Prove que $E$ idempotent is self-adjoint if and only if $EE^=E^E$.

Question

Let $V$ be a finite-dimensional inner product space, and let $E$ be a idempotent linear operator om $V$, i.e. $E^2 = E$. Prove que $E$ is self-adjoint if and only if $EE^*=E^*E$.

This question has already been answered here. – Chirag Jain Apr 18 '14 at 08:30 — Chirag Jain, Apr 18 '14 at 08:30

score 0 · Answer 1 · answered Feb 20 '14 at 17:15

0

Hint: Write $V=U\oplus W$ where $E|_U$ is the identity and $E|_W$ is the zero map. Then compare $\langle u+w,E(u'+w')\rangle$ with $\langle E^*(u+w),u'+w'\rangle$. What can you conclude about $E^*u$? $E^*w$?

answered Feb 20 '14 at 17:15

Hagen von Eitzen

374,180

Prove que $E$ idempotent is self-adjoint if and only if $EE^=E^E$.

1 Answers1

Linked

Prove que $E$ idempotent is self-adjoint if and only if $EE^*=E^*E$.

1 Answers1

Linked

Prove que $E$ idempotent is self-adjoint if and only if $EE^=E^E$.