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

Printable View

- 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.

$\displaystyle det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$

$\displaystyle =det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$

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 $\displaystyle \lambda_{1} $ be an eigenvalue of $\displaystyle AB$

then $\displaystyle ABx=\lambda_{1} x$

$\displaystyle BABx = \lambda_{1} B x $

which means that $\displaystyle \lambda_{1} $ is an eigenvalue of $\displaystyle BA $ with associated eigenvector $\displaystyle Bx $

Now let $\displaystyle \lambda_{2} $ be an eigenvalue of $\displaystyle BA $

then $\displaystyle BAx=\lambda_{2}x $

$\displaystyle ABAx = \lambda_{2}Ax$

which means that $\displaystyle \lambda_{2} $ is an eigenvalue of $\displaystyle AB $ with associated eigenvector $\displaystyle Ax$

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 $\displaystyle ABx= \lambda x$ 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 $\displaystyle C=S^{-1}DS$.

$\displaystyle p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$

$\displaystyle =det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$

$\displaystyle =det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$