Page 294 - Matrix Analysis & Applied Linear Algebra

290 Chapter 5 Norms, Inner Products, and Orthogonality

Parallelogram Identity
Fora given norm on a vector space V, there exists an inner product
2
on V such that = if and only if the parallelogram identity

2 2 2 2
x + y + x − y =2 x + y (5.3.7)
holds for all x, y ∈V.

Proof. Consider real spaces—complex spaces are discussed in Exercise 5.3.6. If
2
there exists an inner product such that = , then the parallelogram
identity is immediate because x + y x + y + x − y x − y =2 x x +2 y y .
The difficult part is establishing the converse. Suppose satisfies the paral-
lelogram identity, and prove that the function

1 2 2
x y = x + y − x − y (5.3.8)
4
2
is an inner product for V such that x x = x for all x by showing the four
defining conditions (5.3.1) hold. The first and fourth conditions are immediate.
To establish the third, use the parallelogram identity to write

2 2 1 2 2
x + y + x + z = x + y + x + z + y − z ,
2
2 2 1 2 2
x − y + x − z = x − y + x − z + z − y ,
2
and then subtract to obtain
2 2
2 2 2 2 2x +(y + z) − 2x − (y + z)
x + y − x − y + x + z − x − z = .
2
Consequently,

1 2 2 2 2
x y + x z = x + y − x − y + x + z − x − z
4
1 2 2
= 2x +(y + z) − 2x − (y + z)
8 (5.3.9)

2

2
1
y + z

y + z
y + z
=
x +
− x −
=2 x ,

2
2

2
2
and setting z = 0 produces the statement that x y =2 x y/2 for all y ∈V.
Replacing y by y + z yields x y + z =2 x (y + z)/2 , and thus (5.3.9)