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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?

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What is your module? – Mariano Suárez-Alvarez Jan 23 '12 at 2:53
up vote 8 down vote accepted

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 $\mathbb Q$...)

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Just in case it isn't clear, the reason the eigenvalues are $0$ and $1$ is because all the eigenvalues are roots of the minimal polynomial, which is $x^2-x$. Because the minimal polynomial has no repeated roots, it is diagonalizable, and thus has a basis of eigenvectors. Writing the vector space as $V_0\oplus V_1$, the transformation is projection onto $V_1$, and so the rank is the dimension of $V_1$. – Aaron Jan 23 '12 at 3:49
Just for the record, you don't need minimal polynomials or eigenvectors. Let $A: V \to V$ be idempotent, let $V_0 = \mathrm{Ker}(A)$ and $V_1 = \mathrm{Im}(A)$. If $u \in V_0 \cap V_1$ then $u = Au = 0$, so $V_0 \cap V_1 = \{ 0 \}$. For any $v$, we have $v = Av + (v-Av)$, and $Av \in V_1$, $v-Av \in V_0$, so $V = V_0 + V_1$. So we have $V =V_0 \oplus V_1$. (Probably not the right route for most students, but I happen to be teaching a class at the moment where I want this fact and we haven't hit Jordan canonical form yet.) – David Speyer Sep 12 '14 at 13:03

Hint: what are the eigenvalues of an idempotent matrix?

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I came to this page by accident but just wanted to note that the statement above that

"the multiplicity of one as an eigenvalue is precisely the rank"

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.

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