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In Axler's linear algebra book, 3rd edition, in page 229, the author states the following theorem:

7.42 Characterization of isometries
Suppose $S \in \mathcal{L}(V)$. Then the following are equivalent:
(a) $S$ is an isometry;
(b) $\langle S u, S v \rangle = \langle u, v \rangle$ for all $u, v \in V$;
(c) $S e_1, \dotsc, S e_n$ is orthonormal for every orthonormal list of vectors $e_1, \dotsc, e_n$ in $V$;
(d) there exists an orthonormal basis $e_1, \dotsc, e_n$ of $V$ such that $S e_1, \dotsc, S e_n$ is orthonormal;

In proving that (c) $\implies$ (d), Axler just states that it's obvious. However, I don't find it... In (c) there could be no orthonormal basis, but still exist orthonormal lists. In (d) we affirm the existence of one. This could be easy if we were assuming that $V$ was finite dimensional(Gram-Schmidt Procedure), however the theorem is silent in that regard.

Any help would be appreciated.

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    $\begingroup$ Excellent observation. In fact, (d) explictly claims that $\dim V=n$ $\endgroup$ – Hagen von Eitzen Oct 16 '16 at 16:13
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    $\begingroup$ From the beginning of chapter $\;7\;$ Axler remarks that $\;V\;$ is a finite dimensional linear space $\endgroup$ – DonAntonio Oct 16 '16 at 16:22
  • $\begingroup$ For whatever it’s worth (and I’m not convinced it’s much), you could accommodate the infinite-dimensional case by changing (d) to say “for every finite-dimensional subspace $W$ of $V$ there exists an orthonormal basis $e_1,\dotsc,e_n$ of $W$ such that $Se_1,\dotsc,Se_n$ is orthonormal.” $\endgroup$ – Branimir Ćaćić Oct 16 '16 at 16:23
  • $\begingroup$ @DonAntonio Many thanks ;) $\endgroup$ – An old man in the sea. Oct 16 '16 at 19:17

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