Editing Matrix norm (section)

==Equivalence of norms==
{{See also|Equivalent norms}}

For any two matrix norms <math>\|\cdot\|_{\alpha}</math> and <math>\|\cdot\|_{\beta}</math>, we have that:

:<math>r\|A\|_\alpha\leq\|A\|_\beta\leq s\|A\|_\alpha</math>

for some positive numbers ''r'' and ''s'', for all matrices <math>A\in K^{m \times n}</math>. In other words, all norms on <math>K^{m \times n}</math> are ''equivalent''; they induce the same [[topology (structure)|topology]] on <math>K^{m \times n}</math>. This is true because the vector space <math>K^{m \times n}</math> has the finite [[dimension (mathematics)|dimension]] <math>m \times n</math>.

Moreover, for every matrix norm <math>\|\cdot\|</math> on <math>\R^{n\times n}</math> there exists a unique positive real number <math>k</math> such that <math>\ell\|\cdot\|</math> is a sub-multiplicative matrix norm for every <math>\ell \ge k</math>; to wit,
:<math>k = \sup\{\Vert A B \Vert \,:\, \Vert A \Vert \leq 1, \Vert B \Vert \leq 1\}. </math>

A sub-multiplicative matrix norm <math>\|\cdot\|_{\alpha}</math> is said to be ''minimal'', if there exists no other sub-multiplicative matrix norm <math>\|\cdot\|_{\beta}</math> satisfying <math>\|\cdot\|_{\beta} < \|\cdot\|_{\alpha}</math>.

===Examples of norm equivalence===
Let <math>\|A\|_p</math> once again refer to the norm induced by the vector ''p''-norm (as above in the Induced norm section).

For matrix <math>A\in\R^{m\times n}</math> of [[Rank (linear algebra)|rank]] <math>r</math>, the following inequalities hold:<ref>
[[Gene Golub|Golub, Gene]]; [[Charles Van Loan|Charles F. Van Loan]] (1996). Matrix Computations – Third Edition. Baltimore: The Johns Hopkins University Press, 56–57. {{ISBN|0-8018-5413-X}}.</ref><ref>Roger Horn and Charles Johnson. ''Matrix Analysis,'' Chapter 5, Cambridge University Press, 1985. {{ISBN|0-521-38632-2}}.</ref>

*<math>\|A\|_2\le\|A\|_F\le\sqrt{r}\|A\|_2</math>
*<math>\|A\|_F \le \|A\|_{*} \le \sqrt{r} \|A\|_F</math>
*<math>\|A\|_{\max} \le \|A\|_2 \le \sqrt{mn}\|A\|_{\max}</math>
*<math>\frac{1}{\sqrt{n}}\|A\|_\infty\le\|A\|_2\le\sqrt{m}\|A\|_\infty</math>
*<math>\frac{1}{\sqrt{m}}\|A\|_1\le\|A\|_2\le\sqrt{n}\|A\|_1.</math>