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If I have a quadratic form $x^T A x$, with $A$ idempotent ($A \times A = A$), rank($A$) = $r$, $x$ is a Gaussian multivariate random vector with $\mu = 0, C = I$, why is the distribution of $x^T A x$ equal to the central chi-square random variable with degrees of freedom = $r$?

Suppose $A$ is not symmetric.

If $A$ was symmetric, then the eigenvalue decomposition procedure could be used, and we would reach $x^T A x = y^T y$, $y$ is multivariate Gaussian with $\mu_y = 0, C_y = D$, where $D$ is the diagonal matrix resulting from the eigenvalue decomposition of $A$. Also, the degrees of freedom in the chi-square variable is equal to the rank of $A$, since the diagonal matrix limits the non-zero entries.


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up vote 1 down vote accepted

If A is the matrix

 ( 1 0 )
 ( a 0 )

(where a != 0), then A*A = A and A has rank 1 but

 (x y) * A * (x )  == x*x + a*x*y
             (y )

which isn't a chisquare(1) variable -- it's variance exceeds the variance of a chisquare by a*a.

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You are absolutely correct. In the problem statement, it was stated $A$ was symmetric, and I overlooked that statement. – jrand Jun 17 '11 at 17:05

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