If X~$N(0,I_n)$ and A is a symmetric nxn matrix
Let $Q=X^TAX$. Then we know Q is chi-square iff A is idempotent matrix. However can Q be weight chi square if A be any symmetric nxn matrix?
Weight Chi squares is $\sum_{I}a_iz_i^2$ where $z_i$ ~ N(0,1)
I try to express the $Q=X^TAX = \sum_{i=1}x_i^2A_{ii}+2\sum_{i<j}x_ix_jA_{ij}$ Then first sum will be weight Chi-square. The second sum will be some non-central chi-square. Then I got lost.