Talk:Matrix norm

From Wikipedia, the free encyclopedia
Jump to: navigation, search

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 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

German Link[edit]

This site needs to be linked to http://de.wikipedia.org/wiki/Matrixnorm

--91.113.18.247 (talk) 19:03, 5 January 2011 (UTC)

What's wrong with Frobenius norm?[edit]

Why does the article say that Frobenius norm is not sub-multiplicative? It does satisfy the condition \|A B\|\leq \|A\| \|B\|, which can be easily proved as follows:  \|A B\|^2_F = \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\|^2_F \|B\|^2_F . --Igor 21:21, Feb 18, 2005 (UTC)

Is it true that the Frobenius norm is \|A\|_p when p=2. It seems to me that it is the  \|A\|_2 norm that is mentioned earlier in the article.  \|A\|=\sqrt{\lambda_{max}A^HA} . Also it is also called the Hilbert-Schmidt norm, because the page for Hilbert-Schmidt norm says that it is only analogous to the Frobenius norm.--kfrance 13:40, Oct 9, 2007 (MST)

@KFrance, That is not true. The Frobenius norm is the Hilbert-Schmidt norm, but it is not the same as  \|A\|=\sqrt{\lambda_{max}A^HA} (this is the 'spectral norm'). For vectors,  \|a\|_2 is the Euclidean norm which is the same as the Frobenius norm if the input vector is treated like a matrix, but when the input is a matrix, the notation  \|A\|_2 usually denotes spectral norm, which is not the Frobenius norm. @Igor, that is true. Lavaka (talk) 17:54, 9 July 2014 (UTC)

What happened to the article?[edit]

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[edit]

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)

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)

Submultiplicativity[edit]

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{\Bbb{C}}^{m\times n} and B\in{\Bbb{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)

Bad Notation[edit]

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 (talkcontribs) 02:53, 24 December 2006 (UTC).

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)

Matrix Norm not Vector Norm[edit]

*\|A\|_1\le \sqrt n \|A\|_2
*\|A\|_1\le n \|A\|_\infty
*\|A\|_2\le \sqrt n \|A\|_1
*\|A\|_2\le \sqrt n  \|A\|_\infty
*\|A\|_\infty \le n \|A\|_1
*\|A\|_\infty \le \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)

equivalence of norms[edit]

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)

trace norm vs. Frobenius norm[edit]

it isn't true that the trace norm, sum(sigma), is <= the Frob. norm, sum(sigma^2); e.g. suppose all sigma<1. Lpwithers 16:34, 8 October 2007 (UTC)

trace norm[edit]

The article doesn't explain why the "trace norm" is an "entry-wise norm". sattath (talk) 14:49, 23 July 2008 (UTC) Fixed. --sattath (talk) 13:02, 27 April 2011 (UTC)

Gradient of the Norm[edit]

I'm interested in learning about the gradient of the matrix norm but I can't seem to find this information within wikipedia. I guess I'm requesting a new article and I don't know where to do that, but it seems logical for this article to point me to the gradient of the norm (maybe under see also). —Preceding unsigned comment added by Arbitrary18 (talkcontribs) 01:00, 23 September 2008 (UTC)

Matrix Norm Definition[edit]

Matrix norm on the set of all nxn matrices is a real value function, ||.|| defined on this set, satisfying for all nxn matrices A and B and all real number \alpha:

  • \|A\|> 0 if A\ne0 and \|A\|= 0 if and only if A=0
  • \|\alpha A\|=|\alpha| \|A\| for all \alpha in K and all matrices A in K^{m \times n}
  • \|A+B\| \le \|A\|+\|B\| for all matrices A and B in K^{m \times n}.
  • \|AB\| \le \|A\|\|B\|

@@@@

Matrix Norm Example[edit]

The two following functions are two examples of matrix norm:

 \left \| A \right \| _1 = \max \limits _{1 \leq j \leq n} \sum _{i=1} ^m | a_{ij} |,

and

 \left \| A \right \| _\infty = \max \limits _{1 \leq i \leq m} \sum _{j=1} ^n | a_{ij} |,


For examples: With matrix A:


 
      \begin{bmatrix}
           3 & 5 & 7 \\
           2 & -6 & 4 \\
           0 & 2 & 8 \\
        \end{bmatrix}


We have:    \left \| A \right \| _1 = |7|+|4|+|8|=19

And:

 
 \left \|A \right \| _\infty = |3|+|5|+|7|=15

Note: In the above example  \left \| A \right \| _1 is the maximum absolute column sum of the matrix, and  
 \left \|A \right \| _\infty is the maximum absolute row sum of the matrix. In addition both  \left \| A \right \| _1 and  
 \left \|A \right \| _\infty are the special norm of a general norm called p-norm for vectors

@@@@

max?[edit]

In some of the definitions I wasn't sure if max should actually be the supremum. I thought a maximum is guaranteed to exist for compact sets of real numbers, but not necessarily for open sets. In the case of linear, finite-dimensional operators(open sets are mapped to open sets) wouldn't this be equivalent to the domain being compact? In the case of the induced norm that would imply (from my perspective) max in the case abs(x)<=1 and supremum in the case x not equal to zero. I am not sure if it is actually an issue or not because at least in case of the induced 2 norm, the supremum is actually part of the range. That in turn implies to me that the supremum is reached for any similarly defined induced norm because of the equivalence of norms in finite dimensional spaces. Can someone with experience maybe point out the disconnect I seem to be having? —Preceding unsigned comment added by 79.235.159.125 (talk) 18:49, 19 July 2010 (UTC)

The domain is usually a sphere. These are closed and bounded, and thus compact by Heine-Borel. — Preceding unsigned comment added by 79.131.226.245 (talk) 17:06, 1 August 2011 (UTC)

spectral radius[edit]

There is a statement in the article: "For a symmetric or hermitian matrix A, we have equality for the 2-norm, since in this case the 2-norm is the spectral radius of A"

I guess the equality actually holds for more general case: It holds for any diagonalizable A. (Note that symmetric/hermitian is a special case of diagonalizable matrices when the diagonalizing matrix are unitary, which in turn, is a special case of normal matrices. All these are diagonalizable.)

Trivial proof: Let A = P D P-1. Then \|A\| = \sqrt{\lambda_{max} (A^{*} A) } = \sqrt{\lambda_{max} ((P^{-1})^{*} D^{*} P^{*} P D P^{-1}) } = \sqrt{\lambda_{max} ((P^{-1})^{*} P^{*} D^{*} D P P^{-1}) } = \sqrt{\lambda_{max} (D^{*} D) } (since the set of eigenvalues of AB is same as the set of eigenvalue of BA)

Does anybody see any problem with this argument? - Subh83 (talk | contribs) 18:47, 7 February 2013 (UTC)

That argument was wrong. If \lambda(A) gives the set of eigenvalues of matrix A, then \lambda(A_1 A_2 \cdots A_n) = \lambda(A_{\sigma(1)} A_{\sigma(2)} \cdots A_{\sigma(n)}) if \sigma is a cyclic permutation. - Subh83 (talk | contribs) 04:23, 8 February 2013 (UTC)

Thank you[edit]

This article was very useful. I was getting confused with that double-meaning notation and this article clarified it. Sorry for my English.--147.83.79.107 (talk) 15:31, 19 October 2013 (UTC)