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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.

2 Answers 2

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$||T(v)||^2 = = == ==||T^*v||^2$

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    Why $=$?-------I got it, because $=$ right?2012-12-01
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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.