# Show that $A$ and $A^{-1}$ have same eigenvalues?

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?

• If $\lambda$ is an eigenvalue of $A$, then $Ax = \lambda x$, so... – ua11 Feb 16 '14 at 15:19

Hint: note that if $\lambda_1$ is an eigenvalue of $A$, then $1/\lambda_1$ is an eigenvalue of $A^{-1}$ (why?).
• 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 $A^{-1}$. – Emin Feb 16 '14 at 15:27
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 $A$.