2
$\begingroup$

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

$\endgroup$
  • 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
7
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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. $~~\blacksquare$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +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
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.