Let A and Beta be eta chi eta matrices and assume that Beta

Let A and Beta be eta chi eta matrices, and assume that Beta is invertible. How are the eigenvalues of B and B^-1 related? Show that A and A^1 have the same eigenvalues. Do they necessarily have the same eigenvectors? Show that AB and BA have the same eigenvalues.

Solution

a)

LEt, Bv=tv

Left multiplying by B^-1 gives

B^{-1}Bv=tB^{-1}v

B^{-1}v=1/t v

So eigenvalues are related as if t is eigenvalue of B then 1/t is eigenvalue of B^{-1} and vica versa

b.

Eigenvalues of A are given by

det(A-tI)=0

det(A-tI)=det((A-tI)^T)=det(A^T-tI)=0

HEnce, A and A^T have the same characteristic polynomial and hence same eignevalues

No they need not have the same eigenvectors

c)

Eigenvalues of AB are given by

det(AB-tI)=0

A and B are square matrices and B is invertible

det(AB-tI)=det(AB-tB^{-1}B)=det(A-tB^{-1})det(B)=det(B)det(A-tB^{-1})=det(BA-tI)=0

Hence, AB and BA have the same characteristic polynomial