Determinants and diagonalizable matrices

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Homework Help Overview

The discussion revolves around proving that the determinant of a matrix A is equal to the determinant of the transformed matrix P^-1AP, where P is an invertible nxn matrix. The subject area includes linear algebra concepts related to determinants and diagonalizable matrices.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the determinants of matrices and question the conditions under which P^-1AP results in a diagonal matrix. There are attempts to apply known determinant properties, such as det(AB) = det(A)det(B), and discussions about the rigor of the proof.

Discussion Status

The discussion is active, with participants offering various insights and questioning assumptions about the diagonalization of matrices. Some express uncertainty about the rigor of the proof and the necessity of certain identities, while others provide counterexamples to challenge claims made earlier in the thread.

Contextual Notes

There is mention of a potential misunderstanding regarding the diagonalization of matrix A by matrix P, which is not explicitly stated in the problem. This raises questions about the assumptions underlying the proof and the definitions being used.

Jennifer1990
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Homework Statement


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


Homework Equations


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
 
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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
 
Jennifer1990 said:
=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 [itex]\det(P^{-1})=\det(P)^{-1}[/itex].
You can use [itex]\det(P^{-1})\det(A)\det(P)=\det(P^{-1})\det(P)\det(A)=\det(P^{-1}P)\det(A)=\det(I)\det(A)[/itex]

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?
 
Cyosis said:
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.
 
Last edited:
For example:

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

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

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