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# Matrix norm

A matrix norm is a norm on the vector space of all real or complex m-by-n matrices. These norms are used to measure the "sizes" of matrices, and allow to talk about limits of sequences and infinite series of matrices. Several different matrix norms ||.|| are in common use. The more important ones in the case m = n are compatible with matrix multiplication in the sense that
$\|AB\|\le\|A\| \|B\|$
The set of all n-by-n matrices, together with such a sub-multiplicative norm, is a Banach algebra.

Suppose A=(aij) is an m-by-n matrix with entries from the field K (which is either R or C). The Frobenius norm of A is defined as

$\|A\|_F=\sqrt{\sum_{i=1}^m\sum_{j=1}^n |a_{ij}|^2}=\operatorname{trace}(AA^*)$
where A* denotes the conjugate transpose of A and the trace function is used. This norm is very similar to the Euclidean norm on Kn and comes from an inner product on the space of all matrices; however, it is not sub-multiplicative for m=n.

If norms on Km and Kn are given, then one defines the corresponding operator norm[?] on the space of m-by-n matrices as the following suprema:

$\|A\|=\sup\{\|Ax\| : x\in K^n \mbox{ with }\|x\|\le 1\}$
$= \sup\{\|Ax\| : x\in K^n \mbox{ with }\|x\| = 1\}$
$= \sup\left\{\frac{\|Ax\|}{\|x\|} : x\in K^n \mbox{ with }x\ne 0\right\}$
If m = n and one uses the same norm on domain and range, then these operator norms are all sub-multiplicative and give rise to Banach algebras.

The most "natural" of these operator norms is the one which arises from the Euclidean norms ||.||2 on Km and Kn. It is unfortunately relatively difficult to compute; we have

$\|A\|_2=\mbox{ the largest singular value of } A$
(see singular value). If we use the taxicab norm ||.||1 on Km and Kn, then we obtain the operator norm
$\|A\|_1=\max_{1\le j\le n} \sum_{i=1}^m |a_{ij}|$
and if we use the maximum norm ||.|| on Km and Kn, we get
$\|A\|_\infty=\max_{1\le i\le m} \sum_{j=1}^n |a_{ij}|$
The following inequalities obtain among the various discussed matrix norms for the m-by-n matrix A:
$\frac{1}{\sqrt{n}}\Vert\,A\,\Vert_\infty \leq \Vert\,A\,\Vert_2 \leq \sqrt{m}\Vert\,A\,\Vert_\infty$
$\frac{1}{\sqrt{m}}\Vert\,A\,\Vert_1 \leq \Vert\,A\,\Vert_2 \leq \sqrt{n}\Vert\,A\,\Vert_1$
$\Vert\,A\,\Vert_2 \leq \Vert\,A\,\Vert_F\leq\sqrt{n}\Vert\,A\,\Vert_2$

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