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Let $T$ be a linear operator on a finite diemnsional inner product space $V$.
Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$.

Proof
Let $\beta=\{v_1,\dots,v_n\}$ be an orthonormal basis for $V$ so $T(\beta)$.
$\langle T(v_i),T(v_j)\rangle=\langle v_i,v_j\rangle =\delta_{ij}$ , therefore $T(\beta)$ is an orthonormal basis for $V$.
The first equality is from the equivalent theorem $\langle T(x),T(y)\rangle=\langle x,y\rangle$.
I wonder why the last sentence means that $T(\beta)$ is an orthonormal basis for $V$.
They are linearly independent, of course, but they do span $V$? How can show that?


Sorry to make you be confused. I omitted this information : $TT^*=T^*T=I$.

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    $\begingroup$ This result is clearly false for a general linear operator $T$. Are you perhaps assuming $T$ is unitary (or an orthogonal map in the real case)? $\endgroup$ May 29, 2013 at 14:28
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    $\begingroup$ The claim is wrong, for example if $T$ is not injective or not orthogonal $\endgroup$ May 29, 2013 at 14:29
  • $\begingroup$ Given your condition that $<T(v_i), T(v_j)> = <v_i, v_j>$, what properties of $T$ are there? Do you want $T$ to be a unitary operator? $\endgroup$
    – Calvin Lin
    May 29, 2013 at 14:34
  • $\begingroup$ The claim is precisely the characterization of orthogonal maps (which, of course, are bijective) so it can't be true in general. $\endgroup$
    – DonAntonio
    May 29, 2013 at 14:35
  • $\begingroup$ @DonAntonio You're right. Then why $T(\beta)$ can be a basis in here? $\endgroup$
    – noname
    May 29, 2013 at 14:44

2 Answers 2

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So, now that you've added the hypothesis that $T$ is unitary, it makes sense. You've shown $T$ takes an orthonormal set of vectors to another orthonormal set. If $\{v_j\}$ is a basis, then $\dim V=n$, and so the $n$ linearly independent vectors $T(v_j)$ must likewise be a basis, as they span an $n$-dimensional subspace of $V$.

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The statement ⟨T(x),T(y)⟩=⟨x,y⟩ means that T is an isomorphism, the result isn't true for an arbitrary linear operator T.

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  • $\begingroup$ In fact that statement gives way more, @Adolfo : $\,T\,$ the is an orthogonal map (unitary in the complex case). $\endgroup$
    – DonAntonio
    May 29, 2013 at 14:37

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