Linear Algebra Determinants Proof

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The discussion focuses on proving that for an n x n matrix A with the property A^3 = A, the determinant det(A) must equal -1, 0, or 1. Participants explore applying the determinant property det(AB) = det(A)det(B) to derive the equation (det(A))^3 = det(A). This leads to the polynomial equation det(A)(det(A)^2 - 1) = 0, revealing that det(A) can be 0 or ±1. Clarifications are made on manipulating determinants as real numbers, aiding in building a proof format. The conversation emphasizes understanding the properties of determinants to complete the proof effectively.
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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?
 
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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.
 
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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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