# Linear Algebra Determinants Proof

1. Feb 12, 2015

### B18

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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. Im just unsure on how to achieve the proof of those solutions to the property.
Am I missing something simple here?

Last edited: Feb 12, 2015
2. Feb 12, 2015

### Dick

Did you apply the property that det(AB)=det(A)det(B)? A^3=AAA.

3. Feb 12, 2015

### B18

Hi Dick, sorry. I did apply this property. I I'm just trying to get past this step.

4. Feb 12, 2015

### Dick

Then show me what you got after you applied the property.

5. Feb 12, 2015

### B18

det(A)*det(A)*det(A)=det(a)
(det(A))^3=det(A)

6. Feb 12, 2015

### Dick

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?

7. Feb 12, 2015

### B18

(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

8. Feb 12, 2015

### Dick

Much better.

9. Feb 12, 2015

### B18

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. Feb 12, 2015

### Dick

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.