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Why is it important that a basis be orthonormal? The requirement for Orthonormal basis is so often repeated in linear algebra that it seems linear algebra depends on it as requirement. What is gained (or lost) if the basis are not orthonormal? Is it just for convention and convenience or is it just to restrict to the type of problems that assumption of orthonormality is beneficial?

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Important for what? –  Pedro Tamaroff May 22 '13 at 23:56
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Do you think it is important? Regardless of yes or no: Why? –  Zev Chonoles May 23 '13 at 0:00
    
Also: how familiar are you with linear algebra? Explaining your current knowledge will help people write answers that motivate the concept in ways you will appreciate. –  Zev Chonoles May 23 '13 at 0:01
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"Have your say" sounds like you're looking for opinions or trying to generate discussion, neither of which are this site's focus. Also, your question is extremely vague. Please provide some context. –  kigen May 23 '13 at 0:03
    
proximal, yes, indeed. I am asking what is the importance in different subfields in mathematics, physics or others(Fourier Series for example). I suppose that there are people with many different backgrounds and interests here, so if everyone could come up with something, it would be very informative. –  Rayhunter May 23 '13 at 0:14
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4 Answers

Often orthogonality represents a form of independence (as in statistics, where it says that there is a lack of covariance, or linear algebra where direct sums of vector spaces have a canonical inner product for which the sum is orthogonal). Orthogonal basis then means the ability to decompose an effect into separate, independent, non-interacting parts that simply add up to form the whole effect. This kind of decomposition is hugely important in situations where it can be done.

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When a basis is orthonormal, then a vector is merely the sum of its orthogonal projections onto the various members of the basis.

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If I may, this is true for any Hamel basis $(e_i)$: $x=\sum e_i^*(x)e_i=\sum p_i(x)$. The great advantage of orthonormal bases is that they allow to reconstruct the norm from the projections $p_i(x)$. And that the latter are easy to compute. –  1015 May 23 '13 at 0:51
    
@julien : I've edited it to say "orthogonal projections". If a basis is not orthonormal, then you can't tell which projection to use instead of the orthogonal projection unless you take the whole basis into account. –  Michael Hardy May 23 '13 at 19:32
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Calculating scalar product -- hence length and angle -- of vectors given with coordinates, is much simpler w.r.t. an orthonormal basis.

If $b_1,..,b_n$ is any basis, and a scalar product $\langle-,-\rangle$ is given in the vector space (else it wouldn't even make sense to speak about 'orthogonality'), then for the general formula for scalar products with given coordinates, we have to form the so called Gram-matrix of the basis: $$\Gamma:=(\langle b_i,b_j\rangle)_{i,j}$$ And then $$\langle\sum_i\alpha_ib_i,\,\sum_i\beta_ib_i\rangle=\sum_{i,j}\alpha_i\,\langle b_i,b_j\rangle\,\beta_j={\bf\alpha}^T\Gamma{\bf\beta}\,,$$ where ${\bf\alpha}^T=(\alpha_1,..,\alpha_n)$.

If $(b_i)$ is an orthonormal basis, then $\Gamma$ is the identity matrix.

Why would one carry $\Gamma$ in all formulas which uses coordinates and scalar products, if we can always construct (by Gram-Schmidt process) an orthonormal basis starting out from arbitrary basis?

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The question is somewhat equivalent to asking "why is it important to use the standard basis in $\mathbb R^n$". The answer is that sometimes it is important and sometimes it is not, but clearly it is convenient to use the standard basis. So, what do you do if you have a vector space that does not have a standrad basis? or, more fundamentally, what is so standard about the standard basis?

Well, the standard basis is an orthonormal basis with respect to a very familiar inner product space. And any orthonormal basis has the same kind of nice properties as the standard basis has.

As with everything, the choice of the basis should be made with consideration to the problem one is trying to solve. In some cases, orthonormal bases will help. If there is no particular reason to prefer one base over another, then choosing an orthonormal one is likely to be easier to compute with.

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