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What are some disadvantages of using a basis whose elements are not orthogonal? (The set of vectors in a basis are linearly independent by definition.) One disadvantage is that for some vector $\vec v$, it involves more computation to find the coordinates with respect to a non-orthogonal basis. Are there any other undesirable outcomes?

I'm working on a simple model where it's more intuitive to use several physical but non-orthogonal entities as the basis.

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    $\begingroup$ In an orthogonal basis $(E_a)$ of a f.d. vector space, say of dimension $n$, the inner product has a particularly simple form, namely, $\langle A^a E_a, B^b E_b \rangle = \sum_{a = 1}^n \lambda^a A^a B^a$, where $\lambda^a = \langle E_a, E_a \rangle = ||E_a||^2$. For an orthonormal basis, we furthermore have $\lambda^1 = \cdots = \lambda^n = 1$. For a general basis, we have $\frac{1}{2} n (n + 1)$ independent coefficients instead of the $n$ coefficients $\lambda^a$. (NB I have used the Einstein summation convention throughout.) $\endgroup$ Commented Sep 30, 2015 at 6:52
  • $\begingroup$ @Travis I was staring at this for a while and I think it's finally making sense now. Two questions: 1. I'm interesting in reading more about the derivation of $\lambda^a=||E_a||^2$ for the orthogonal case, and the $\frac{1}{2} n (n + 1)$ coefficients in the non-orthogonal case. Is this something I would find in an abstract algebra text? 2. For a non-orthogonal basis, computing the inner product does not help in finding the coordinates $\vec x$ of some vector $\vec v$, right? I would think I have to solve $A\vec x=\vec v$, using Gaussian elimination for example (where A contains the basis). $\endgroup$
    – TSJ
    Commented Sep 30, 2015 at 8:18
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    $\begingroup$ Since inner products are symmetric, we can specify one (w.r.t. any basis) with a symmetric $n \times n$ matrix, and in turn this requires specifying the number of entries on and above the diagonal (by symmetry, the upper triangular entries determine the lower triangular ones), and there are $\frac{1}{2} n (n + 1)$ of these. If we are working with an orthogonal basis $(E_a)$ for the inner product, then for $a \neq b$ the $(a, b)$ entry of the matrix representation of the inner product is $\langle E_a, E_b \rangle = 0$, leaving only the diagonal entries, which I called $\lambda^a$. $\endgroup$ Commented Sep 30, 2015 at 12:00
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    $\begingroup$ For (2), yes, for nonorthogonal bases the components w.r.t. a basis are in general not given by (normalized) orthogonal projections---after all, they aren't orthogonal. Any computation of the coordinates is essentially equivalent to the one you mention. $\endgroup$ Commented Sep 30, 2015 at 12:08

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I would like to mention advantage of orthogonal basis in following way: For simplicity, consider vector space to be real vector space $\mathbb{R}^n$.

  • We know that $\mathbb{R}^n$, considered as abstract space, has some basis, say $\{v_1,v_2,\cdots,v_n\}$. Please don't consider that this is standard basis, think that it is any basis. What can we do with this? Well! Given any $v\in \mathbb{R}^n$, there exists $a_1,a_2,\cdots,a_n\in\mathbb{R}$ such that $v=\sum_i a_iv_i$. We don't know here how to determine these $a_i$'s? We know only that there are such $a_i$'s.

  • However, if we have an orthonormal (orthogonal) basis then we can determine the coefficients $a_i$'s involved in the expression of $v$.

For example, if we consider the standard dot product, and if the above basis of $\mathbb{R}^n$ is orthonormal, then $a_i$ are determined by $$a_i=v{\circ}v_i.$$

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  • $\begingroup$ Isn't this what OP already said in the question? $\endgroup$ Commented Sep 30, 2015 at 7:23
  • $\begingroup$ yes. this is almost similar. But while teaching inner product spaces in college, I was stressing on orthonormal basis through "this" elementary distinction between "usual basis" and "orthonormal basis". As $\endgroup$
    – Groups
    Commented Sep 30, 2015 at 7:27
  • $\begingroup$ @Groups Thanks for your answer. I'm unclear on the second point - can't we determine the coefficients $a_i's$ even if the basis is not orthogonal? On another note, your comment above appears to be truncated. $\endgroup$
    – TSJ
    Commented Sep 30, 2015 at 8:24
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    $\begingroup$ given abstract vector space $V$ and a basis $\{v_i\}_i$ of it, we always have that for any vector $v$, there exists scalars $a_i$'s s.t. $v=\sum_i a_iv_i$. We don't how to express these coefficients in terms of $v$ and basis elements. But, if the basis would be orthonormal, we can always express these coefficients in terms of $v$ and basis elements. $\endgroup$
    – Groups
    Commented Sep 30, 2015 at 8:46
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    $\begingroup$ @TSJ dot-products are easier than solving systems of equations $\endgroup$ Commented Sep 30, 2015 at 16:23

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