= Solution
Let $r=\operatorname{rank}(A)$. If $r=0$, the <matrix> is zero and the identity is immediate. Otherwise choose the <rank factorization> $A=BC$ from part b. The <column space> of $B$ is the image of $A$.
An <idempotent matrix> acts as the identity on its image: if $u=Av$, then $Au=A^2v=Av=u$. Applying this to each column of $B$ gives $AB=B$. Substituting $A=BC$ yields $B(CB-I_r)=0$. Since $B$ has <full column rank>, the represented map is injective, so each column of $CB-I_r$ is zero. Thus $CB=I_r$.
Now the rectangular <cyclic property of the trace> gives
$$
\operatorname{tr}(A)=\operatorname{tr}(BC)=\operatorname{tr}(CB)
=\operatorname{tr}(I_r)=r=\operatorname{rank}(A).
$$
This proves that the <trace of an idempotent equals its rank> using the supplied factorization method. Orthogonality of the projection is unnecessary: the argument applies to every real idempotent <matrix>.
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