Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$.
Let $N$ be a linear transformation with $N^*N = NN^*$. $T$ is a linear transformation. Show that if $NT = TN$ then $T^*N = NT^*$.
Here is what I did, is it correct?
Since $NN^* = N^*N$ , there is an orthonormal basis $v_1,...,v_n$ such that $v_1,...,v_n$ are eigenvectors of $N$ (Theorem).
It is enough to prove $T^*N(v_i) = NT^*(v_i)$ for all $i$. Let's try to prove it for example for $v_1$.
Suppose first $N(v_1) = av_1$. Then $N^*(v_1) = \bar av_1$. We have that $T^*N(v_1) = NT^*(v_1)$ is equivilent to $aT^*(v_1)=N(T^*(v_1))$ which is the same as saying $T^*v_1 $ is an eigenvecor for $N$ with eigenvalue $a$. Now: $NT = TN \rightarrow (NT)^* = (TN)^* \rightarrow T^*N^* = N^*T^* \rightarrow T^*N^*(v_1) = N^*T^*(v_1) \rightarrow \bar aT^*(v_1) =N^*(T^*(v_1))$.
Therefore $T^*v_1$ is an eigenvector for $N^*$ with eigenvalue $\bar a$. Therefore it is an eigenvector for $N$ with eigenvalue $a$ , which is what we wanted. Is the proof correct? If so, is there an easier proof? I feel like the Spectral Theorem is "Too strong" for this question and there should be something simpler
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Omer
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Why is $V$ finite-dimensional? – user10354138 Jun 01 '19 at 16:08
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@user10354138 It is given, I forgot to mention that. I'll edit the post. Given that $V$ is finite-dimensional, is the proof correct? – Omer Jun 01 '19 at 16:09
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This post seems relevant – Ben Grossmann Jun 01 '19 at 16:41
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By the way, the infinite-dimensional version is known as Fuglede's theorem – Ben Grossmann Jun 01 '19 at 16:42
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@Omnomnomnom Thanks, but It doesn't really help me to know if the proof I presented in this post is correct, That is my question – Omer Jun 01 '19 at 16:45