Why is the Inner Product Induced by a Gaussian Matrix a Gaussian Process?

Question

Let $A \in \mathbb{R}^{m \times n}$ be a Gaussian matrix such that $A_{ij} \sim N(0, 1)$ i.i.d. We define $$ X_{uv} := \langle Au, v \rangle. $$ Then it is claimed that $\{ X_{uv} \}_{(u,v) \in T}$, where $T := S^{n - 1} \times S^{m - 1}$ is a Gaussian process. This should be easily observable, I am not sure I am seeing how. It is understandable that $X_{uv} \sim N(0, 1)$ through computation, but in order for $\{ X_{uv} \}_{(u,v) \in T}$ be a Gaussian process, we need to show $\sum_{(u, v) \in T_0} a_{uv}X_{uv}$ is a normal distribution for any $T_0 \subseteq T$ finite set. I do not see why this is true as $X_{uv}$ are not independent across different $u$ and $v$.

Answer 1

With your notation $X_{uv}$ is a linear combination of $A_{ij}$, namely $X_{uv}=\sum_{1\le i\le m,1\le j\le n}v_iu_jA_{ij}$. So $$ \sum_{(u,v)\in T_0}a_{uv}X_{uv}=\sum_{(u,v)\in T_0}\sum_{1\le i\le m,1\le j\le n}a_{uv}v_iu_jA_{ij}=\sum_{1\le i\le m,1\le j\le n}\sum_{(u,v)\in T_0}a_{uv}v_iu_jA_{ij} $$ which is a sum of independent Gaussian variables. Hence it is Gaussian.

Answer 2

$$\sum_{(u,v)\in T_0}a_{uv}X_{uv}=\mathrm{trace}(A\sum_{(u,v)\in T_0}a_{uv}u\otimes v).$$