# Proof of normal operator and self-adjoint operator

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.

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

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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.

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