Talk:Matrix norm

The following page will be replaced by a table.--wshun 01:34, 8 Aug 2003 (UTC)

The most "natural" of these operator norms is the one which arises from the Euclidean norms ||.||₂ on K^m 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 K^m and Kⁿ, then we obtain the operator norm

\|A\|_{1}=\max _{1\leq j\leq n}\sum _{i=1}^{m}|a_{ij}|

and if we use the maximum norm ||.||_∞ on K^m and Kⁿ, we get

\|A\|_{\infty }=\max _{1\leq i\leq 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}

What's wrong with Frobenius norm?

Why does the article say that Frobenius norm is not sub-multiplicative? It does satisfy the condition $\|AB\|\leq \|A\|\|B\|$ , which can be easily proved as follows: $\|AB\|_{F}^{2}=\sum _{i,j=1}^{n}|\sum _{k=1}^{n}a_{i,k}b_{k,j}|^{2}\leq \sum _{i,j=1}^{n}{\Big (}\sum _{k=1}^{n}|a_{i,k}|^{2}{\Big )}{\Big (}\sum _{l=1}^{n}|b_{l,j}|^{2}{\Big )}=$ $=(\sum _{i,k=1}^{n}|a_{i,k}|^{2})(\sum _{j,l=1}^{n}|b_{l,j}|^{2})=\|A\|_{F}^{2}\|B\|_{F}^{2}$ . --Igor 21:21, Feb 18, 2005 (UTC)

What happened to the article?

The above discussion suggests that the article used to be more extensive. However, the revision history of the current article shows only one edit, by CyborgTosser on 25 Feb 2005. Did something drastic happen to the article? -- Jitse Niesen 11:36, 2 Mar 2005 (UTC)

I'm not quite sure what happened. Apparently there used to be an article here, but the content must have been moved. I'm not sure where and I'm not sure why, but a lot of articles link here, so I figured we needed the article. Hopefully whoever moved the content will replace whatever is relevant. CyborgTosser (Only half the battle) 03:21, 11 Mar 2005 (UTC)

I don't know either. I couldn't find the old page on wikipedia with google, but I've put a copy (from a wikipedia clone) at Matrix norm/old. Lupin 13:50, 11 Mar 2005 (UTC)

It seems that User:RickK deleted this page after it had been vandalised. Idiot. I've asked him to restore it with edit history to a subpage if possible. Lupin 14:10, 11 Mar 2005 (UTC)

Induced norm

I'm a little confused where the article says that "any induced norm satisfies the inequality ...". Is the intended meaning that the operator norm satisfies that inequality, or are there other norms which are also known as induced norms which satisfy that inequality? If the former, it should be rephrased as "the induced norm satisfies..." and if the latter, an explanation of what is meant by an induced norm should be given. Lupin 01:24, 11 May 2005 (UTC)[reply]

The terms "induced norm" and "operator norm" are synonymous. I used "any induced norm" instead of "the induced norm" because there are several operator norms. One example is the spectral norm, another example arises when one takes the ∞-norm on Kⁿ, defined by

\|v\|_{\infty }=\max _{i}|v_{i}|;

the resulting operator norm is

\|A\|_{\infty }=\max _{i}\sum _{j}|a_{ij}|.

I hope this resolves the confusion; feel free (of course) to edit the article to make it clearer. -- Jitse Niesen 10:23, 11 May 2005 (UTC)[reply]

Submultiplicativity

I feel that this article is quite unclear about when submultiplicativity applies. In particular, it should be made clear that for matrix norms based on vectors p-norms that for $A\in {\mathbb {C} }^{m\times n}$ and $B\in {\mathbb {C} }^{n\times q}$ that $\|AB\|_{p}\leq \|A\|_{p}\|B\|_{q}$ . This is shown in Proposition 2.7.2 on the following page [1].

You are right that this could be added. So, why don't you change the article to include this? You can edit the article by clicking on "edit this page", see How to edit a page for details. Don't worry about making mistakes; you will be corrected if necessary. I look forward to your contributions, Jitse Niesen (talk) 11:24, 12 August 2005 (UTC)[reply]

Bad Notation

Resolved

Moreover, when m = n, then for any vector norm | · |, there exists a unique
positive number k such that k| · | is a (submultiplicative) matrix norm.

A matrix norm || · || is said to be minimal if there exists no other matrix norm
| · | satisfying |A|≤||A|| for all |A|.

Doesn't |A| specify the absolute value? Using the correct notation yields ||A||≤||A|| for all ||A||. Isn't that self evident? Furthermore m and n are not specified. Therefore I have removed this section till someone can clarify this content. It looks as if though someone partially moved content such that it's meaning was lost. —The preceding unsigned comment was added by ANONYMOUS COWARD0xC0DE (talk • contribs) 02:53, 24 December 2006 (UTC).[reply]

So sorry; don't know what I was thinking. I will just change |A| to ||A||_q and ||A|| to ||A||_p, it's clear from the sentence what |A| refereed to. I was reading a book earlier and |A| was refereed to as the determinant of A. More-over I will just add these statements back in and reword them. --ANONYMOUS COWARD0xC0DE 01:06, 29 December 2006 (UTC)[reply]

Matrix Norm not Vector Norm

Resolved

* $\|A\|_{1}\leq {\sqrt {n}}\|A\|_{2}$ 
* $\|A\|_{1}\leq n\|A\|_{\infty }$ 
* $\|A\|_{2}\leq {\sqrt {n}}\|A\|_{1}$ 
* $\|A\|_{2}\leq {\sqrt {n}}\|A\|_{\infty }$ 
* $\|A\|_{\infty }\leq n\|A\|_{1}$ 
* $\|A\|_{\infty }\leq {\sqrt {n}}\|A\|_{2}$

These are properties of vectors of the form $A\in \mathbb {R} ^{n}$ and not of the form $A\in \mathbb {R} ^{m\times n}$ . --ANONYMOUS COWARD0xC0DE 03:38, 24 December 2006 (UTC)[reply]

equivalence of norms

article is not really clear about the equivalence of norms: since we are talking about matrices of finite size, all vector norms should be equivalent. the bunch of inequalities in the bottom could (mis)lead the reader into thinking otherwise. if, in addition, submultiplicativity is required, does this change? (apparently so, the article seems to imply the Banach algebra topology is not unique.) Mct mht 14:08, 13 February 2007 (UTC)[reply]