# Determinants and diagonalizable matrices

• Jennifer1990

## Homework Statement

Let P be an invertible nxn matrix. Prove that det(A) = det(P^-1 AP)

none

## The Attempt at a Solution

P^-1 AP gives me a diagonal matrix so to find the determinant , i just multiply the entry in the diagonal. However, i don't understand why P^-1 AP gives such a matrix

Use the fact that det(P-1AP) = det(P-1)det(A)det(P)

Can you not just use the fact that det(AB)=det(A)det(B)?

hmm so can i say :
det(P-1 AP)
= det(P^-1)det(A)det(P)
= 1/det(P) det(A) det(P)
=det(A)

=S, for some reason, i don't think this proof is rigorous enough

=S, for some reason, i don't think this proof is rigorous enough

I guess it depends on your lecturer. I would say that det(AB)=det(A)det(B) is a standard relationship that one can use, unless told explicitly to prove it.

Jennifer1990 said:
P^-1 AP gives me a diagonal matrix so to find the determinant

This is not true. I can think of a matrix P and a matrix A such that P^-1AP is not a diagonal matrix.

Secondly if you're afraid of using identities you can easily cut down on one identity you used. That is $\det(P^{-1})=\det(P)^{-1}$.
You can use $\det(P^{-1})\det(A)\det(P)=\det(P^{-1})\det(P)\det(A)=\det(P^{-1}P)\det(A)=\det(I)\det(A)$

Thirdly rigorous is not synonym to beating around the bush as long as possible, if you can prove it in one line by all means do so.

i can't think of a counterexample such that a matrix P and a matrix A whose P^-1AP is not a diagonal matrix.

What is the counterexample that u know?

This is not true. I can think of a matrix P and a matrix A such that P^-1AP is not a diagonal matrix.

I believe it is given that P diagonalizes A as the OP says in the solution, he just didn't mention it in the problem statement. Although I'm not really sure why it matters... it's important that the determinant doesn't change when you diagonalize a matrix but it seems as if the concept of denationalization isn't being emphasized here.

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For example:

$$P=\left(\begin{matrix} 2 & 5 \\ 2 & 1 \end{matrix}\right),\;\;\; A=\left(\begin{matrix} 2 & 5 \\ 2 & 8 \end{matrix}\right)$$

Edit: To Pengwuino: Ah of course if it is given that P diagonalizes A the problem is nonexistent.