Answer to Question #210854 in Linear Algebra for sabelo Zwelakhe Xu

Question

Suppose T€ L(V ) is invertible.

(a) Suppose lemtha € F with lemtha not equal to 0. Prove that lemtha is an eigenvalue of T if and only if 1/lemtha is an eigenvalue of T^-1.

(b) Prove that T and T^-1 have the same eigenvectors.

Answer 1

$\texttt{(a)}\\ \texttt{If }\lambda \texttt{ is a Eigen value of T then, }\\\exists v(\neq 0)\in V \ni Tv=\lambda v\\ \texttt{Now as T is invertible, } \\ Tv=\lambda v\\ \Rightarrow T^{-1}(T(v))=T^{-1}(\lambda v)=\lambda T^{-1}(v)\\ \Rightarrow T^{-1}(v)=\dfrac{1}{\lambda}v\\ \texttt{Hence } 1/\lambda \texttt{ is the eigen value of }T^{-1}.\\ \texttt{And } v \texttt{ is also the Eigen vector of }T^{-1}. \texttt{(i)}$

$\texttt{Conversely, }\\ \texttt{If } 1/\lambda \texttt{ is a Eigen value of } T^{-1} \texttt{ then, }\\\exists w(\neq 0)\in V \ni T^{-1}w=w/\lambda\\ \texttt{Now as T is invertible, } \\ T^{-1}w=w/\lambda\\ \Rightarrow T(T^{-1}(w))=T(w)/\lambda\\ \Rightarrow T(w)=\lambda w\\ \texttt{Hence }\lambda \texttt{ is the eigen value of }T.\\ \texttt{And } w \texttt{ is also the Eigen vector of }T.~ \texttt{(ii)}$

$\;\\\texttt{(b)}\\ \texttt{From (i) and (ii),}\\ \texttt{We can see that both } T \texttt{ and } T^{-1} \texttt{ have same eigen vectors.}$