Question

Suppose that A = PDP^-1. Prove that det(A) = det(D). 6. Suppose that A and B are nxn matrices that can be diagonalized with the same invertible matrix P (but with possibly different diagonal matrices D.D.). Prove that AB = BA. .. hank

Answer 1

Answer:

a) The main idea to solve this exercise is to use the identity $\det(AB)=\det(A)\det(B)$, where $A$ and $B$ are two square matrices.

Then, $\det(A) = \det(PDP^{-1}) =\det(P)\det(D)\det(P^{-1})$. Now, recall that [\det(Id) = \det(P)\det(P^{-1})[/tex], where $Id$ stands for the identity matrix. But $\det(Id)=1$, thus $\det(P)$ and $\det(P^{-1})$ are reciprocal to each other.

Hence,

$\det(A) =det(P)\det(D)\det(P^{-1}) = det(P)\det(P^{-1})\det(D) = \det(D).$

b) Let us write $A = PD_AP^{-1}$ and $B = PD_BP^{-1}$. Then

$AB = (PD_AP^{-1})(PD_BP^{-1}) = PD_AD_BP^{-1}$

$BA = (PD_BP^{-1})(PD_AP^{-1}) = PD_BD_AP^{-1}$

But the product of two diagonal matrices is commutative, so $D_AD_B = D_BD_A$, from where the statement readily follows.

Step-by-step explanation: