# eigenvalues of AB and of BA

• May 14th 2010, 04:28 PM
math8
eigenvalues 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 PM
dwsmith
Quote:

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.
• May 14th 2010, 05:23 PM
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.
• May 14th 2010, 05:27 PM
dwsmith
Quote:

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
• May 14th 2010, 05:55 PM
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 $Bx$

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
• May 14th 2010, 06:03 PM
roninpro
Quote:

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.
• May 15th 2010, 03:38 AM
HallsofIvy
Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.
• May 15th 2010, 04:47 AM
NonCommAlg
Quote:

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 $Bx=0.$
• May 15th 2010, 07:39 AM
Random Variable
Quote:

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 $\lambda = 0$
• May 15th 2010, 02:54 PM
HallsofIvy
Quote:

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.
• May 15th 2010, 03:13 PM
Random Variable
Quote:

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.(Doh)
• May 15th 2010, 03:20 PM
dwsmith
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 $C=S^{-1}DS$.

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