See pic. 46. Let A and B be similar matrices. Prove that the geometric multiplicities of the eigenvalues of A and Bare the same. Hint: Show that, if B P 1 AP, then every eigenvector of B is of the form P iv for some eigenvector v of A Solution suppose ? is an eigenvalue of A, so for some vector v, Av = ?v, and that B = P^(-1)AP. suppose P(u) = v (we know we can find u, it is P^(-1)(v)). then B(u) = P^(-1)AP(u) = P^(-1)(AP(u)) = P^(-1)A(P(u)) = P^(-1)(Av) = P^(-1)(?v) = ?(P^(-1)(v)) = ?u, so ? is an eigenvalue of B. now if ? is an eigenvalue of multiplicity k, then (x - ?)^k is a factor of det(A - xI) = (1/det(P))(det(A - xI))(det(P)) = (det(P^(-1)))(det(A - xI))(det(P)) = det(P^(-1)(A - xI)P) = det(P^(-1)AP - P^-(1)(xI)P) = det(B - x(P^-(1)P) = det(B - xI) so ? is an eigenvalue of at least multiplicity k for B. the reverse argument shows that if ? is an eigenvalue of multiplicity m for B, it is an eigenvalue of at least multiplicity m for A. hence k = m. suppose ? is an eigenvalue of A, so for some vector v, Av = ?v, and that B = P^(-1)AP. suppose P(u) = v (we know we can find u, it is P^(-1)(v)). then B(u) = P^(-1)AP(u) = P^(-1)(AP(u)) = P^(-1)A(P(u)) = P^(-1)(Av) = P^(-1)(?v) = ?(P^(-1)(v)) = ?u, so ? is an eigenvalue of B. now if ? is an eigenvalue of multiplicity k, then (x - ?)^k is a factor of det(A - xI) = (1/det(P))(det(A - xI))(det(P)) = (det(P^(-1)))(det(A - xI))(det(P)) = det(P^(-1)(A - xI)P) = det(P^(-1)AP - P^-(1)(xI)P) = det(B - x(P^-(1)P) = det(B - xI) so ? is an eigenvalue of at least multiplicity k for B. the reverse argument shows that if ? is an eigenvalue of multiplicity m for B, it is an eigenvalue of at least multiplicity m for A. hence k = m..