Answer to Question #213781 in Linear Algebra for Simphiwe Dlamini

Question

Suppose V is finite-dimensional and S,T $\isin$ L (V). Prove that ST and TS have the same eigenvalues

Answer 1

let V is finite dimensional

let S.T are L.T on V

ST and LT are also L.T on V

Let $\lambda$ is not equal to 0

$\lambda$ an eigen values of S

$\therefore$ STx=$\lambda$ x; x is not equal to 0

Let y=Tx

-sy=$\lambda$ x

-T(sy)=T($\lambda$ x)

-TSY=$\lambda$Tx

-TSY=$\lambda$y; y is not equal to 0

$\therefore$ ST & TS have same non zero eigen values

Suppose 'o' is eigen value of ST

-ST is non invertible

-either S or T is non invertible

-TS is non invertible

-O is eigen value of TS

hence ST & TS have same eigen values over a finite dimensional vector space V