For a non-singular matrix, its pretty straightforward to prove that $\lambda$ is eigenvalue of $A$ if and only if $\frac{1}{\lambda}$ is eigenvalue of $A^{-1}$. Let $A$ be a non-singular matrix, $x$ an eigenvector of $A$ and $\lambda \neq 0$ its eigenvalue :
$$Ax = \lambda x \iff A^{-1}Ax = \lambda A^{-1}x \iff x \frac{1}{\lambda} = A^{-1}x$$
from this approach it seems to me that inverse matrix has reciprocal eigenvalues, but same eigenvectors.
However, consider the following matrix $$ A = \begin{pmatrix}1 & 0 & 0\\
-2 & -1 & -1\\
4 & 2 & 1
\end{pmatrix} $$
whose eigenvalues are $\left\{ 1,i,-i \right\} $. So eigenvalues of $A^{-1}$ are $\left\{1,\frac{1}{i},\frac{1}{-i} \right\} = \left\{ 1,-i,i \right\} $.
I tried putting this matrix to MATLAB and computing eigenvectors of it's inverse, but the elements in eigenvectors of $A^{-1}$ seem to be complex conjugates of the elements in eigenvectors of $A$. Where is the hidden flaw in the proof?
Best Answer
There is now hidden flow in your proof, I'm sure you misinterpreted MATLAB's results. MATLAB will sort the eigenvalues and -vectors for $A$ and $A^{-1}$ in a particular order, $\{1, i, -i\}$, say. Note moreover that, as $A$ is real, if $v$ is an eigenvector corresponding to $\lambda$, its complex conjugate $\bar v$ is an $\bar \lambda$-eigenvector, as $$ A\bar v = \overline{Av} = \overline{\lambda v} = \bar\lambda \bar v. $$ So the complex conjugate eigenvectors correspond to the complex conjugate eigenvalues of $A^{-1}$.