If $T$ is an invertible linear transformation and $\vec{v}$ is an eigenvector of $T$, then $\vec{v}$ is an eigenvector of $T^{-1}$

Question

I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation. How do I prove an invertible linear transformation has the same eigenvectors as its inverse?

Answer 1

OK, so Iv'e found a solution!

\begin{align*}\\ Tv &= \lambda v \ \ \vert *T^{-1} \\ T^{-1}Tv &= T^{-1}\lambda v \\ v &= \lambda T^{-1}v \ \ \vert *\lambda^{-1} & \text{(Invertible transformation, $\lambda\neq 0$)} \\ \lambda^{-1}v &= T^{-1}v \end{align*}

For $T^{-1}$, eigenvalue $\lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

Answer 2

Observe that for an invertible matrix $A$ with eigenvector $\mathbf v$ and corresponding eigenvalue $\lambda\neq 0$, you have that \begin{align*} \mathbf v &= I \mathbf v\\&=A^{-1}A \mathbf v\\&=\lambda A^{-1} \mathbf v \end{align*} Hence $A^{-1} \mathbf v = \lambda^{-1}\mathbf v$