Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $V$ be a complex vector space over $\mathbb{C}$ with inner product $\langle ,\rangle$. Let $E$ be an linear operator on $V$ such that $E^2=E$ with adjoint $E^*$. Show that $E$ is autoadjoint if and only if $EE^*=E^*E$. Show that in this case $E$ is the orthogonal projection on $W=ImE$.

If $E^*=E$ then it's clear that $E^*E=EE=EE^*$
Also, in this case, $Ev$ is the orthogonal projection of $v$ on $W$. Let $u\in W$. Then $u=Ew$ for some $w\in W$. Then: $\langle v-Ev, Ew\rangle=\langle v, Ew\rangle-\langle Ev, Ew\rangle=\langle v, Ew \rangle-\langle v, E^2w\rangle=0$. This show us that $v-Ev \in W^\perp$, therefore $Ev$ is the orthogonal projection of $v$ on $W$.

It remains to show that if $EE^*=E^*E$ then $E=E^*$.

$\langle Eu - E^*u, Eu - E^*u\rangle=\langle Eu - E^*u, Eu\rangle-\langle Eu - E^*u, E^*u\rangle=\langle E^*Eu - E^*E^*u, u\rangle-\langle EE^*u - EEu, u\rangle=\langle Eu, u\rangle-\langle u, Eu\rangle=2img(\langle Eu, u\rangle)$.
I would like to get a 0. Can someone give a tip?

share|cite|improve this question
Is $V$ of finite dimension ? – Bebop Jun 9 '13 at 19:55
A short answer is ($V$ Hilbert space of any dimension): the eigenspaces of a normal ($T^*T=TT^*$) operator are pairwise orthogonal. So $\ker E$ and $\ker(I-E)=\mbox{im} E$ give an orthogonal decomposition of $V$ where $E$ and $E^*$ are both equal to $0\oplus Id$. No matter what, all this relies on $\ker T=\ker T^*T$ (in general) whence $\ker T=\ker T^*$ if $T$ is normal. Whence $\ker E=\ker E^*$ and $\ker (I-E)=\ker (I-E^*)$ in your case. – 1015 Jun 9 '13 at 19:59
Actually, the answers on the duplicate exhibited by anon give interesting alternatives. – 1015 Jun 9 '13 at 20:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.