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Statement: If $ \lambda$ is an eigenvalue of $AB^{-1}$, then $ \lambda$ is an eigenvalue of $ B^{-1}A$ and vice versa.

One way of the proof.

We have $B(B^{-1}A ) B^{-1} = AB^{-1}. $ Assuming $ \lambda$ is an eigenvalue of $AB^{-1}$ then we have, $\begin{align*} \det(\lambda I - AB^{-1}) &= \det( \lambda I - B( B^{-1}A ) B^{-1} )\\ &= \det( B(\lambda I - B^{-1}A ) B^{-1})\\ &= \det(B) \det\bigl( \lambda I - B^{-1}A \bigr) \det(B^{-1})\\ &= \det(B) \det\bigl( \lambda I - (B^{-1}A )\bigr) \frac{1}{ \det(B) }\\ \ &= \det( \lambda I - B^{-1}A ). \end{align*}$ It follows that $ \lambda$ is an eigenvalue of $ B^{-1}A.$ The other side of the lemma can also be proved similarly.

Is there another way how to prove the statement?

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    @Arturo Thanks a lot. Cheers, susso.2011-12-29

2 Answers 2

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A shorter way of seeing this would be to observe that if $ (AB^{-1})x=\lambda x $ for some non-zero vector $x$, then by multiplying that equation by $B^{-1}$ (from the left) we get that $ (B^{-1}A)(B^{-1}x)=\lambda (B^{-1}x). $ In other words $(B^{-1}A)y=\lambda y$ for the non-zero vector $y=B^{-1}x$. This process is clearly reversible.

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    +1 Nice answer. // @Suso Note that essentially the same idea works to show that any pair of similar matrices have the same set of eigenvalues. So hidden inside this answer is (essentially) your observation that the two matrices are similar.2011-12-29
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Even if $A$ is $n\times m$ and $B$ is $m\times n$ with $m\le n$, we have $ \det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag{1} $ Proof:

For $\lambda>0$, $ \hspace{-1cm}\small\begin{bmatrix}I_n&A/\sqrt{\lambda}\\0&I_m\end{bmatrix}\begin{bmatrix}\lambda I_n-AB&0\\\sqrt{\lambda}B&\lambda I_m\end{bmatrix}=\begin{bmatrix}\lambda I_n&\sqrt{\lambda}A\\\sqrt{\lambda}B&\lambda I_m\end{bmatrix}=\begin{bmatrix}I_n&0\\B/\sqrt{\lambda}&I_m\end{bmatrix}\begin{bmatrix}\lambda I_n&\sqrt{\lambda}A\\0&\lambda I_m-BA\end{bmatrix}\tag{2} $ Taking the determinant of $(2)$, we get $ \lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)\tag{3} $ For $\lambda\le0$, note that $(3)$ is a polynomial in $\lambda$.

In the case of square matrices, since the characteristic polynomials are the same, the eigenvalues are the same.


As julien points out, there is a proof that doesn't require $\sqrt{\lambda}$ : $ \begin{bmatrix}I_n&-A\\0&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n-AB&0\\\lambda B&\lambda I_m\end{bmatrix}\tag{4} $ $ \begin{bmatrix}I_n&0\\-B&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n&A\\0&\lambda I_m-BA\end{bmatrix}\tag{5} $ Since the determinants on the left sides of $(4)$ and $(5)$ are equal, the determinants on the right side prove $(3)$.

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    @Guy: It *is* Sylvester's determinant formula. The proof is even similar.2013-03-30