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

1) Let $T∈L(V,V)$ be a normal operator. Prove that $||T(v)||=||T^*(v)||$ for every $v∈V$. ($T^*$ is the adjoint of $T$)

2) Let $T$ be an operator on the finite dimensional inner product space $(V,<,>)$ and assume that $TT^*=T^2$. Prove that T is self-adjoint. (Can I simple get $T=T^*$ from $TT^*=T^2$? So there is nothing to prove)

Thank you for this two questions.

share|cite|improve this question
up vote 0 down vote accepted


$||T(v)||^2 = <Tv,Tv> = <v,T^*Tv>=<v,TT^*v>=<T^*v,T^*v> =<T^*v,T^*v>=||T^*v||^2$

share|cite|improve this answer
Why $<v,TT^∗v>=<T^∗v,T^∗v>$?-------I got it, because $<v,Tv>=<T^*v,v>$ right? – i_a_n Dec 1 '12 at 6:21

For 1: $$ \|Tv\|^2=\langle Tv,Tv\rangle =\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle=\|T^*v\|^2. $$

For 2, what you say would work if $T$ is invertible, but no one is saying it is. And you wouldn't be using the finite-dimension hypothesis.

If you look at the Schur decomposition of $T$, you have $T=VXV^*$, with $V$ a unitary and $X$ upper triangular. The equality $TT^*=T^2$ implies $XX^*=X^2$.

The diagonal entries of $XX^*$ are non-negative, and they agree with the diagonal entries of $X^2$, which are $X_{kk}^2$ (since $X$ is triangular). So the numbers $X_{kk}^2$ are non-negative, which implies that $X_{kk}$ is real for all $k$. The diagonal entries of $X^2$ are $$ X_{11}^2,X_{22}^2,\ldots,X_{nn}^2; $$ and the diagonal entries of $XX^*$ are $$ X_{11}^2,|X_{12}|^2+X_{22}^2, |X_{13}|^2+|X_{23}|^2+X_{33}^2,\ldots,|X_{11}|^2+\cdots+|X_{1,n-1}|^2+X_{nn}^2. $$ So the equality $XX^*=X^2$ implies that $X_{kj}=0$ if $j>k$. That is, $X$ is diagonal with real diagonal, so it is selfadjoint. Then $T$ is selfadjoint.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.