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

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.

Explanation: