Determinants and diagonalizable matrices

In summary: I guess I just didn't read carefully enough.In summary, the conversation discusses how to prove that the determinant of an invertible nxn matrix, P, is equal to the determinant of its inverse, P^-1, multiplied by the determinant of the original matrix, A, multiplied by the determinant of P. This can be proven using the identity det(AB)=det(A)det(B) and the fact that det(P^-1)=1/det(P). The conversation also mentions that this proof may not be rigorous enough and that the concept of diagonalizing a matrix is important in this context. A counterexample is also given to show that P^-1AP is not always a diagonal matrix.
  • #1
Jennifer1990
55
0

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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  • #2
Use the fact that det(P-1AP) = det(P-1)det(A)det(P)
 
  • #3
Can you not just use the fact that det(AB)=det(A)det(B)?
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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:
  • #9
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.
 

1. What is a determinant?

A determinant is a numerical value that can be calculated from a square matrix. It is used to determine properties of the matrix, such as whether it is invertible or singular.

2. How is the determinant of a matrix calculated?

The determinant of a 2x2 matrix can be calculated by multiplying the values on the main diagonal and subtracting the product of the values on the opposite diagonal. For larger matrices, there are various methods such as cofactor expansion or using row operations.

3. What is the significance of a matrix being diagonalizable?

A diagonalizable matrix is one that can be written as a product of three matrices: P, D, and P^-1, where D is a diagonal matrix. This allows for easier calculation of powers and inverses of the matrix.

4. How do you determine if a matrix is diagonalizable?

A matrix is diagonalizable if and only if it has n linearly independent eigenvectors, where n is the size of the matrix. This can be determined by calculating the eigenvalues and eigenvectors of the matrix.

5. Can a non-square matrix be diagonalizable?

No, a non-square matrix cannot be diagonalizable because it does not have eigenvalues or eigenvectors. Diagonalization is only applicable to square matrices.

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