Question

Suppose V is the finite-dimensional and S; T€ L(V ). Prove that ST and T S have the same eigenvalues.

Answer 1

$\texttt{Let}\;\; \lambda \;\; \texttt{be the eigen value of } ST.\\ \texttt{So, } \exists v(\neq 0) \in V \ni STv=\lambda v \;\;\;\;\;\;~~~~~~~ -(i)\\ \texttt{Let } w= Tv\\ \texttt{From (i), } T(STv)=T(\lambda v) \\\Rightarrow TS(Tv)=\lambda \times (Tv)\\ \Rightarrow TSw=\lambda w\\$

$\texttt{Case I :- }(\lambda\neq 0)\\ \texttt{If } \lambda \neq 0, \texttt{then } \lambda v \neq 0\\ \Rightarrow T(\lambda v)\neq 0 \Rightarrow \lambda w \neq0 \Rightarrow w\neq 0\\ \texttt{So, } \lambda \texttt{ is the eigen value of }TS.$

$\texttt{Case II :- }(\lambda=0)\\ \lambda =0 \\\Rightarrow ST \texttt{ is not invertible} \\\Rightarrow S \texttt{ or } T \texttt{ is not invertibe} \\\Rightarrow TS \texttt{ is not invertible} \\\Rightarrow \lambda =0\texttt{ is a eigen value of }TS.$

$\texttt{Hense } ST \texttt{ and }TS \texttt{ have the same eigen values.(Proved)}$