Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $j$.

share|cite|improve this question

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.

It's a bit worrying that I haven't used the symmetry of the matrices. Are you sure this is necessary?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.