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

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?

• Can't you just fix bases and consider your linear transformation as a matrix? – dcolazin Apr 13 at 7:22
• yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution. – DGR Apr 13 at 7:26
• Wouldn’t the proof be exactly the same as the one for matrices? – amd Apr 13 at 18:42

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 $$A^{-1} \mathbf v = \lambda^{-1}\mathbf v$$

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!