Kenneth Kuttler

2.8. NORMS ON LINEAR MAPS 65

It is an easy exercise to verify that ∥·∥ is a norm on L (Rn,Rm) and it is always thecase that ∥Ax∥Rm ≤ ∥A∥∥x∥Rn . This is shown next. Furthermore, you should verify thatyou can replace ≤ 1 with = 1 in the definition. Thus ∥A∥ ≡ sup{∥Ax∥Rm : ∥x∥Rn = 1} .

It is necessary to verify that this norm is actually well defined.

Lemma 2.8.5 The operator norm is well defined. Let A ∈L (Rn,Rm).

Proof: We can use the matrix of the linear transformation with matrix multiplicationinterchangeably with the linear transformation. This follows from the above considerations.Suppose limk→∞ vk = v in Rn. Does it follow that Avk→ Av? This is indeed the case withthe usual Euclidean norm and therefore, it is also true with respect to any other norm bythe equivalence of norms (Theorem 2.7.4). To see this,

∣∣∣Avk−Av∣∣∣≡( m

∑i=1

∣∣∣(Avk)

i− (Av)i

∣∣∣2)1/2

≤

 m

∑i=1

∣∣∣∣∣ n

∑j=1

Ai j

(vk

j− v j

)∣∣∣∣∣21/2

≤

 m

∑i=1

∑j=1

∣∣Ai j∣∣ ∣∣∣vk

j− v j

∣∣∣)21/2

≤∣∣∣vk−v

∣∣∣ m

∑i=1

∑j=1

∣∣Ai j∣∣)21/2

Thus A is continuous. Then also v→∥Av∥Rm is a continuous function by the triangleinequality. Indeed,

|∥Av∥−∥Au∥| ≤ ∥Av−Au∥Rm

Now let D be the closed ball of radius 1 in V . By Theorem 2.7.3, this set D is compact andso

max{∥Av∥Rm : ∥v∥Rn ≤ 1} ≡ ∥A∥< ∞.■

Then we have the following theorem.

Theorem 2.8.6 Let Rn and Rm be finite dimensional normed linear spaces of di-mension n and m respectively and denote by ∥·∥ the norm on either Rn or Rm. Then if Ais any linear function mapping Rn to Rm, then A ∈L (Rn,Rm) and (L (Rn,Rm) ,∥·∥) is acomplete normed linear space of dimension nm with

∥Ax∥ ≤ ∥A∥∥x∥ .

Also if A ∈L (Rn,Rm) and B ∈L (Rm,Rp) where Rn,Rm,Rp are normed linear spaces,

∥BA∥ ≤ ∥B∥∥A∥

Proof: It is necessary to show the norm defined on linear transformations really is anorm. Again the triangle inequality is the only property which is not obvious. It remainsto show this and verify ∥A∥< ∞. This last follows from the above Lemma 2.8.5. Thus thenorm is at least well defined. It remains to verify its properties.

∥A+B∥ ≡ sup{∥(A+B)(x)∥ : ∥x∥ ≤ 1}

≤ sup{∥Ax∥ : ∥x∥ ≤ 1}+ sup{∥Bx∥ : ∥x∥ ≤ 1} ≡ ∥A∥+∥B∥ .

2.8. NORMS ON LINEAR MAPS 65It is an easy exercise to verify that ||-|| is a norm on # (R",R”) and it is always thecase that ||AX||jgn < ||A]| ||x||,jyx. This is shown next. Furthermore, you should verify thatyou can replace < | with = 1 in the definition. Thus ||A|| = sup {||AX||pm : ||x||jgn = 1}.It is necessary to verify that this norm is actually well defined.Lemma 2.8.5 The operator norm is well defined. Let A € & (IR",R™).Proof: We can use the matrix of the linear transformation with matrix multiplicationinterchangeably with the linear transformation. This follows from the above considerations.Suppose lim;_,.. Vv‘ = v in R”. Does it follow that Av’ —> Av? This is indeed the case withthe usual Euclidean norm and therefore, it is also true with respect to any other norm bythe equivalence of norms (Theorem 2.7.4). To see this,m V2 am | vielav'—av = (s | (av) - (Av), ‘ < vai (v4 -»))n/n a\ 1/2 n/n 5\ 1/2< E (Slay +») <|v'—y| ¥ (Ela)i=l \j=l i=1 \j=1Thus A is continuous. Then also v + ||Av||gm is a continuous function by the triangleinequality. Indeed,|||Av|] — ||Aull] < |Av—AullgnNow let D be the closed ball of radius | in V. By Theorem 2.7.3, this set D is compact andsomax {||AV||pm : ||V|l gn < 1} = ||Al| < co.Then we have the following theorem.Theorem 2.8.6 Let R” and R” be finite dimensional normed linear spaces of di-mension n and m respectively and denote by ||-|| the norm on either R" or R". Then if Ais any linear function mapping R" to R", then A € & (IR",R”) and (£ (R",R”), ||-||) isacomplete normed linear space of dimension nm withI|Ax|| < [IA] IIx||-Also if A € & (IR",R”) and B € & (R”,R”) where R",R”,R? are normed linear spaces,\|BAl| < ||B]| ||A]Proof: It is necessary to show the norm defined on linear transformations really is anorm. Again the triangle inequality is the only property which is not obvious. It remainsto show this and verify ||A]| < oo. This last follows from the above Lemma 2.8.5. Thus thenorm is at least well defined. It remains to verify its properties.||A + Bl] = sup{||(A + B) (x)|] = |Ixl] < 1}< sup{||Ax| : [|x|] < 1} + sup{||Bx|] : |Ix|] <1} = |All + |B 1.