2.8. NORMS ON LINEAR MAPS 67

and so if ∥v∥ ≤ 1,

δ

∥∥∥Bkv∥∥∥≤ ∆

∑i

(∑

j

∣∣∣Bki j

∣∣∣)21/2

and so ∥∥∥Bk∥∥∥≤ ∆

δ

∑i

(∑

j

∣∣∣Bki j

∣∣∣)21/2

the quantity on the right converging to 0. ■More generally, you might want to consider linear transformations L (V,W ) where

V,W are finite dimensional normed linear spaces. In this case, the operator norm definedabove is the same and it is well defined.

Proposition 2.8.8 Let L ∈L (V,W ) where V,W are normed linear spaces with V finitedimensional. Then the operator norm defined above as

∥L∥ ≡ sup∥v∥≤1

∥Lv∥W

is finite and satisfies all the axioms of a norm. Also ∥Lv∥ ≤ ∥L∥∥v∥.

Proof: I won’t bother with the subscript on the norms and allow this to be determinedby context in what follows. Let (v1, ...,vn) be a basis for V . For v ∈V, let v = ∑

nj=1 a jv j so

the a j are the coordinates of v in F. Let h(v)≡ a where a = (a1, ...,an) with v = ∑nj=1 a jv j.

Thus h(v)i = ai. Obviously h is linear, one to one, and onto. Define ∥|v∥| ≡ |h(v)| wherethe last is the usual Euclidean norm on Fn. Then this is obviously a norm because ∥|v∥|= 0if and only if h(v) = 0 if and only if v= 0. It is also clear that ∥|αv∥| ≡ |h(αv)|= |α| |h(v)|whenever α is a scalar. It only remains to show the triangle inequality. However, this isalso easy because we know it for |·| .

∥|v+w∥| ≡ |h(v+w)| ≤ |h(v)|+ |h(w)| ≡ ∥|v∥|+∥|w∥| .

It follows the two norms are equivalent. Thus ∥|v∥| ≤ ∆∥v∥ for some ∆. Hence

sup∥v∥≤1

∥Lv∥W ≤ sup∥|v∥|≤∆

∥Lv∥W = sup|h(v)|≤∆

∥∥∥∥∥L∑i

h(v)i vi

∥∥∥∥∥≤ sup

|h(v)|≤∆

∑i|h(vi)|∥Lvi∥W

≤ sup|h(v)|≤∆

(∑

i|h(v)i|

2

)1/2(∑

i∥Lvi∥2

)1/2

≤ ∆

(∑

i∥Lvi∥2

)1/2

< ∞

Thus the operator norm is well defined. That it satisfies the axioms of a norm on L (V,W )follows in the same way as above for the case where V = Rn and W = Rm.