# Prove that $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$ if $T$ is an invertible linear operator on a finite dimensional vector space $V$

Let $T$ be an invertible linear operator on the finite dimensional $F$-vector space $V$. Prove that if $\lambda$ is an eigenvalue of $T$, then $\lambda^{-1}$ is an eigenvalue of $T^{-1}$ and that the eigenspaces $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$.

I have shown the first part of the proof, but it remains to show that $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$ is true. I'm thinking that this needs to be shown by proving the dimension of each of the eigenspaces, or can it be done through just by showing that all elements of one eigenspace are in the other, and vice versa.

• What does the symbol $E_\lambda$ mean? – user1551 Nov 29 '12 at 17:27
• Sorry I should have been clear. It is the eigenspace associated with a particular eigenvalue – tk2 Nov 29 '12 at 17:28
• Nope, you've said that in the text. I just have overlooked it. – user1551 Nov 29 '12 at 17:31

How about $Tv = \lambda v$ iff $\frac{1}{\lambda} v = T^{-1} v$ (Multiply the first equation by $T^{-1}$ and divide through by $\lambda$. $T$ is invertible, so all eigenvalues are $\neq 0$).
Explicitly, if $v \in E_\lambda(T)$ then $Tv = \lambda v$, hence $v = \lambda T^{-1} v$ and $\frac{1}{\lambda} v = T^{-1} v$. Hence $v \in E_{\frac{1}{\lambda}}(T^{-1})$. This gives $E_\lambda(T) \subset E_{\frac{1}{\lambda}}(T^{-1})$.
Since the same applies to $E_{\frac{1}{\lambda}}$ (and $(T^{-1})^{-1} = T$), we also have $E_{\frac{1}{\lambda}}(T^{-1}) \subset E_\lambda(T)$.