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Here is my question, which is part of the beginning of the chi-squared decomposition theorem proof : Suppose that $P_1, ....P_k $ are symmetric projection matices with sum the identity:

$I = P_1 + P_2 + ... + P_k $

Then squaring,

$ I = I^2 = \sum_i P_i^2 + \sum_{i<j} P_i P_j $

My question is why after squaring, it yields the term $\sum_{i <j} P_i P_j$ , how is it derived ???

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Should it be $2 \sum_{i < j} P_{i} P_{j}$? –  Isaac Solomon Nov 25 '12 at 4:25
    
Yes, I think so, but my lecture notes dont have the coefficient 2, which puzzled me da whole night ==. –  user1769197 Nov 25 '12 at 4:36

1 Answer 1

up vote -1 down vote accepted

Actually, under your conditions, $$I=I^2=\sum_{_i}P_i + 2\sum_{i<j}P_i P_j$$ because $P_i^2 = P_i$ (all projection matrices are idempotent) and because $P_i$ , $P_j$ are symmetric with the same dimensions, so $P_i P_j + P_j P_i = 2 P_i P_j$.

This is simple algebra $(P_1+...+P_n)^2=(P_1+...+P_n)(P_1+...+P_n)$.

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Ok. Because in my lecture notes, the coefficient 2 is missing. So , it must be wrong then. Thanks –  user1769197 Nov 25 '12 at 4:31
    
I don't see why any two symmetric matrices should commute. –  user18197 Aug 5 at 18:05

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