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Orthogonal Bases

Module by: Mark A. Davenport. E-mail the author

Definition 1

A collection of vectors BB in an inner product space VV is called an orthogonal basis if

  1. span ( B ) = V span ( B ) = V
  2. vi⊥vjvi⊥vj (i.e., 〈vi,vj〉=0〈vi,vj〉=0) ∀i≠j∀i≠j

If, in addition, the vectors are normalized under the induced norm, i.e., ∥vi∥=1∀i∥vi∥=1∀i , then we call VV an orthonormal basis (or “orthobasis”). If VV is infinite dimensional, we need to be a bit more careful with 1. Specifically, we really only need the closure of span (B) span (B) to equal VV. In this case any x∈Vx∈V can be written as

x = ∑ i = 1 ∞ c i v i x = ∑ i = 1 ∞ c i v i (1)

for some sequence of coefficients {ci}i=1∞.{ci}i=1∞.

(This last point is a technical one since the span is typically defined as the set of linear combinations of a finite number of vectors. See Young Ch 3 and 4 for the details. This won't affect too much so we will gloss over the details.)

Example 1

  • V=R2V=R2, standard basis
    v1=10v2=01v1=10v2=01(2)

Example 2

  • Suppose V={ piecewise constant functions on [0,14),[14,12),[12,34),[34,1]}V={ piecewise constant functions on [0,14),[14,12),[12,34),[34,1]}. An example of such a function is illustrated below.
    Figure 1
    An example of a piecewise constant function.
    Consider the set
    Figure 2
    (a)
    Illustrations of the Haar basis functions. v_1 is the all constant function, i.e., it is 1 on [0,1].
    (b)
    Illustrations of the Haar basis functions. v_2 is 1 on [0,0.5) and -1 on [0.5,1].
    (c)
    Illustrations of the Haar basis functions. v_3 is sqrt(2) on [0,0.25), -sqrt(2) on [0.25,0.5), and 0 on [0.5,1].
    (d)
    Illustrations of the Haar basis functions. v_4 is 0 on [0,0.5), sqrt(2) on [0.5,0.75), and -sqrt(2) on [0.75,1].
    The vectors {v1,v2,v3,v4}{v1,v2,v3,v4} form an orthobasis for VV.
  • Suppose V=L2[-π,π]V=L2[-π,π]. B={12πejkt}k=-∞∞B={12πejkt}k=-∞∞, i.e, the Fourier series basis vectors, form an orthobasis for VV. To verify the orthogonality of the vectors, note that:
    12πejkt,12πejkt=12π∫-ππej(k1-k2)t=12πej(k1-k2)tj(k1-k2)-ππ=12π·-1+1j(k1-k2)=0(k1≠k2)12πejkt,12πejkt=12π∫-ππej(k1-k2)t=12πej(k1-k2)tj(k1-k2)-ππ=12π·-1+1j(k1-k2)=0(k1≠k2)(3)
    See Young for proof that the closure of BB is L2[-π,π]L2[-π,π], i.e., the fact that anyf∈L2[-π,π]f∈L2[-π,π] has a Fourier series representation.

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