Gaussians are rotationally invariant I am trying to understand why the following statements are true:


*

*If $g \sim N(0, I_n)$ and $u \in \mathbf{R}^n$ is deterministic, then 
$\langle g, u \rangle \sim N(0, \|u\|_2^2)$. 

*If $G$ is an $m \times n$ guassian random matrix with i.i.d entries distributed 
$N(0, 1)$, and if $u \in \mathbf{R}^n$ is a fixed unit vector then 
$Gu \sim N(0, I_m)$


I tried to make progress using the fact that $g \sim N(0, I_n)$ and $U \in O(n)$ means $Ug \sim N(0, I_n)$. 
 A: $\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$I see you have an answer for the first part. I'll help you for the second part. 
We need to show that for every $0\neq b\in\mathbb{R}^m$ we have $b^TGu$ has univariate normal distribution. 
\begin{align}
b^TGu=b^T
\begin{pmatrix}
\sum_{i=1}^n u_i g_{1i}\\
\vdots\\
\sum_{i=1}^n u_i g_{1m}
\end{pmatrix}=\sum_{k=1}^m \sum_{i=1}^n b_k u_i g_{ki}
\end{align}
That is just a sum of i.i.d. normally distributed random variables so that means that $b^TGu$ is normally distributed. So $Gu$ has multivariate normal distribution. The mean is clearly zero. Let's find the variance-covariance matrix. Let $i\neq j$. 
\begin{align}
\Cov((Gu)_i, (Gu)_j)= 0
\end{align}
simply because  the random variables in $G$ are all independent and if $i\neq j$ we don't encounter the same random variable in both $(Gu)_i$ and $(Gu)_j$. If $i=j$ we have:
\begin{align}
\Cov((Gu)_i, (Gu)_j)=\Var((Gu)_i)=\sum_{l=1}^n u_l^2 \Var(g_{il})=\sum_{i=1}^nu_i^2=1
\end{align}
since $u$ is a unit vector. This is the result we wanted: $Gu\sim \mathcal{N}(0, I_m)$.
A: Any linear combination of independent normals is normally distributed. For a proof see here. Since $g\sim N(0, I_n)$, $g_i$ are i.i.d $N(0,1)$ whence
$$
E<g,u>=\sum Eg_iu_i=\sum u_iEg_i=0
$$
and
$$
\text{Var}<g,u>=\sum \text{Var}(g_iu_i)=\sum u_i^2\text{Var} g_i=\lVert u\rVert^2
$$
by indepdence. Hence $<g,u>\sim N(0, \lVert u\rVert^2)$
A: Actually, on second thought this is easier than anticipated. 
The first part can easily be seen: 
$$
\mathbf{E}\langle g, u \rangle^2 = 
\mathbf{E} u^T gg^T u  = 
u^T I_n u = \|u\|_2^2. 
$$
Also, 
$$
\mathbf{E} \langle g, u\rangle = \langle u, \mathbf{E} g\rangle = 0.
$$
