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.
Thanks.