Suppose $T \in L(V)$ is invertible. Prove that $G(\lambda, T)=G(\frac{1}{\lambda},T^{-1})$ for every $\lambda \in \mathbf{F}$ with $\lambda \neq 0$.
It is a problem from Ex8.A, Linear Algebra Done Right 3rd.
The $G(\lambda, T)$ represents the generalized eigenspace.
I tried to express T and T inverse in matrixes, but I'm confused.
So, please help me! Thank you a lot. :)
The definitions are here:From Linear Algebra Done Right 3rd, Page 245
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$\begingroup$HINT: Note that for $\lambda\neq0$ and any integer $k$ you have $$(T^{-1}-\tfrac{1}{\lambda}I)^k=(-1)^k\frac{1}{\lambda^k}T^{-k}(T-\lambda I)^k.$$
In more detail; if $\lambda\neq0$ and $v\in G(\lambda,T)$ then there is some integer $k$ such that $(T-\lambda I)^kv=0$. Multiplying by $\lambda^{-k}T^{-k}$ then shows that $$0=\lambda^{-k}T^{-k}(T-\lambda I)^kv=(\lambda^{-1}T^{-1}(T-\lambda I))^kv=(\lambda^{-1}I-T^{-1})^kv,$$ which implies that $v\in G(\lambda^{-1},T^{-1})$. This shows that $G(\lambda,T)\subseteq G(\lambda^{-1},T^{-1})$. The proof of the other inclusion is identical.
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