it seems to me that one of the main advantages an orthogonal basis has is how much easier it is to find vector coordinates than in a non orthogonal basis, does that make it always favored over non orthogonal vectors and basis? and are there any any situations where orthogonality is not only unimportant but also not favorable?

  • $\begingroup$ Worse, in some situations the notion of orthogonal is meaningless: Many vector spaces don't come equipped with a natural inner product. $\endgroup$ – Travis Apr 22 '16 at 13:16
  • $\begingroup$ @Travis Just curious: what sense of 'natural' are you thinking of? $\endgroup$ – rschwieb Apr 22 '16 at 13:24
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    $\begingroup$ As far as I know it's always favorable to work with an orthogonal basis, you simply have more tools than without it. Gramm-Schmidt allows you to transform any set of linearly independent vectors into an orthogonal set. Also, since any finite-dimensional real vector space is isomorphic to $\mathbb{R}^n$, you can always pull-back the standard inner product. $\endgroup$ – Mathematician 42 Apr 22 '16 at 13:33
  • $\begingroup$ @rschwieb I didn't mean 'natural' in any technical sense here. What I meant is that, when one encounters a vector space in practice, often it comes with an inner product (e.g., the tangent space to a point on a Riemannian manifold, or the adjoint representation of a semisimple Lie group)---of course in general there is no preferred inner product on a vector space, in which case orthogonality isn't even defined. $\endgroup$ – Travis Apr 22 '16 at 14:02
  • $\begingroup$ @Mathematician42 It's worth noting that, when pulling back to a vector space $\Bbb V$ an inner product from $\Bbb R^n$, (1) usually there is no canonical isomorphism $\Bbb V \to \Bbb R^n$, and (2) the pullback inner product---and hence the notion of orthogonality---depends on that (arbitrary) choice of isomorphism. $\endgroup$ – Travis Apr 22 '16 at 14:05

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