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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.

Question:

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.

(This is essentially exercise 5(a) in sec. 80 on p.162 of Paul R. Halmos, Finite-Dimensional Vector Spaces, but Halmos didn't assume that the dimension of $V$ is finite.)

user1551
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providence
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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 03 '13 at 05:32
  • Is this true also for infinite dimensional spaces? – ric.san Aug 31 '21 at 07:10

2 Answers2

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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 $E = E^*$.

Robert Israel
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  • How do you get $ (I−E)Ex=0$ implies $(I−E^∗)Ex=0 $ ? – A. Napster Dec 01 '17 at 13:18
  • @Napster: Zero is the unique element with zero norm, so if you know that one is zero, then by preservation of the norm, the other has to be zero as well – John Oct 18 '18 at 13:15
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Update. As $E$ is normal and idempotent, direct calculation shows that $$ (E-E^\ast E)^\ast(E-E^\ast E)=0=(E^\ast-E^\ast E)^\ast(E^\ast-E^\ast E).\tag{1} $$ Note that for any linear operator $A$, if $A^\ast A=0$, then $\langle Ax,Ax\rangle=\langle x,A^\ast Ax\rangle=0$ for every vector $x$, so that $A=0$. It follows from $(1)$ that $E-E^\ast E=0=E^\ast-E^\ast E$. Hence $E=E^\ast E=E^\ast$.

**(Old answer:)**

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.

user1551
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