Possible Values of Determinant for Idempotent Matrix

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For an idempotent matrix A, where A^2 = A, the determinant must satisfy the equation det(A) = det(A)^2. This leads to two possible cases: if det(A) is non-zero, then det(A) equals 1. The only other solution occurs when det(A) equals 0. Therefore, the possible values for det(A) of an idempotent matrix are 0 and 1. This conclusion is derived from the equation x^2 - x = 0, confirming that x can only be 0 or 1.
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Homework Statement


If A is an idempotent matrix (A^2 = A), find all possible values of det(A).


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The Attempt at a Solution


I'm not sure if this is the proper way to show it, but here's what I did:

Since A = A^2, det(A)=det(A^2)
So det(A) = det(A)*det(A)

Considering two cases:
If det(A) is not 0, then
det(A) = 1.

The only other way to satisfy det(A) = det(A)*det(A)
is if det(A) = 0.

So det(A) is either 1 or 0.
 
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Sure. If det(A)=x then you have x^2=x so x^2-x=0 so x*(x-1)=0. And, yes, that means x=0 or x=1.
 
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