Mathematics of Physics and Engineering

Inner Product 15

While an inner product defines a norm, other norms in En exist that are
not inner product-based; see Problem 1.8 on page 411.
An orthonormal basis in K™ is a basis consisting of pair-wise orthog-
onal vectors of unit length.

EXERCISE 1.2.5.^ Verify that, under definition (1.2.8), the corresponding

basis U\,..., un is necessarily orthonormal. Hint: argue that the basis vector

Ufc is represented by an n-tuple with zeros everywhere except the position k.

EXERCISE 1.2.6.B Prove the parallelogram law;

||u + vf + ||u - vf = 2\\u\\^2 + 2\\v\\^2. (1.2.10)

Show that in R^3 this equality can be stated as follows: in a parallelogram,

the sum of the squares of the diagonals is equal to the sum of the squares

of the sides (hence the name "parallelogram law").

Theorem 1.2.1 The norm defined by (1.2.9) satisfies the triangle

inequality

l|ti + w||<H| + ||t;|| (1.2.11)

and the Cauchy-Schwartz inequality

|ii-«|<||tt||.||t»||. (1.2.12)

Proof. We first show that (1.2.11) follows from (1.2.12). Indeed,

\\u + v\\^2 = {u + v)-{u + v) = \\u\\^2 + 2(u • v) + \\v\\^2

< \\uf + 2|u • »| + IMI^2 < ||M||^2 + 2||u||.||v|| + ||v||^2 = (||u|| + IMI)^2.

To prove (1.2.12), first suppose u and v are unit vectors. By properties

(II)—(13) of the inner product, for any scalar A,

0 < (u + Xv) • (u + Xv) = u • u + 2A« -v + X^2 v-v = l + 2Xu -v + X^2.

Now, take A = -{u • v). Then 0 < 1 - 2(u • v)^2 + {u • v)^2 = 1 - (u • v)^2.

Hence, \u • v\ < 1. On the other hand, for every non-zero vectors u and

v, u = \\u\\ -u/\\u\\ and v = ||w|| -«/||«||. Since u/||«|| and v/\\v\\ are unit

vectors, we have \u • v|/(||ii|| ||w||) < 1, and so |u • i;| < ||u|| ||«||.

If either u or v is a zero vector, then (1.2.12) trivially holds. Theorem

1.2.1 is proved. •

Remark 1.2 Analysis of the proof of Theorem 1.2.1 shows that equality

in either (1.2.11) or (1.2.12) holds if and only if one of the vectors is a