Page 114 - Linear Algebra Done Right

Inner Products
1

p(x)q(x) dx,
6.2
p, q =
0
as you should verify. Once again, if F = R, then the complex conjugate 101
is not needed.
Let’s agree for the rest of this chapter that
V is a finite-dimensional inner-product space over F.
In the definition of an inner product, the conditions of additivity
and homogeneity in the first slot can be combined into a requirement
of linearity in the first slot. More precisely, for each fixed w ∈ V, the
function that takes v to v, w is a linear map from V to F. Because
every linear map takes 0 to 0, we must have
0,w = 0

for every w ∈ V. Thus we also have

w, 0 = 0
for every w ∈ V (by the conjugate symmetry property).
In an inner-product space, we have additivity in the second slot as
well as the first slot. Proof:

u, v + w = v + w, u
= v, u + w, u
= v, u + w, u
= u, v + u, w ;

here u, v, w ∈ V.
In an inner-product space, we have conjugate homogeneity in the
second slot, meaning that u, av = ¯ a u, v for all scalars a ∈ F.
Proof:

u, av = av, u
= a v, u
= ¯ a v, u
= ¯ a u, v ;

here a ∈ F and u, v ∈ V. Note that in a real vector space, conjugate
homogeneity is the same as homogeneity.