6.11. THE SPACE L (Fn,Fm) 115

The first two properties are obvious but you should verify them. It remains to verify thenorm is well defined and also to verify the triangle inequality above. First if |x| ≤ 1, and(Ai j) is the matrix of the linear transformation with respect to the usual basis vectors, then

||A|| = max

(

∑i|(Ax)i|

2

)1/2

: |x| ≤ 1

= max

∑

i

∣∣∣∣∣∑jAi jx j

∣∣∣∣∣21/2

: |x| ≤ 1

which is a finite number by the extreme value theorem.

It is clear that a basis for L (Fn,Fm) consists of linear transformations whose matricesare of the form Ei j where Ei j consists of the m×n matrix having all zeros except for a 1 inthe i jth position. In effect, this considers L (Fn,Fm) as Fnm. Think of the m×n matrix asa long vector folded up.

If x ̸= 0,

|Ax| 1|x|

=

∣∣∣∣A x|x|

∣∣∣∣≤ ||A|| (6.11.12)

It only remains to verify completeness. Suppose then that {Ak} is a Cauchy sequence inL (Fn,Fm) . Then from 6.11.12 {Akx} is a Cauchy sequence for each x ∈ Fn. This followsbecause

|Akx−Alx| ≤ ||Ak−Al || |x|

which converges to 0 as k, l→ ∞. Therefore, by completeness of Fm, there exists Ax, thename of the thing to which the sequence, {Akx} converges such that

limk→∞

Akx = Ax.

Then A is linear because

A(ax+by) ≡ limk→∞

Ak (ax+by)

= limk→∞

(aAkx+bAky)

= a limk→∞

Akx+b limk→∞

Aky

= aAx+bAy.

By the first part of this argument, ||A||<∞ and so A∈L (Fn,Fm) . This proves the theorem.

Proposition 6.11.4 Let A(x) ∈ L (Fn,Fm) for each x ∈ U ⊆ Fp. Then letting (Ai j (x))denote the matrix of A(x) with respect to the standard basis, it follows Ai j is continuous atx for each i, j if and only if for all ε > 0, there exists a δ > 0 such that if |x−y|< δ , then||A(x)−A(y)||< ε . That is, A is a continuous function having values in L (Fn,Fm) at x.