First here is a definition which applies in all cases, even if X,Y are infinite dimensional.
Definition 11.6.1 Let X and Y be normed linear spaces with norms

Then
It is an easy exercise to verify that

Furthermore, you should verify that you can replace ≤ 1 with = 1 in the definition. Thus

In the case that the vector spaces are finite dimensional, the situation becomes very simple.
Lemma 11.6.2 Let V be a finite dimensional vector space with norm
Proof: Suppose lim_{k→∞}v^{k} = v in V. Let

which converges to ∑ _{k=1}^{n}α_{j}Av_{j} = Av as k →∞. Thus A is continuous. Then also v →

Then we have the following theorem.
Theorem 11.6.3 Let X and Y be finite dimensional normed linear spaces of dimension n and m respectively and denote by

Also if A ∈ℒ

Proof: It is necessary to show the norm defined on linear transformations really is a norm. Again the triangle inequality is the only property which is not obvious. It remains to show this and verify


Next consider the assertion about the dimension of ℒ

Thus
Consider the last claim.

Note by Theorem 11.5.4 you can define a norm any way desired on any finite dimensional linear space which has the field of scalars ℝ or ℂ and any other way of defining a norm on this space yields an equivalent norm. Thus, it doesn’t much matter as far as notions of convergence are concerned which norm is used for a finite dimensional space. In particular in the space of m×n matrices, you can use the operator norm defined above, or some other way of giving this space a norm. A popular choice for a norm is the Frobenius norm.
Definition 11.6.4 Define A^{∗} as the transpose of the conjugate of A. This is called the adjoint of A. Make the space of m × n matrices into a inner product space by defining

This is clearly a norm because, as implied by the notation, A,B →
+ class=”left” align=”middle”(X,Y )11.7. LIMITS OF A FUNCTION