# I A real matrix and its inverse share the same eigenvectors?

1. Apr 6, 2016

### Happiness

Suppose $v_i$ is an eigenvector of $A$ with eigenvalue $\lambda_i$ and multiplicity $1$.

$AA^{-1}v_i=A^{-1}Av_i=A^{-1}\lambda_iv_i=\lambda_iA^{-1}v_i$

Thus $A^{-1}v_i$ is also an eigenvector of $A$ with the same eigenvalue $\lambda_i$.

Since the multiplicity of $\lambda_i$ is $1$,

$A^{-1}v_i=k_iv_i$, where $k_i$ is a constant.

Thus $v_i$ is also an eigenvector of $A^{-1}$ with the same eigenvalue $\lambda_i$.

What's wrong with this proof?

Counterexample:

$A=\begin{pmatrix}\frac{\sqrt3}{2}&\frac{1}{2}&0\\-\frac{1}{2}&\frac{\sqrt3}{2}&0\\0&0&1\end{pmatrix}$

The eigenvector corresponding to eigenvalue $\frac{\sqrt3}{2}+\frac{1}{2}i$ is $\begin{pmatrix}-\frac{\sqrt2}{2}i\\ \frac{\sqrt2}{2}\\0\end{pmatrix}$, but that for $A^{-1}$ is $\begin{pmatrix}\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}i\\0\end{pmatrix}$.

Last edited: Apr 6, 2016
2. Apr 6, 2016

### Samy_A

If $A v= \lambda v$, then $A^{-1} v= \frac{1}{\lambda} v$.

(Of course assuming here that $A^{-1}$ exists.)

3. Apr 6, 2016

### Happiness

Thanks! I've found the mistake.