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Let $V$ be a finite dimensional vector space over the field $K$ with a scalar product. Show that $V$ has an orthogonal basis. Furthermore, does the statement valid if $V$ is infinite dimensional?

Cagan Ozdemir
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    For the finite case, one can explicitly construct a ONB via Gram-Schmidt. In the infinitive case Zorn's Lemma garanties only the existence of an ONB. – Michael Hoppe Nov 13 '16 at 11:28
  • How does Zorn's Lemma garanties the existence of ONB? In the proof of existence of any basis of infinite dimensional v.s. we just extend any linearly independent subset of $V$ to the maximal linearly independent subset of $V$. – Cagan Ozdemir Nov 13 '16 at 11:39
  • I don't remember exactly, but they are useless anyhow. In a Hilbert space one deal with orthonormal systems, that is a set of orthonormal vectors which is dense in the space. – Michael Hoppe Nov 13 '16 at 11:42
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    @user251257 In a Hilbert space, Zorn's Lemma guarantees the existence of maximal orthonormal set, and such a set spans a dense subspace. See this proof. – Anne Bauval Dec 15 '24 at 16:29

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When the characteristic of your field $K$ is $\ne2$, the finite-dimensional space "$V$ always has an orthogonal basis. This can be proven by induction", but I prefer Gauss's reduction of the associated quadratic form. (For more details about the induction proof, and about the case $\operatorname{char}K=2$, see - after reduction to the non-degenerate case - this answer.)

When $K=\Bbb R$ or $\Bbb C$ and $V$ is a Hilbert space, or a separable inner product space, it has so-called "Hilbert bases", which are orthonormal. But if $V$ is infinite-dimensional, a Hilbert basis is not a true basis: it spans a subspace which is only dense in $V$ (the "orthogonal complement" of this subspace is $\{0\}$).

Anne Bauval
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