# Independence of quadratic forms

Let us consider the quadratic form

$$q_1 = \mathbf{x}_1^\mathrm{H} \mathbf{A}\,\mathbf{x}_1$$

and the quadratic form $$q_2= \mathbf{x}_2^\mathrm{H} \mathbf{A}\, \mathbf{x}_2$$

where $\mathbf{x}_1\in\mathbb{C}^{N\times 1}$, $\mathbf{x}_2\in\mathbb{C}^{N\times 1}$ and the elements of both $\mathbf{x}_1$ and $\mathbf{x}_2$ are independent and identically distributed circularly-symmetric Gaussian random variables. The matrix $\mathbf{a}\in\mathbb{C}^{N\times N}$ is singular and idempotent. The superscript $(.)^H$ is the conjugate transpose.

I am confused about the independence of $q_1$ and $q_2$. For example if I apply Criag's theorem for the independence of quadratic forms, then $q_1$ and $q_2$ are not independent since $\mathbf{A}\mathbf{A}=\mathbf{A}\ne \mathbf{O}_N$, where $\mathbf{O}_N$ is the $N\times N$ null matrix. So, the question is are they independent?

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## 1 Answer

Yes. $q_1$ and $q_2$ are independent because they are functions of independent random variables, respectively $x_1$ and $x_2$. You don't need Craig's theorem, which is about independence of different quadratic functions of the same random variable.

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Thanks Jonathan –  Sophie Oct 28 '12 at 15:26