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Question:

Let $V$ be a finite dimensional vector space with $\langle\ ,\, \rangle$ inner product and let $T$ be a linear operator on $V$. Suppose $\|u\| = \|v\| \implies \|Tu\| = \|Tv\|$ $\forall u, v \in V$.

Show that $\langle Tu, Tv\rangle = \langle u, v\rangle\ \forall u, v \in V$.

I think I almost got it. Let $B=\{v_1, ..., v_n\}$ be an orthonormal basis for $V$. Then all I must do is to show that $\langle Tv_i, Tv_j\rangle = \delta_{ij}$. I have done it for $i\neq j$ (using the polarization identity) but i'm having trouble on showing that $\forall j\ \|Tv_j\| = 1$.

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    $\begingroup$ I think you mean you must show that $\langle Tv_i, Tv_j \rangle = \delta_{ij}$. $\endgroup$ – al0 Jun 7 '13 at 18:55
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    $\begingroup$ As written, this cannot be right: Consider the operator defined by $Tv=2v$ for all $v$. This is clearly a linear operator, and from $||u||=||v||$ it follows that $||2u||=||2v||$. However we do not have $\langle 2u,2v\rangle=\langle u,v\rangle$ unless $\langle u,v\rangle=0$. $\endgroup$ – celtschk Jun 7 '13 at 18:58
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    $\begingroup$ The conclusion of the question is incorrect. You can only expect that there exists $c\ge 0$, such that $\langle Tu,Tv\rangle=c\langle u,v\rangle$, $\forall u,v\in V$. $\endgroup$ – 23rd Jun 7 '13 at 18:59
  • $\begingroup$ Note that $||v_1||=||v_2||=\ldots$, so the last condition can be applied quite directly. $\endgroup$ – celtschk Jun 7 '13 at 19:02
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    $\begingroup$ Yes, I think that the question is incorrect :S $\endgroup$ – Vinicius Rodrigues Jun 7 '13 at 19:41
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As written, this cannot be right: Consider the operator defined by $Tv=2v$ for all $v$. This is clearly a linear operator, and from $||u||=||v||$ it follows that $||2u||=||2v||$. However we do not have $⟨2u,2v⟩=⟨u,v⟩$ unless $⟨u,v⟩=0$.

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