@zareen said in MTH603 Mid Term Past and Current Solved Paper Discussion:

If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities).


Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. If B = PAP−1 and v = 0 is an eigenvector of A (say Av = λv) then B(Pv) = PAP−1(Pv) = PA(P−1P)v = PAv = λPv. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.