Let $A$ and $ B$ be two nonsingular matrices. Show that $AB^{-1}$ and $B^{-1}A$ have the same eigenvalues

My attempt: $$ \begin{align} f(\lambda) &= | I\lambda -AB^{-1}| \\ &= |(I\frac{1}{\lambda}-B^{-1}A)^{-1}| \\ &= \dfrac{1}{|I\frac{1}{\lambda}-B^{-1}A|} \\ \end{align} $$

Not sure how to complete the final step. Any help would be appreciated

  • 3
    $\begingroup$ I think a better idea is using the eigenvectors. Take an eigenvector $x$ of $AB^{-1}$. You know that $AB^{-1}x = \lambda x$. Now, show that there exists a vector $y$ such that $B^{-1}A y = \lambda y$ $\endgroup$ – 5xum Aug 8 '14 at 6:59

It's a bit tricky: $$ \begin{align} &f(\lambda)\\ &=\det(AB^{-1}-\lambda I_n)\\ &=\det((\color{#00A000}{BB^{-1}})AB^{-1}-\lambda (\color{#00A000}{BB^{-1}}))\\ &=\det(B(B^{-1}A)B^{-1}- B(\lambda I_n)B^{-1})\\ &=\det(\color{#00A000}B(B^{-1}A-\lambda I_n)\color{#00A000}{B^{-1}})\\ &=\det(BB^{-1})\det(B^{-1}A-\lambda I_n)\\ &=\det(B^{-1}A-\lambda I_n) \end{align} $$

In a more general fashion , you can relate the char polynomials of $AB$ and $BA$ even when $A$ and $B$ are not square matrices, see my answer.

  • $\begingroup$ Thank you, this was exactly what I was looking for. I will accept your answer whenever it lets me. $\endgroup$ – MathMajor Aug 8 '14 at 7:07

For completeness, I'll expand on 5xum's comment.

Choose any eigenvalue $\lambda$ of $AB^{-1}$. Then we know that there exists some nonzero vector $x$ such that: $$ (AB^{-1})x = \lambda x $$ We want to show that $\lambda$ is also an eigenvalue of $B^{-1}A$. That is, we seek some nonzero vector $y$ such that: $$ (B^{-1}A)y = \lambda y $$ To this end, consider $y = B^{-1}x$ (which is a nonzero vector, since otherwise $x$ would be an eigenvector whose corresponding eigenvalue is $0$, contradicting the invertibility of $B^{-1}$). Observe that: $$ (B^{-1}A)y = (B^{-1}A)(B^{-1}x) = B^{-1}((AB^{-1})x) = B^{-1}(\lambda x) = \lambda ( B^{-1}x) = \lambda y $$ as desired. $~~\blacksquare$

  • $\begingroup$ +1 It is short, it uses the definition of eigenvalue/eigenvector and it does not use the fact that $B$ has an inverse $B$. So it immediatly proofs that $AB$ and $BA$ have the same eigenvalues. $\endgroup$ – miracle173 Aug 8 '14 at 7:35

Suppose $CD$ has a non-zero eigenvalue, then $CDv = \lambda v$ for some $v \neq 0$.

Then $D (CD)v = \lambda Dv = (DC) Dv$, and we know that $Dv \neq 0$ (otherwise $\lambda =0$), hence $\lambda$ is an eigenvalue of $DC$.

Since $A,B$ are non-singular, all eigenvalues of matrices $AB^{-1},B^{-1}A$ are non-zero, hence the have the same eigenvalues.

Alternatively, \begin{eqnarray} \det (sI-AB^{-1}) &=&\det ((sB-A) B^{-1}) \\ &=& \det (sB-A) \det B^{-1} \\ &=& \det B^{-1} \det (sB-A) \\ &=& \det (B^{-1}(sB-A) ) \\ &=& \det (sI-B^{-1}A) \end{eqnarray}

And finally, the simplest answer:

Since $AB^{-1}$ and $B^{-1} (A B^{-1}) B$ are similar matrices, they have exactly the same eigenvalues.

  • $\begingroup$ Thanks for the well thought out answer. I saw G.T.R's method using the characteristic polynomial first so naturally I accepted his answer. But thanks for this, I will upvote you. $\endgroup$ – MathMajor Aug 8 '14 at 7:17
  • $\begingroup$ Glad to help! Slow to type :-). $\endgroup$ – copper.hat Aug 8 '14 at 7:18

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