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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?

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  • $\begingroup$ You are aware that if $\mathbf v$ is an eigenvector of $\mathbf A$ corresponding to an eigenvalue $\lambda$, then $c\mathbf v$ ($c\neq 0$) is an equally valid eigenvector as well? $\endgroup$ Jun 8, 2013 at 13:42
  • $\begingroup$ @J.M. - but in general, the complex conjugate of a vector is not a multiple of the vector. $\endgroup$ Jun 8, 2013 at 13:47
  • $\begingroup$ @Sharkos, certainly true, but I suspect that OP might not have noticed that his eigenvectors can look a bit different, since MATLAB does no normalization of eigenvectors, if memory serves. $\endgroup$ Jun 8, 2013 at 13:50
  • $\begingroup$ @J.M. You may be right in that, but I thought it was worth clarifying that this isn't really the answer to the problem. $\endgroup$ Jun 8, 2013 at 14:05

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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 $A^{-1}$.

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I suspect you've got your vectors switched around. Note that the $+i$ eigenvector of $A$ should correspond to the $1/(+i)=-i$ eigenvector of $A^{-1}$.

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