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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?

Zizo
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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. ain't a mathematician Dec 29 '11 at 13:56
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    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
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    @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
  • @Suso: You don't need to assume that $A$ is invertible. Why did you add that assumption? – Arturo Magidin Dec 29 '11 at 19:48
  • @Arturo, I hope I'm right now. Please correct me if not, thanks. – Zizo Dec 29 '11 at 20:03
  • @Suso: The problem had been that somewhere along the line you exchanged the roles of $A$ and $B$; the statement assumed something about $AB^{-1}$ and asked about $B^{-1}A$. Your proof worked on $BA^{-1}$ and $A^{-1}B$ (which implied $A$ was invertible, which was neither assumed nor required); I fixed that with an edit replacing all your $A$s with $B$s and vice-versa. Then you added the assumption on $A$ (now removed) to account for my original question (before cardinal pointed out what the issue was). It's correct now. – Arturo Magidin Dec 29 '11 at 20:07
  • @Arturo Thanks a lot. Cheers, susso. – Zizo Dec 29 '11 at 20:13

2 Answers2

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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.

Jyrki Lahtonen
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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
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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. – Srivatsan Dec 29 '11 at 20:34
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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:

Drawing from an answer of julien's, $$ \begin{bmatrix}I_n&-A\\0&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n-AB&0\\\lambda B&\lambda I_m\end{bmatrix}\tag{2} $$ $$ \begin{bmatrix}I_n&0\\-B&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n&A\\0&\lambda I_m-BA\end{bmatrix}\tag{3} $$ Since the determinants on the left sides of $(2)$ and $(3)$ are equal, the determinants on the right side prove $$ \lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)\tag{4} $$ In the case of square matrices, since the characteristic polynomials are the same, the eigenvalues are the same.

robjohn
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  • This is a direct corollary of Sylvester's determinant formula (http://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem) – 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