[wanderlust.xx]

B is similar to A if there exists P such that .

Let's say we have a matrix M which has the eigenvectors of A as the columns, and we say .

Then M isn't unique since we could swap the columns around and yield the same J. However, does this mean that M and P are the same matrix which means B=J?

Does this also mean that J is a unique matrix similar to A, ie there are no other matrices that are constructed differently also similar to A?

[sputum]

(Original post by wanderlust.xx)
does this mean that M and P are the same matrix which means B=J?

I'm pretty sure that isn't the case

Does this also mean that J is a unique matrix similar to A, ie there are no other matrices that are constructed differently also similar to A?

nor this

Are there no restrictions on A,B or P other than as stated?

[wanderlust.xx]

(Original post by sputum)
I'm pretty sure that isn't the case



nor this

Are there no restrictions on A,B or P other than as stated?

They're both square matrices. I'm also doing the 2x2 case. So if A is a 2x2 matrix doesn't that mean that there are at most 2 ways of constructing P?

[nuodai]

(Original post by wanderlust.xx)
B is similar to A if there exists P such that .

Let's say we have a matrix M which has the eigenvectors of A as the columns, and we say .

Then M isn't unique since we could swap the columns around and yield the same J. However, does this mean that M and P are the same matrix which means B=J?

Does this also mean that J is a unique matrix similar to A, ie there are no other matrices that are constructed differently also similar to A?

You seem to know the definitions of the things involved, but you've mixed up what they do in your head a bit.

A matrix is similar to a matrix if there is an invertible matrix such that . There is no condition put on other than that it should be invertible, so need not have columns which are eigenvectors of .

If is a matrix whose columns are eigenvectors of , and it is invertible, then will be diagonal and its diagonal entries will be the eigenvalues of . This representation will be unique up to reordering the eigenvalues.

So for instance, take . This has eigenvectors and with eigenvalues 1 and -1, respectively. So we have only two choices for . (But we have infinitely many choices for , corresponding to different scalings of the columns.)

If then .

If then .

But now say we take an invertible matrix , for instance . Then . This is a matrix which is similar to , but it certainly isn't equal to either of the possibilities for .