How could we prove that the "The trace of an idempotent matrix equals the rank of the matrix"?
This is another property that is used in my module without any proof, could anybody tell me how to prove this one?
An idempotent has two possible eigenvalues, zero and one, and the multiplicity of one as an eigenvalue is precisely the rank. Therefore the trace, being the sum of the eigenvalues, is the rank (assuming your field contains $\mathbb Q$...)
I came to this page by accident but just wanted to note that the statement above that
is non-trivial and is not true for general matrices. You still need to prove that algebraic multiplicity equals geometric multiplicity (in other words, that the number of linearly independent eigenvectors equals the multiplicity of one)
The fact that "since y = Px = P(Px) therefore members of an orthogonal basis of the range of P are also eigenvectors of P " is the missing piece. Because of this we can comfortably say that the rank is at least equal to the multiplicity. After that we need to state that none of the eigenvectors whose eigenvalue is zero could contribute to the range (though that one might omit because it's trivial.) @DavidSpeyer said similar things in his comment.