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If $A$ is a square matrix of order $2$, and determinant of $A$ is $1$, then prove that $A$ and its inverse have the same eigenvalues.

So, let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Since determinant is $1$, it means $\lambda_1\,\lambda_2 = 1$. What do I do after this to reach the conclusion?

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  • $\begingroup$ If $\lambda$ is an eigenvalue of $A$, then $Ax = \lambda x$, so... $\endgroup$ – ua11 Feb 16 '14 at 15:19
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Hint: note that if $\lambda_1$ is an eigenvalue of $A$, then $1/\lambda_1$ is an eigenvalue of $A^{-1}$ (why?).

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  • $\begingroup$ why ? Because it is a standard property $\endgroup$ – Little Child Feb 16 '14 at 15:18
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    $\begingroup$ I'm not sure what you mean by "standard property". But if that's something you happen to know from before, then use it. $\endgroup$ – Omnomnomnom Feb 16 '14 at 15:19
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    $\begingroup$ @LittleChild "Standard property" or not, it is useful to know how to prove that result $\endgroup$ – ua11 Feb 16 '14 at 15:21
  • $\begingroup$ Let $\lambda_1$ be a eigenvalue of$A$, than $Ax=\lambda_1x$. This is equivalent with $x=A^{-1}\lambda_1x$ or $\lambda_1^{-1}x=A^{-1}x$. The last equality tells us that $\lambda_1^{-1}$ is an eigenvalue for $A^{-1}$. $\endgroup$ – Emin Feb 16 '14 at 15:27
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    $\begingroup$ To whoever down-voted my answer: I would appreciate an explanation as to why you decided this was necessary or how my answer could be improved. $\endgroup$ – Omnomnomnom Feb 16 '14 at 15:38
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If $A$ is 2 by 2 and has determinant $1$, then its eigenvalues are $\lambda$ and $\frac{1}{\lambda}$. If you invert $A$, the $\lambda$ eigenvalue maps to $\frac{1}{\lambda}$, and the $\frac{1}{\lambda}$ eigenvalue maps to $\frac{1}{\frac{1}{\lambda}} = \lambda$. Thus, they have the same eigenvalues.

This follows from $A x = \lambda x \iff \frac{1}{\lambda} A x = x \iff \frac{1}{\lambda}x = A^{-1} x \iff A^{-1} x = \frac{1}{\lambda} x$ for invertible $A$.

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