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Let $ V $ be an finite-dimensional inner product space over $ \mathbb{C} $.

Let $ T : V \to V $ be a linear transformation and assume that for every $ v $, $ \lVert Tv \rVert =\lVert T^*v\rVert$. I want to prove that $ T $ is a normal transformation.

My effort:

Suppose, to the contrary, that $ T $ is not normal and therefore there exists a $ \vec{u} $ such that $ TT^*(\vec{u}) \neq T^*T(\vec{u}) $.

$ \lVert Tv\rVert^2= \langle Tv, Tv \rangle = \langle v, T^*Tv\rangle$

$ \lVert T^*v \rVert^2=\langle T^*v, T^*v\rangle=\langle v, TT^*v\rangle$

which means $ \lVert Tv\rVert^2 - \lVert T^*v \rVert^2==\langle v, (T^*Tv -TT^*v) \rangle$

I now want to show that $ T^*Tv- TT^*v \not \perp u $, but I don't really know how? Could you give me a $ \textbf{hint} $?

talopl
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  • $A=T^* T- T T^*$ is self-adjoint, so choose $u$ to be an eigenvector of $A$ and not just any $u$. – Chad K Apr 23 '23 at 21:16

1 Answers1

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From the last equation you have derived you get that the RHS inner product is zero for all $v\in V$.

The hint is the term "Polarisation" to finish your proof,
cf wikipedia or math SE's "Polarization identity in Hilbert space" if you'd want to stay on this site.

Furthermore, the post If $(Tx \mathbin{|} x) = 0$ for all $x$ then $T = 0$ is a seamless sequel of your post.

Hanno
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  • My linear algebra course is not in English, and so I am not familiar with the term "Polarization" $ \textbf{ in English } $. Could you, please, direct me to a page where I could read about it? – talopl Apr 23 '23 at 21:48
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    @talopl Polarization Identity is a standard result. You can find it in Wikipedia. – Kavi Rama Murthy Apr 23 '23 at 23:07
  • As I see no reaction may I ask @talopl if this is a satisfactory hint/answer ? – Hanno Apr 30 '23 at 17:56
  • @Hanno Unfortunately, no. I tried to expand what I got using the Polarization Identity, and yet all I got were messy calculations. – talopl Apr 30 '23 at 18:54
  • @talopl I'd propose you first get yourself acquainted with the polarisation identity itself, e.g. by going through the given references/links, for which you asked in your first comment. Secondly, you will see that it directly applies to your reasoning above as $0=\lVert Tv\rVert^2 - \lVert T^v \rVert^2=\langle ,v,, (T^T -TT^*)v,\rangle$ for all $v\in V$. – Hanno May 01 '23 at 04:20