# Similar matrices = Same Eigenvalues (NO DETERMINANTS!)

1. Oct 26, 2009

### brru25

1. The problem statement, all variables and given/known data

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.

2. Oct 26, 2009

### lanedance

$$B = P^{-1}AP$$

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

3. Oct 27, 2009

### Staff: Mentor

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?

4. Oct 27, 2009

### 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

5. Oct 27, 2009

### Staff: Mentor

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?

6. Oct 27, 2009

### brru25

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

7. Oct 27, 2009

### Staff: Mentor

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.

8. Oct 27, 2009

### brru25

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

9. Oct 27, 2009

### Staff: Mentor

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?

10. Oct 27, 2009

### brru25

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.

11. Oct 28, 2009

### HallsofIvy

Staff Emeritus
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"!)

12. Oct 28, 2009

### brru25

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

13. Oct 28, 2009