Associated eigenvector of the inverse of an eigenvalue

Question

Let A be an invertible $n×n$ matrix over the field of reals. Suppose $(a,v)$ is an eigenpair such that $Av=av$. Then $1/a$ is an eigenvalue of $A$. But is there a property or way of knowing the eigenvector of $1/a$ given those limited information above?

Answer 1

What is true is that $\frac{1}{a}$ is an eigenvalue for $A^{-1}$, so if you want to know the eigenvector of $A^{-1}$ associated to the $\frac{1}{a}$

$Av=av\rightarrow A^{-1}Av=A^{-1}av\rightarrow v=a A^{-1}v\rightarrow A^{-1}v=\frac{1}{a}v$,

so the eigenvectors for $A^{-1}$ will be the same as those for $A$, while their associated eigenvalues will be $\frac{1}{a},\forall a$ that is eigenvalue of $A$.

Answer 2

No, there isn't. Consider the matrix$$A=\begin{pmatrix}2&a\\0&\frac12\end{pmatrix}.$$It is invertible and it has two eigenvalues: $2$ and $\frac12$. The eigenvectors corresponding to $2$ are the multiples of $(1,0)$. But the eigenvectors corresponding to $(0,1)$ are the multiples of $(-2a,3)$.

Answer 3

If $A=\begin{pmatrix}2&0\\0&4711\end{pmatrix},$ then $A$ has the eigenvalues $2$ and $4711$. But $\frac12$ and $\frac{1}{4711}$ are not eigenvalues of $A$.