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I have been trying to solve this problem for quite a while. I am still unsure of whether any of the avenues I have pursued have been of any use. Any advice will be much appreciated.


Let $V$ be a finite-dimensional inner product space, and let $E$ be an idempotent linear operator on $V$. Prove that if $EE^* = E^*E$ then $E$ is self-adjoint.

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Well, if you want to use a big hammer, the spectral theorem says that normal operators are unitarily diagonalizable. So it suffices to consider the case where $E$ is diagonal, which should be pretty easy. – Nate Eldredge Mar 3 '13 at 5:32
@NateEldredge Is there a smaller hammer? – providence Mar 3 '13 at 5:49
up vote 8 down vote accepted

If $E$ is normal, $\| E x \| = \|E^* x \|$ for all $x$. Similarly, $I-E$ is normal, so $\|(I-E) x\| = \|(I - E^*)x \|$. In particular, since $(I-E)Ex = 0$, $(I-E^*)Ex = 0$, i.e. $E^* E = E$, and similarly since $E(I-E)x = 0$, $E^*(I-E)x = 0$, i.e. $E^* E = E^* $. But those together say $E = E^*$.

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The equality $EE^\ast=E^\ast E$ implies that $(E+E^\ast-I)(E-E^\ast)=0$. If you can show that $E+E^\ast-I$ is invertible, you are done.

Suppose $(E+E^\ast -I)v=0$. Left-mulitply the equation by $E^\ast$, we get $E^\ast Ev=0$. Hence show that $Ev=0$. Since $EE^\ast=E^\ast E$, similarly, show that $E^\ast v=0$. Now, consider the equation $(E+E^\ast -I)v=0$ again and show that $E+E^\ast -I$ is invertible.

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How do you get $Ev=0$? I don't see that... – user59083 Jun 5 '14 at 0:10
@5space $E^\ast Ev=0\Rightarrow v^\ast E^\ast Ev=0$. – user1551 Jun 5 '14 at 9:49

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