Suppose $A$ is a $n\times n$ matrix with complex entries and $A^*A=A^2$.
Does it imply $A=A^*.$
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gopal dutta
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This is necessarily true if $A$ is invertible. This is also true if $A$ is normal, by the spectral theorem. – Ben Grossmann Jun 23 '15 at 17:15
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We can rewrite the equality as $$ (A^* - A)A = 0 $$ Perhaps this is helpful. – Ben Grossmann Jun 23 '15 at 17:17
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but $AB=0$ does not imply $A=0$ or $B=0$. – gopal dutta Jun 23 '15 at 17:26
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I know, but it's a start. Just adding some thoughts here for what to try. – Ben Grossmann Jun 23 '15 at 17:27
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Yes.
If $A$ is normal and satisfies the above equality, then $A$ is Hermitian, by the spectral theorem.
In the case that $A$ is non-normal, we have $$ \operatorname{trace}(A^*A) > \sum_{j=1}^n |\lambda|^2 \geq \left|\sum_{j=1}^n \lambda^2 \right| = \left|\operatorname{trace}(A^2)\right| $$ So, if $A$ is not normal, it is impossible to have $A^*A = A^2$.
We conclude that if $A^*A = A^2$, then $A$ is Hermitian, as desired.
Ben Grossmann
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