Talk:Matrix similarity

Examples

Examples would be useful of permutation-similar and unitarily equivalent matrices, plus a pair that is neither of those but is still similar. 128.83.138.141 (talk) 16:55, 29 January 2010 (UTC)

Determining similarity

I came to this article looking for an answer to the question

Is there a procedure to determine if two arbitrary matrices are similar? More strongly, given two matrices $A$ and $B$, is it possible to find $P$ such that $B=P^{-1}AP$?

Obviously if $B$ is diagonalizable then diagonalization will find $A$ and $P$, but I'm wondering about the general case for arbitrary $A$ and $B$. Could this information be added to the article? — Unbitwise (talk) 15:21, 23 April 2010 (UTC)

Just find Jordan canonical form of each matrix and check that they're the same. -- X7q (talk) 16:07, 23 April 2010 (UTC)

And btw this is already in the article: "By looking at the Jordan forms or rational canonical forms of A and B, one can immediately decide whether A and B are similar". -- X7q (talk) 16:12, 23 April 2010 (UTC)

Title

The title "similar matrix" is incorrect. A matrix can not be similar. Two of more matrices may be similar.82.75.67.221 (talk) 22:48, 23 January 2009 (UTC)

I don't think so; what if I take a matrix A and look at the set of all matrices similar to A. I would call all of those matrices "similar matrices", and I would call one in particular a "similar matrix". Suppose you had matrix trading cards. Fred says, "I have a matrix such-and-such!" Then Joe overhears and says "Oh, I have a similar matrix". I rest my case. --Fusionshrimp 128.194.39.250 (talk) 18:56, 4 April 2009 (UTC)

It looks rather silly to speak of (a) similar matrix, as if a single matrix may be called similar. Nijdam (talk) 18:20, 1 September 2011 (UTC)

It would be ideal to call this article similarity relation, but that namespace is taken by an article on music. The similarity concept is very important in linear algebra. It must be distinguished from similarity (geometry) where shape is preserved.Rgdboer (talk) 22:19, 5 August 2012 (UTC)

Why not: "similarity (matrix)"? Nijdam (talk) 11:09, 6 August 2012 (UTC)

I second the proposal to the name change. There is no "similar matrix" after all; unlike, say, orthogonal matrix. similarity (matrix) sounds good to me. -- Taku (talk) 12:17, 6 August 2012 (UTC)

The article has been moved to Matrix similarity since it is about a relation, not a matrix. The name of the relation is now the article title. Two important redirects have been updated: similar (linear algebra) and Similar matrices. There are many links to this article; I will try to fix them. The clear interest to correct the title commends the progressive nature of our project.Rgdboer (talk) 21:52, 7 August 2012 (UTC)

Links edited: 39. Several cases of similarity (geometry) corrected.Rgdboer (talk) 00:35, 8 August 2012 (UTC)

What is S as a function of T?

If $T=P^{-1}SP$ then what is S as a function of T? Just granpa (talk) 21:33, 16 May 2019 (UTC)

If $T=P^{-1}SP$ then $S=PTP^{-1}$.—Anita5192 (talk) 22:33, 16 May 2019 (UTC)

This might be wrong, some expert needs to look into it.

Look at what wolfgram says I also think $P^{-1}$ is used to change the basis from {$e_{i}$} to {$p_{i}$}. So we would have $x^{'}=P^{-1}x$ and $y^{'}=P^{-1}y$, in this new basis we can use the simpler rotation matrix. $y^{'}=Ax^{'}$ So this means,

$P^{-1}y=AP^{-1}x$

$y=Tx=PAP^{-1}x$

I think what I have written above is right since $Py^{'}=y$ means that while the scalar values in y can be used with the standard basis vectors to find the vector y. The scalar values in $y^{'}$ is used to choose a linear combination of the columns of P which is the same as y. So $y^{'}$ is y written in the language of P. As 3b1b clearly puts it here the matrix P constructed with the new basis vectors as columns multiplied by a vector $x^{'}$ in that language brings it over to our $e_{i}$ language, i.e. we get x. — Preceding unsigned comment added by Aditya8795 (talk • contribs) 16:47, 30 November 2019 (UTC)

The two definitions are equivalent, as $\exists P\mid PAP^{-1}=B$ and $\exists Q\mid Q^{-1}AQ=B$ are equivalent assertions; it suffices to set $Q=P^{-1}.$ So, the choice of the side where the exponent –1 appears is arbitrary. D.Lazard (talk) 17:55, 30 November 2019 (UTC)