0
$\begingroup$

Normally an orthogonal basis of a finite vector space is referred as a basis that contains many vectors, i.e. 2 or more.

Consider a vector space that its dimension is 1 - does it have an orthogonal basis?
Is it true to refer to all the bases of that vector space as "orthogonal"?

I didn't find a reference for that in Wikipedia.

$\endgroup$
  • $\begingroup$ Indeed, that WP article on orthogonal basis is very stubby and does not discuss existence at all. However, the Gram-Schmidt process works in any finite dimensional inner product space, including dimension 1 (and even 0) $\endgroup$ – Hagen von Eitzen Feb 6 '15 at 19:23
3
$\begingroup$

You are correct. Any basis for a one dimensional inner product space is an orthogonal basis because the orthogonality condition is vacuously true, i.e. there are no pairs which must be orthogonal.

$\endgroup$
  • 2
    $\begingroup$ The same applies to zero-dimensional space, "even more vacuously". $\endgroup$ – Hagen von Eitzen Feb 6 '15 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.