Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?

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- May 14th 2010, 04:28 PMmath8eigenvalues of AB and of BA
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?

- May 14th 2010, 04:57 PMdwsmith
- May 14th 2010, 05:23 PMmath8
I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.

- May 14th 2010, 05:27 PMdwsmith
Sorry for the mistake at first.

I am pretty sure I have proving it here as well as some other proofs that may help in Linear:

http://www.mathhelpforum.com/math-he...rexamples.html - May 14th 2010, 05:55 PMRandom Variable
It's easy to prove.

Let be an eigenvalue of

then

which means that is an eigenvalue of with associated eigenvector

Now let be an eigenvalue of

then

which means that is an eigenvalue of with associated eigenvector

QED - May 14th 2010, 06:03 PMroninpro
- May 15th 2010, 03:38 AMHallsofIvy
Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.

- May 15th 2010, 04:47 AMNonCommAlg
- May 15th 2010, 07:39 AMRandom Variable
- May 15th 2010, 02:54 PMHallsofIvy
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA

**is**0. - May 15th 2010, 03:13 PMRandom Variable
- May 15th 2010, 03:20 PMdwsmith
I know I have this proof covering all cases, and once I find it, I will post it.

I am not sure if the red is necessarily correct.

AB and BA have the same eigenvalues iff. AB is similar to BA.

AB=C

BA=D

If C and D are similar matrices, then there exist S such that .