Suppose that A PBP1 we say that A and B are similar matrice

Suppose that A = PBP^1 (we say that A and B are similar matrices). Show that A and B have the same characteristic polynomials. [Hint: You might factor out P, P^1 at some point and use a determinant properties]

Solution

Since A B, there is an invertible P such that A = PBP^-1 . Then det(A) = det(PBP^-1 ) = det(P) det(B) det(P^-1 ) = det(P) det(B) 1 /det(P) = det(P)/ det(P) det(B) = det(B). Similarly,

det(A I) = det(PBP^-1 I) = det(PBP^-1 (PIP^-1 )) = det[P(B I)P^-1 ] = det(P) det(B I) det(P^-1 ) =               det(P) det(B I) 1/ det(P) = det(B I). The characteristic polynomials of A and B are det ( A – I) and det ( B- I) respectively. Thus A and B have the same characteristic polynomials.

Note:We have used the following 2 properties of determinants: