Cochran's Theorem - Lemma involving symmetric idempotent matrices

I am writing notes for a reading class and I've decided to add a proof of Cochran's theorem in order to show that a statistic is $\chi^2$. I am struggling for the proof of a particular lemma but the rest is just peachy.

Lemma: Let $A$ be a real real symmetric idempotent matrix of order $n$ with rank $r$. Suppose $A=A_1+\cdots+A_k$ with rank $A_i = r_i$ and $A_i$ symmetric. Additionally, $r_1+\cdots+r_k=r$. Then each $A_i$ is idempotent.

I've been really struggling with this. It seems fairly obvious that $A_i$ gives an orthogonal decomposition of $A$. I've tried a lot of methods, looking at $A$'s and $A_i'$s spectral decomposition so I can show that the eigenvalues of $A_i$ must be $1$. Tried showing that $A_iA_j =0$ for $i$ not equal to $j$.

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The ranks of the matrices are the dimensions of their column spaces. Since the ranks of the $A_i$ add up to the rank of $A$, the column space $V$ of $A$ must be the direct sum of the column spaces $V_i$ of the $A_i$. Since $A$ is idempotent, it acts as the identity on $V$, and thus in particular on the $V_i$. That is, applying $A$ to a vector $v_i$ in $V_i$ leaves that vector invariant. Since each $A_i$ only contributes a component in its column space, it follows that $A_iv_i=v_i$ and $A_jv_i=0$ for $i\ne j$. Thus each $A_i$ is the identity on its column space and zero on its complement, and thus idempotent.
I am not certain. In the context of statistics the $A_i$'s are quadratic forms of random variables. I wonder if, at any rate, they are necessarily symmetric. My linear algebra is a bit rusty. – user18263 Oct 26 '11 at 7:05