. – Ben Grossmann Oct 14 '21 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 $r$. – Ben Grossmann Oct 14 '21 at 23:50
  • See this post for a proof of the diagonalizability of the product $AB$. – Ben Grossmann Oct 14 '21 at 23:54
  • Great answer. Understood your point. Thank you so very much for your elegant approach. – Lawrence Mano Oct 15 '21 at 00:26