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In this question, $A^\ast$ is the conjugate transpose of $A$. I am asked to show that if $A$ is a projection matrix, that $A$ is Hermitian if and only if $A=AA^\ast A$. One direction is easy--if $A$ is Hermitian, the result is trivial. So what about the converse? The assumption that $A=AA^\ast A$ and that $A$ is a projection gives us some identities: $$A^2=A,$$ $$A^\ast=A^\ast AA^\ast,$$ $$A=A(A^\ast)^2A,$$ etc.... But I am at a loss. How can I use these to prove the converse: If $A=AA^\ast A$ and $A$ is a projection, then $A$ is Hermitian?

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    Wild idea: maybe take the singular value decomposition of $A$? – 2017-02-15
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    Presumably: $A$ is a projection means that $A^2 = A$. Is that so? – 2017-02-15
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    @Omnomnomnom, indeed it is one of my identities. – 2017-02-15
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    @CheeHan that's doomed to fail; SVD doesn't play nice with squaring. – 2017-02-15
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    Schur decomposition seems like a good idea here, since it plays well with both squaring and conjugate-transposition. – 2017-02-15

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Consider the inner product $\langle C,D\rangle=tr(CD^*)$.

Now $AA^*=AA^*AA^*$ and $A^*A=A^*AA^*A$. Thus, $A$, $A^*$, $AA^*$ and $A^*A$ are projections. So their traces are equal to their ranks, which are also equal.

Finally, $\langle A-A^*,A-A^*\rangle=\langle A,A\rangle-\langle A,A^*\rangle-\langle A^* ,A\rangle+\langle A^*,A^*\rangle=$

$=tr(AA^*)-tr(AA)-tr(A^*A^*)+tr(A^*A)=0$. So $A=A^*$.

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    Nifty! I can't believe I missed that first trick. Also, a much neater use of the trace there. – 2017-02-15
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    @Omnomnomnom Don't worry, you rarely miss a trick. Best Regards. – 2017-02-16
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Using $A=AA^*A$, we can say that the restriction of $AA^*$ to the image of $A$ is the identity map. On the other hand, the kernel of $AA^*$ is the kernel of $A^*$, which is the orthogonal complement to the image of $A$. Conclude that $AA^*$ is the orthogonal projection onto the image of $A$.

Now, we may note that both $AA^*$ and $A$ are projections onto an $r$-dimensional subspace (where $r = rk(A)$). It follows that $tr(A) = tr(AA^*) = r$. However, since $A$ satisfies $tr(AA^*) = tr(A)$, $A$ must be normal (this follows from Schur decomposition). Since $A$ is normal with real eigenvalues, it must be Hermitian (this follows from the spectral theorem).

Thus, $A$ is Hermitian.

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    Can I get some feedback with that downvote? – 2017-02-15