Matrix A and its inverse have the same eigenvectors

[Mr Davis 97]

Homework Statement

T/F: Each eigenvector of an invertible matrix A is also an eignevector of A^-1

Homework Equations

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 $P^{-1}$?

 

[fresh_42]

Homework Statement

T/F: Each eigenvector of an invertible matrix A is also an eignevector of A^-1

Homework Equations

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 $P^{-1}$?

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?