Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
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
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
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.
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 $\displaystyle C=S^{-1}DS$.
$\displaystyle p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$
$\displaystyle =det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$
$\displaystyle =det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$