Determinant of diagonal matrix D

Question

Suppose $A$ is $n\times n$ and $P^{-1}AP = D$ for some invertible $P$. Does $\det(A) = \det(D)$?

I'm pretty sure the answer is yes (just naively guessing and checking), but I don't know how to prove it.

Answer 1

det has the following property: $$\det(AB) = \det(A)\det(B)$$ So $$\det(D) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = (\det(P))^{-1}\det(A)\det(P) = \det(A)$$

$$1 = \det(\textrm{Id}) = \det(P^{-1}P) = \det(P^{-1})\det(P)$$

and so

$$\frac{1}{\det(P)} = \det(P^{-1})$$

Answer 2

If we are in some field $F$, the following is true: Using the product rule and $\det P\neq 0$ since it has an inverse, we get

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

Answer 3

The two matrices are similar, and thus their determinant is the same.