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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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What made you say that that's "long"? You seem to have already shown that your two matrices are similar... – J. M. Dec 29 '11 at 13:56
One simplification: once you observe that $A^{-1}(AB^{-1})A = B^{-1}A$, you can directly conclude that $B^{-1}A$ and $AB^{-1}$ are similar, and hence they have the same set of eigenvalues (also the same characteristic equation, the same trace, the same determinant, ...). – Srivatsan Dec 29 '11 at 14:15
Is it me or did the roles of $A$ and $B$ get swapped between the statement and the proof? – cardinal Dec 29 '11 at 14:32
@Arturo, I think if you swap $A$ and $B$, i.e., $(A,B) \mapsto (B,A)$ following "One way of the proof", then it will make more sense. That was the point of my previous comment. – cardinal Dec 29 '11 at 17:27
@cardinal: Ah, yes; it does make a lot more sense that way. I'm going to edit, this is way too confusing as it stands. – Arturo Magidin Dec 29 '11 at 17:57

2 Answers 2

up vote 19 down vote accepted

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 That's exactly what I was going to post as an answer but you beat me to it! – Dilip Sarwate Dec 29 '11 at 15:04
+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. – Srivatsan Dec 29 '11 at 20:34

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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This is a direct corollary of Sylvester's determinant formula ( – Guy Mar 30 '13 at 14:20
@Guy: It is Sylvester's determinant formula. The proof is even similar. – robjohn Mar 30 '13 at 14:41

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