Linear Algebra, Theory And Applications, 2012a

338 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
Let X be a finite dimensional normed linear space with norm ||·|| where the field of
scalars is denoted by F and is understood to be either R or C. Let{v 1 ,···, v n } be a basis
for X. If x ∈ X, denote by x i the i th component of x with respect to this basis. Thus
x =
n∑
x i v i .
Definition 14.0.5 For x ∈ X and {v 1 , ··· , v n } a basis, define a new norm by
where
i=1
( n
) 1/2
∑
|x| ≡ |x i | 2 .
x =
i=1
n∑
x i v i .
i=1
Similarly, for y ∈ Y with basis {w 1 , ··· , w m }, and y i its components with respect to this
basis,
( m
) 1/2
∑
|y| ≡ |y i | 2
For A ∈L(X, Y ) , the space of linear mappings from X to Y,
i=1
||A|| ≡ sup{|Ax| : |x| ≤1}. (14.2)
The first thing to show is that the two norms, ||·|| and |·| , are equivalent. This means
the conclusion of the following theorem holds.
Theorem 14.0.6 Let (X, ||·||) be a finite dimensional normed linear space and let |·| be
described above relative to a given basis, {v 1 , ··· , v n } . Then |·| is a norm and there exist
constants δ, Δ > 0 independent of x such that
δ ||x|| ≤ |x|≤Δ ||x|| . (14.3)
Proof: All of the above properties of a norm are obvious except the second, the triangle
inequality. To establish this inequality, use the Cauchy Schwarz inequality to write
|x + y| 2 ≡
n∑
|x i + y i | 2 ≤
i=1
n∑
|x i | 2 +
i=1
n∑
n∑
|y i | 2 +2Re x i y i
i=1
( n
) 1/2 (
∑
n
) 1/2
∑
≤ |x| 2 + |y| 2 +2 |x i | 2 |y i | 2
i=1
i=1
= |x| 2 + |y| 2 +2|x||y| =(|x| + |y|) 2
and this proves the second property above.
It remains to show the equivalence of the two norms. By the Cauchy Schwarz inequality
again,
||x||
≡
∣∣ n∑ ∣∣∣∣ ∣∣∣∣ x i v i ≤
∣∣
i=1
≡ δ −1 |x| .
i=1
(
n∑
n
) 1/2
∑
|x i |||v i || ≤ |x| ||v i || 2
i=1
i=1