# Thread: eigenvalues of AB and of BA

1. ## eigenvalues of AB and of BA

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

2. Originally Posted by math8
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
No, multiplication of matrices isn't commutative.

3. 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.

4. Originally Posted by math8
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.
Sorry for the mistake at first.

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

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

5. It's easy to prove.

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

then $ABx=\lambda_{1} x$

$BABx = \lambda_{1} B x$

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

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

then $BAx=\lambda_{2}x$

$ABAx = \lambda_{2}Ax$

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

QED

6. Originally Posted by dwsmith
Sorry for the mistake at first.

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

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
There might be a slight hiccup. You don't know whether or not $A$ or $B$ is invertible.

7. Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.

8. Originally Posted by Random Variable
It's easy to prove.

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

then $ABx=\lambda_{1} x$

$BABx = \lambda_{1} B x$

which means that $\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$

9. Originally Posted by NonCommAlg
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$
Deal with the case of $A$ or $B$ being zero matrices separately.

If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$

10. Originally Posted by Random Variable
Deal with the case of $A$ or $B$ being zero matrices separately.

If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ 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.

11. Originally Posted by HallsofIvy
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ 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.
Yeah, I forgot about that possibility.

12. 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 $C=S^{-1}DS$.

$p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$
$=det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$
$=det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$

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