Discussion is right way to get Solution of the every assignment, Quiz and GDB.
We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
Cyberian Team always happy to facilitate to provide the idea solution. Please don't hesitate to contact us! NOTE: Don't copy or replicating idea solutions. Quiz Copy Solution Mid and Final Past Papers Live Chat
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.