Similar matrices = Same Eigenvalues (NO DETERMINANTS)

[brru25]

Homework Statement

Show that two similar matrices A and B share the same determinants, WITHOUT using determinants

2. The attempt at a solution

A previous part of this problem not listed was to show they have the same rank, which I was able to do without determinants. The problem is I can't think of how to show they have the same eigenvalues without going to the characteristic polynomial (derived from the determinant of |A-lamba*I|. My other idea was to think of both A and B as the same linear map with respect to a different basis. After that I draw a blank.

 

[lanedance]

so as an idea, could you start with the simlarity definition
$$B = P^{-1}AP$$

now multiply by an eigenvector u, of B, if you have a play with the action of P hopefully you could show Pu_i must be an eigenvector of A with same eigenvalue, thus showing the eigenvalues are the same

 

[Mark44]

I'm confused. Do you want to show that two similar matrices have the same eigenvalues (as in the title of this thread) or the same determinant?

 

[lanedance]

though as the determinant can be written as the product of the eigenvalues, showing the eigenvalues would be sufficient

though as another option and i think what Mark is hinting at, is you could just take the determinant of the similarity equation & use the properties of determinants with matrix multiplication & inverses... though that might be liimted by the no determinants clause

 

[Mark44]

No, I wasn't actually hinting at that, but it seems like a good idea. I can't think of how you would show that two matrices have the same determinant without using the determinant in some way. Could it be that the intent of the problem is to show that two similar matrices have the same determinant without calculating the determinant?

 

[brru25]

No, I wasn't actually hinting at that, but it seems like a good idea. I can't think of how you would show that two matrices have the same determinant without using the determinant in some way. Could it be that the intent of the problem is to show that two similar matrices have the same determinant without calculating the determinant?

Yea I was thinking the same thing about showing they have the same determinant, because I would think that would be enough.

 

[Mark44]

If A and B are similar matrices, then there is an invertible matrix P such that B = P^-1AP. |P^-1| = 1/|P|, and since P is invertible, its determinant is nonzero.

 

[brru25]

If A and B are similar matrices, then there is an invertible matrix P such that B = P^-1AP. |P^-1| = 1/|P|, and since P is invertible, its determinant is nonzero.

Only problem is a determinant is being used in the proof which isn't allowed.

 

[Mark44]

Which gets me back to my earlier question: How can you show that two matrices have the same determinant if you can't use a determinant?

 

[brru25]

Which gets me back to my earlier question: How can you show that two matrices have the same determinant if you can't use a determinant?

I agree completely. For now I can use the |P^-1| = 1/|P|. Concept. A second opinion basically said the same things we were saying so that will have to be my route. Thank you for your help.

 

[HallsofIvy]

Is it possible that the original problem was mistated?

Since "show that they share the same determinant without using determinants makes no sense I think it is likely the original problem was "Show that two similar matrices A and B share the same eigenvalues, WITHOUT using determinants".


(Especially since the title of this thread is "Similar matrices= same eigenvalues"!)

 

[brru25]

Is it possible that the original problem was mistated?

Since "show that they share the same determinant without using determinants makes no sense I think it is likely the original problem was "Show that two similar matrices A and B share the same eigenvalues, WITHOUT using determinants".


(Especially since the title of this thread is "Similar matrices= same eigenvalues"!)

Yea you're right. We were just trying to think of different approaches to the problem.

 

[lanedance]

well if that is the case, refer the first reply (post #2)
https://www.physicsforums.com/showpost.php?p=2412400&postcount=2