# Determinants and diagonalizable matrices

1. May 24, 2009

### Jennifer1990

1. The problem statement, all variables and given/known data
Let P be an invertible nxn matrix. Prove that det(A) = det(P^-1 AP)

2. Relevant equations
none

3. 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 dont understand why P^-1 AP gives such a matrix

2. May 24, 2009

### VeeEight

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

3. May 24, 2009

### cristo

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

4. May 24, 2009

### Jennifer1990

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 dont think this proof is rigorous enough

5. May 24, 2009

### cristo

Staff Emeritus
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.

6. May 24, 2009

### Cyosis

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.

7. May 24, 2009

### Jennifer1990

i cant 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?

8. May 24, 2009

### Pengwuino

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.

Last edited: May 24, 2009
9. May 24, 2009

### Cyosis

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.