The squared-norm of the projection of a Gaussian vector onto an independent $d$-dimensional subspace is a $\chi^2_{2d}$ How we can prove that:
The squared-norm of the projection of a $N$-dimensional complex vector with i.i.d. unit-variance and zero mean Gaussian components onto an independent $d$-dimensional subspace is a $\chi^2_{2d}$. This projection can be represented as $|v^* \,h|^2$, where $h$ is the Gaussian vector and $v$ is the vector in  direction of the projection.
 A: A mathematics-inclined proof, may need some background of linear algebra
For clarity, let's assume the whole space is $\mathbb{R}^N$ for which we shall consider as an $N$ dimensional vector space and the subspace is $M$ with $\dim(M) = d \leq N$. Let $\{u_1, \ldots, u_d\}$ be an orthonormal basis of $M$, which may be extended to an orthonormal basis $\{u_1, \ldots, u_d, u_{d + 1}, \ldots, u_N\}$ of $\mathbb{R}^N$. 
Treat $X \sim \mathcal{N}_N(0, I)$ as a vector in $\mathbb{R}^N$, the projection of $X$ onto $M$ can be expressed as 
$$P_M X = \sum_{i = 1}^d (X^T u_i) u_i.$$
By orthogonality of $\{u_1, \ldots, u_d\}$, we have
$$\|P_MX\|^2 = \sum_{i = 1}^d (X^T u_i)^2.$$
Also note that for every $(u_1^T X, \ldots, u_d^T X)^T = (u_1^T, \ldots, u_d^T)^T X \sim \mathcal{N}(0, I)$ by the orthogonality of $\{u_1, \ldots, u_d\}$ and the linear transformation property of normal random vectors, thus $X^T u_i \sim \mathcal{N}(0, 1)$ and are mutually independent, consequently, by definition, 
$$\|P_M X\|^2 = \sum_{i = 1}^d(X^T u_i)^2 \sim \chi_d^2.$$
