6—Vector Spaces 133sinπxLsin^2 πxLsin^3 πxLf
To emphasize the relationship between Fourier series and the ideas of
vector spaces, this picture represents three out of the infinite number of basis
vectors and part of a function that uses these vectors to form a Fourier series.
f(x) =
1
2
sinπx
L
+
2
3
sin2 πx
L
+
1
3
sin3 πx
L
+···
The orthogonality of the sines becomes the geometric term “perpendicular,” and
if you look at section8.11, you will see that the subject of least square fitting
of data to a sum of sine functions leads you right back to Fourier series, and to
the same picture as here.
6.8 Gram-Schmidt Orthogonalization
From a basis that is not orthonormal, it is possible to construct one that is. This device is called the
Gram-Schmidt procedure. Suppose that a basis is known (finite or infinite),~v 1 , ~v 2 ,...
Step 1: normalize~v 1. ~e 1 =~v 1
/√〈
~v 1 ,~v 1
〉
.
Step 2: construct a linear combination of~v 1 and~v 2 that is orthogonal to~v 1 :
Let~e 20 =~v 2 −~e 1
〈
~e 1 ,~v 2
〉
and then normalize it.~e 2 =~e 20
/〈
~e 20 ,~e 20
〉 1 / 2
. (6.18)
Step 3: Let~e 30 =~v 3 −~e 1
〈
~e 1 ,~v 3
〉
−~e 2
〈
~e 2 ,~v 3
〉
etc.repeating step 2.
What does this look like? See problem6.3.
6.9 Cauchy-Schwartz inequality
For common three-dimensional vector geometry, it is obvious that for any real angle,cos^2 θ≤ 1. In
terms of a dot product, this is|A~.B~| ≤AB. This can be generalized to any scalar product on any
vector space: ∣
∣〈~u,~v〉
∣
∣≤‖~u‖‖~v‖. (6.19)
The proof starts from a simple but not-so-obvious point. The scalar product of a vector with itself is
by definition positive, so for any two vectors~uand~vyou have the inequality
〈
~u−λ~v,~u−λ~v
〉
≥ 0. (6.20)
whereλis any complex number. This expands to
〈
~u,~u
〉
+|λ|^2
〈
~v,~v
〉
−λ
〈
~u,~v
〉
−λ*
〈
~v,~u
〉
≥ 0. (6.21)
How much bigger than zero the left side is will depend on the parameterλ. To find the smallest value
that the left side can have you simply differentiate. Letλ=x+iyand differentiate with respect tox
andy, setting the results to zero. This gives (see problem6.5)
λ=
〈
~v,~u
〉/〈
~v,~v
〉
. (6.22)
Substitute this value into the above inequality (6.21)