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I really can't find a reason for going through all the work of the Gram-Schmidt method to make a new orthogonal basis $B'$ given an old basis $B$. If I want to change to an orthogonal basis, the most simple solution to me is just the standard basis.

Rodrigo de Azevedo
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Domien
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    The basis represents a subspace, which often is not some subset of the standard basis. – simonzack Jun 15 '15 at 17:29
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    What's the standard basis of the space of real polynomials of degree $\leq N$? What's the standard basis for the subspace of $\mathbb{R}^3$ spanned by $(1, 1, 0)$ and $(0, 1, -1)$? What if I'm dealing with a matrix that's diagonalizable but isn't diagonal over the standard basis? And so on. – anomaly Jun 15 '15 at 17:30

2 Answers2

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Yes in $\mathbb R^n$ the Gram-Schmidt procedure need not be particularly helpful, since there's often an obvious orthonormal basis.

However, if we start with a different vector space, such as the polynomials $\mathbb R[x]$ over the interval $[0,1]$ with an inner product, the first few elements of an orthonormal basis is not obvious and Gram-Schmidt is helpful.

Simon S
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  • From what I am thinking now from the comments above. The guy said it's not obvious to make a orthonormal basis for a plane in R³ .which I understand indeed. But If you have a subspace W of V and span (W) = span (V) then you can choose the standardBasis. Is that what you meant with your first sentence? – Domien Jun 15 '15 at 17:45
  • I mean it's possible without long calculations in $\mathbb R^3$ (or $\mathbb R^n$) with a little practice to find an orthonormal basis of a subspace. When we first learn GS we typically start with $\mathbb R^n$ because that's what we know best. But don't be fooled by the 'triviality' of that example. – Simon S Jun 15 '15 at 17:56
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$\newcommand{\Reals}{\mathbf{R}}$As others have noted, "choosing the standard basis" only makes sense in $\Reals^{n}$; the "most simple solution [of changing to the standard basis]" doesn't have wide applicability. :)

There are at least three reasons you might need to construct an orthonormal basis from a general basis:

  1. Most of the time you don't have the freedom to choose the vector space where your problem resides. Particularly, if $W$ is a proper, non-trivial subspace of $\Reals^{n}$, such as the orthogonal complement of the span of $(1, 1, \dots, 1)$, it's possible (indeed, generically likely) that no standard basis vector whatsoever lies in $W$.

  2. Even in $\Reals^{n}$ you don't generally have the freedom to choose the standard basis. In differential geometry you might have a curve or surface in $\Reals^{3}$ for which you want to pick, at each point of the object, a basis of $\Reals^{3}$ whose elements are either tangent to the object or normal to the object. In computer graphics you might have a "camera" whose spatial orientation is described by an orthonormal basis, most likely not the standard basis. The eigenspaces of a general symmetric real matrix generically do not contain any standard basis vector (as in anomaly's comment). Etc., etc.

  3. If $\Reals^{n}$ comes to you equipped with a non-standard inner product for which you want an orthonormal basis, the standard basis isn't any help. Similarly, in other spaces where orthonormal bases are useful, such as spaces of polynomials or continuous functions, the inner product might arise from integrating over an interval, in which case the "standard basis" $\{1, t, t^{2}, \dots\}$ isn't (ever) orthonormal.

Tangentially, Gram-Schmidt has useful theoretical consequences. For example:

  • An arbitrary orthonormal basis for a finite-dimensional subspace $W$ gives a formula for orthogonal projection to $W$.

  • Gram-Schmidt tells you that $O(n)$, the (compact) group of orthogonal $n \times n$ real matrices, is a strong deformation retract of $GL(n, \Reals)$, the multiplicative group of invertible real $n \times n$ matrices.

Martin Sleziak
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Andrew D. Hwang
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