Linear Algebra Determinants Proof

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Homework Help Overview

The discussion revolves around proving a property of determinants for an n x n matrix A, specifically that if A satisfies the equation A³ = A, then the determinant of A can only be -1, 0, or 1.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss applying the determinant property to the equation A³ = A and explore the implications of this application. There is a focus on manipulating the resulting equations and questioning the steps taken to reach conclusions about the determinant.

Discussion Status

Participants are actively engaging with the mathematical properties involved, with some providing guidance on the manipulation of determinants. There is a progression in reasoning as they clarify the steps taken and explore the implications of their findings.

Contextual Notes

There is an emphasis on understanding the properties of determinants and how they relate to the proof being constructed. Participants express uncertainty about certain steps and seek clarification on the properties applicable to their situation.

B18
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Homework Statement


Prove that if A is an n x n matrix with the property A3=A, then det(A)=-1, det(A)=0, or det(A)=1

Homework Equations

The Attempt at a Solution


At first I started with the property A3=A
I then applied the determinant to both sides.
From this point I don't really see any properties that I can use to solve this equation for the det(A).
I understand how the solutions work. I am just unsure on how to achieve the proof of those solutions to the property.
Am I missing something simple here?
 
Last edited:
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B18 said:

Homework Statement


Prove that if A is an n x n matrix with the property A3=A, then det(A)=-1, det(A)=0, or det(A)=1

Homework Equations

The Attempt at a Solution


At first I started with the property A3=A
I then applied the determinant to both sides.
From this point I don't really see any properties that I can use to solve this equation for the det(A).
Am I missing something simple here?

Did you apply the property that det(AB)=det(A)det(B)? A^3=AAA.
 
Dick said:
Did you apply the property that det(AB)=det(A)det(B)? A^3=AAA.
Hi Dick, sorry. I did apply this property. I I'm just trying to get past this step.
 
B18 said:
Hi Dick, sorry. I did apply this property. I I'm just trying to get past this step.

Then show me what you got after you applied the property.
 
Dick said:
Then show me what you got after you applied the property.
det(A)*det(A)*det(A)=det(a)
(det(A))^3=det(A)
 
B18 said:
det(A)*det(A)*det(A)=det(a)
3det(A)=det(A)

det(A)*det(A)*det(A) is not equal to 3det(A). It's equal to ##det(A)^3##. Let x=det(A). Then your equation is ##x^3=x##. Can you solve that equation for x?
 
(det(A))3=det(A)
(det(A))3-det(A)=0
det(A)(det(A)2-1)=0
Thus:
det(A)=0
and det(A)2=1
det(A)=+/-1
 
B18 said:
(det(A))3=det(A)
(det(A))3-det(A)=0
det(A)(det(A)2-1)=0
Thus:
det(A)=0
and det(A)2=1
det(A)=+/-1

Much better.
 
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Is there a property that allows us to manipulate determinants this way? Or is it applicable because a determinate is a real number. Just need to build my proof format now. Really appreciate the help.
 
  • #10
B18 said:
Is there a property that allows us to manipulate determinants this way? Or is it applicable because a determinate is a real number. Just need to build my proof format now. Really appreciate the help.

As you said, a determinant is just a number. You treat them just like numbers. That's why I said let x=det(A). ##x^3=x## looks less mysterious.
 

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