Kenneth Kuttler

10.7. NORMS ON L (X ,Y ) 275

Theorem 10.7.3 Let X and Y be finite dimensional normed linear spaces of dimension nand m respectively and denote by ||·|| the norm on either X or Y . Then if A is any linearfunction mapping X to Y, then A ∈ L (X ,Y ) and (L (X ,Y ) , ||·||) is a complete normedlinear space of dimension nm with

||Ax|| ≤ ||A|| ||x|| .

Also if A ∈L (X ,Y ) and B ∈L (Y,Z) where X ,Y,Z 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 remains toshow this and verify ||A|| < ∞. This last follows from the above Lemma 10.7.2. 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|| .

Next consider the assertion about the dimension of L (X ,Y ) . It follows from Theorem5.1.4. By Theorem 10.6.4 (L (X ,Y ) , ||·||) is complete. If x ̸= 0,

||Ax|| 1||x||

∣∣∣∣∣∣∣∣A x

||x||

∣∣∣∣∣∣∣∣≤ ||A||Thus ||Ax|| ≤ ||A|| ||x||.

Consider the last claim.

∥BA∥ ≡ sup∥x∥≤1

∥B(A(x))∥ ≤ ∥B∥ sup∥x∥≤1

∥Ax∥= ∥B∥∥A∥ ■

Note by Theorem 10.6.4 you can define a norm any way desired on any finite dimen-sional linear space which has the field of scalars R or C and any other way of defining anorm on this space yields an equivalent norm. Thus, it doesn’t much matter as far as no-tions of convergence are concerned which norm is used for a finite dimensional space. Inparticular in the space of m×n matrices, you can use the operator norm defined above, orsome other way of giving this space a norm. A popular choice for a norm is the Frobeniusnorm.

Definition 10.7.4 Define A∗ as the transpose of the conjugate of A. This is called theadjoint of A. Make the space of m×n matrices into a inner product space by defining

(A,B)≡ trace(AB∗)≡∑i(AB∗)ii = ∑

i∑

jAi jB∗ji ≡∑

i, jAi jBi j

∥A∥ ≡ (A,A)1/2.

This is clearly a norm because, as implied by the notation, A,B→ (A,B) is an innerproduct on the space of m×n matrices. You should verify that this is the case.

10.7. NORMS ON # (X,Y) 275Theorem 10.7.3 Let X and Y be finite dimensional normed linear spaces of dimension nand m respectively and denote by ||-\| the norm on either X or Y. Then if A is any linearfunction mapping X to Y, then A € £(X,Y) and (L(X,Y),||-||) is a complete normedlinear space of dimension nm with||Aar|| < ||Al| |]al].Also ifA € 2 (X,Y) and B € £ (Y,Z) where X ,Y,Z are normed linear spaces,||BA]| < ||B]|||Al|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 remains toshow this and verify ||A|| < °°. This last follows from the above Lemma 10.7.2. Thus thenorm is at least well defined. It remains to verify its properties.||A + Bl| = sup{||(A +B) (a)|| : [lel] < 1}< sup{||Aa|| : ||a|| <1} +sup{]|Barl| : |]a|] < 1} = |/A]|+||BII-Next consider the assertion about the dimension of @ (X,Y). It follows from Theorem5.1.4. By Theorem 10.6.4 (4 (X,Y),||-||) is complete. If « 4 0,ell = {lap <aThus ||Aa|| < ||A]|||2||.Consider the last claim.|BA|| = sup ||B(A(x))|| < [BUI sup laell = IB ||A|| aax||<llz||<Note by Theorem 10.6.4 you can define a norm any way desired on any finite dimen-sional linear space which has the field of scalars R or C and any other way of defining anorm on this space yields an equivalent norm. Thus, it doesn’t much matter as far as no-tions of convergence are concerned which norm is used for a finite dimensional space. Inparticular in the space of m x n matrices, you can use the operator norm defined above, orsome other way of giving this space a norm. A popular choice for a norm is the Frobeniusnorm.Definition 10.7.4 Define A* as the transpose of the conjugate of A. This is called theadjoint of A. Make the space of m x n matrices into a inner product space by defining(A,B) = trace (AB*) = yi (AB*) .. hh AiBi = ye AiiBiji ijAl] = (4,4),This is clearly a norm because, as implied by the notation, A,B — (A,B) is an innerproduct on the space of m x n matrices. You should verify that this is the case.