Number Theory: An Introduction to Mathematics

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10 Inner Product Spaces 73

The norm in any inner product spaceVsatisfies theparallelogram law:

‖u+v‖^2 +‖u−v‖^2 = 2 ‖u‖^2 + 2 ‖v‖^2 for allu,v∈V.

This may be immediately verified by substituting‖w‖^2 =〈w,w〉throughout and
using the linearity of the inner product. The geometrical interpretation is that in any
parallelogram the sum of the squares of thelengths of the two diagonals is equal to the
sum of the squares of the lengths of all four sides.
It may be shown that any normed vector space which satisfies the parallelogram
law can be given the structure of an inner product space by defining

〈u,v〉={‖u+v‖^2 −‖u−v‖^2 }/4ifF=R,
={‖u+v‖^2 −‖u−v‖^2 +i‖u+iv‖^2 −i‖u−iv‖^2 }/4ifF=C.

(Cf. the argument forF=Qin§4 of Chapter XIII.)
In an arbitrary inner product spaceVa vectoruis said to be ‘perpendicular’ or
orthogonalto a vectorvif〈u,v〉=0. The relation is symmetric, since〈u,v〉= 0
implies〈v,u〉=0. For orthogonal vectorsu,v,thelaw of Pythagorasholds:

‖u+v‖^2 =‖u‖^2 +‖v‖^2.

More generally, a subsetEofVis said to beorthogonalif〈u,v〉=0forall
u,v∈ Ewithu =v.Itissaidtobeorthonormalif, in addition,〈u,u〉=1for
everyu∈E. An orthogonal set which does not containOmay be converted into an
orthonormal set by replacing eachu∈Ebyu/‖u‖.
For example, ifV=Fn, then the basis vectors

e 1 =( 1 , 0 ,..., 0 ), e 2 =( 0 , 1 ,..., 0 ),..., en=( 0 , 0 ,..., 1 )

form an orthonormal set. It is easily verified also that, ifI=[0,1], then inC(I)the
functionsen(t)=e^2 πint(n∈Z) form an orthonormal set.
Let{e 1 ,...,em}beanyorthonormal set in the inner product spaceV and let
U be the vector subspace generated bye 1 ,...,em. Then the norm of a vector
u=α 1 e 1 +···+αmem∈Uis given by


‖u‖^2 =|α 1 |^2 +···+|αm|^2 ,

which shows thate 1 ,...,emare linearly independent.
To fi n d t h ebest approximationinUto a given vectorv∈V, put

w=γ 1 e 1 +···+γmem,

where

γj=〈v,ej〉 (j= 1 ,...,m).

Then〈w,ej〉=〈v,ej〉(j= 1 ,...,m)and hence〈v−w,w〉=0. Consequently, by
the law of Pythagoras,

‖v‖^2 =‖v−w‖^2 +‖w‖^2.
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