274 CHAPTER 11. INNER PRODUCT SPACES

and define the inner product by

(x, y) ≡n∑

i=1

xiyi

where

x =

n∑i=1

xivi, y =

n∑i=1

yivi.

The above is well defined because {v1, · · · , vn} is a basis. Thus the components xiassociated with any given x ∈ V are uniquely determined.

This example shows there is no loss of generality when studying finite dimensional vectorspaces with field of scalars R or C in assuming the vector space is actually an inner productspace. The following theorem was presented earlier with slightly different notation.

Theorem 11.1.7 (Cauchy Schwarz) In any inner product space

|(x, y)| ≤ |x||y|.

where |x| ≡ (x, x)1/2

.

Proof: Let ω ∈ C, |ω| = 1, and ω(x, y) = |(x, y)| = Re(x, yω). Let

F (t) = (x+ tyω, x+ tωy).

Then from the axioms of the inner product,

F (t) = |x|2 + 2tRe(x, ωy) + t2|y|2 ≥ 0.

This yields|x|2 + 2t|(x, y)|+ t2|y|2 ≥ 0.

If |y| = 0, then the inequality requires that |(x, y)| = 0 since otherwise, you could pick largenegative t and contradict the inequality. If |y| > 0, it follows from the quadratic formulathat

4|(x, y)|2 − 4|x|2|y|2 ≤ 0. ■

Earlier it was claimed that the inner product defines a norm. In this next propositionthis claim is proved.

Proposition 11.1.8 For an inner product space, |x| ≡ (x, x)1/2

does specify a norm.

Proof: All the axioms are obvious except the triangle inequality. To verify this,

|x+ y|2 ≡ (x+ y, x+ y) ≡ |x|2 + |y|2 + 2Re (x, y)

≤ |x|2 + |y|2 + 2 |(x, y)|≤ |x|2 + |y|2 + 2 |x| |y| = (|x|+ |y|)2. ■

The best norms of all are those which come from an inner product because of the followingidentity which is known as the parallelogram identity.

Proposition 11.1.9 If (V, (·, ·)) is an inner product space then for |x| ≡ (x, x)1/2, the

following identity holds.

|x+ y|2 + |x− y|2 = 2 |x|2 + 2 |y|2 .