3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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.) $\endgroup$ – Travis Willse Sep 30 '15 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 Sep 30 '15 at 8:18
  • 1
    $\begingroup$ 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 $\lambda^a$. $\endgroup$ – Travis Willse Sep 30 '15 at 12:00
  • 1
    $\begingroup$ 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. $\endgroup$ – Travis Willse Sep 30 '15 at 12:08
1
$\begingroup$

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 $$a_i=v{\circ}v_i.$$

$\endgroup$
  • $\begingroup$ Isn't this what OP already said in the question? $\endgroup$ – Gerry Myerson Sep 30 '15 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 Sep 30 '15 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 Sep 30 '15 at 8:24
  • 1
    $\begingroup$ 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 Sep 30 '15 at 8:46
  • 1
    $\begingroup$ @TSJ dot-products are easier than solving systems of equations $\endgroup$ – Omnomnomnom Sep 30 '15 at 16:23

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.