# Eigenvalues of $AB$ and $BA$ ${}\qquad{}$ [duplicate]

Are the eigenvalues of the matrices $AB$ and $BA$ identical? If yes, why?

From the examples that I have tried I think they are identical but I just can't come up with a formal proof for this.

• You can find more reasons here. – Algebraic Pavel Mar 4 '15 at 13:00

If $\lambda$ is an eigenvalue of $AB$ that is $\exists x\neq 0$ and
$$AB(x)=\lambda x$$ $BAB(x)=BA(Bx)=\lambda (Bx)$
Then $Bx$ is eigenvector for eigenvalue $\lambda$.
Only case which we have to worry about is when $Bx=0$. For that case see that $\lambda x =0$ which contradicts the very starting assumption (except when $\lambda =0$).