# Matrix A and its inverse have the same eigenvectors

1. Dec 7, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
T/F: Each eigenvector of an invertible matrix A is also an eignevector of A-1

2. Relevant equations

3. The attempt at a solution
I know that if A is invertible and $A\vec{v} = \lambda \vec{v}$, then $A^{-1} \vec{v} = \frac{1}{\lambda} \vec{v}$, which seems to imply that A and its inverse have the same eigenvectors. However, what about if A has all distinct eigenvalues, then $A = PDP^{-1}$. From this, we conclude that $A^{-1} = P^{-1} D^{-1} P$. Doesn't this show that A and its inverse have different eigenvectors? Since for A the matrix of eigenvectors is $P$ while for A-1 the matrix of eigenvectors is $P^{-1}$?

2. Dec 7, 2016

### Staff: Mentor

It's $A^{-1}=(PDP^{-1})^{-1}=PD^{-1}P^{-1}$. You have provided the argument that is needed, and you showed, that the eigenvalues are inverse, too, e.g. the elements in $D$ and $D^{-1}$. Why should there be a discrepancy?