# Determinants and diagonalizable matrices

## Homework Statement

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

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## 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

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

cristo
Staff Emeritus
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 dont think this proof is rigorous enough

cristo
Staff Emeritus
=S, for some reason, i dont 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.

Cyosis
Homework Helper
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 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?

Pengwuino
Gold Member
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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Cyosis
Homework Helper
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.