# Eigenvectors of inverse complex matrix

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?

• 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? Jun 8, 2013 at 13:42
• @Ｊ.M. - but in general, the complex conjugate of a vector is not a multiple of the vector. Jun 8, 2013 at 13:47
• @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. Jun 8, 2013 at 13:50
• @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. Jun 8, 2013 at 14:05

## 2 Answers

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 $A^{-1}$.

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 $A^{-1}$.