# Inverse eigenvalue of a linear transformation [duplicate]

T is a linear transformation, and $\lambda$ is N eigenvalue of T. How do I prove that $\lambda^{-1}$ is an eigenvalue for $T^{-1}$?

I know for a matrix, I can use the fact that $Av=\lambda v$, but how does a linear transformation work?

## marked as duplicate by copper.hat, Cameron Buie, Bruno Joyal, Nick Peterson, azimutNov 20 '13 at 0:20

If $\lambda\ne0$ (and we know that the eigenvalues of an invertible matrix are different of $0$) then $$Av=\lambda v\iff A^{-1}A v=A^{-1}\lambda v\iff \frac{1}{\lambda }v=A^{-1} v$$ hence $\frac{1}{\lambda }$ is an eigenvalue for $A^{-1}$.
As $\lambda$ is an eigenvalue of $\textbf T$, $\textbf Tv=\lambda v$ for some $\lambda$. Therefore $\textbf T^{-1} \textbf Tv=\textbf T^{-1}\lambda v$ or $v=\lambda \textbf T^{-1} v$. It follows $\textbf T^{-1}v=\lambda^{-1}v$
Linear transformations, as they were taught to me take "lines to lines". In $\mathbb{F}^n$, this means all vectors are transformed continuously under the rules that, if $t$ is the linear transformation and $\vec v\in\mathbb{F}^n$, then $t(\lambda \vec v)=\lambda t(\vec v)$ and $t(\vec v+\vec w)=t(\vec v)+t(\vec w)$ for all $\lambda\in \mathbb{F}$, $\vec w\in \mathbb{F}^n$. You can prove from this information that $t$ can be written as a matrix transformation (try it yourself!).
For any linear transformation $f$ on a finite dimensional vector space, there is a matrix such that for all $x$, $f(x)=Ax$.
You don't even need this. The proof works the same with $f$.