Projection of a multivariate normal into a univariate normal; calculating variance from covariance matrix On the PDF of a multivariate gaussian, we have $e^{-\frac{1}{2}(x-\mu)^t\Sigma^{-1}(x-\mu)}$
This leads me to assume that if I want restrict my distribution to a given direction, I will have $(x-\mu)^t\Sigma^{-1}(x-\mu)=(x-\mu)^2/\sigma^2$ where $\sigma^2$ is a variance in the direction of $(x-\mu)$.
Maybe not exactly this, but at least up to a constant.
Note that this is not an empty affirmation ("this number is another number up to a constant"), but rather that for any $x$ such that $(x-\mu)$ still lies in the same direction, we have a density compatible with the normal distribution.
I have, unfortunately, been unable to find a proof of this fact. How can it be proven?
 A: Let $X\sim\mathcal N_n(\mu,\Sigma)$ and suppose we have a unit vector $u \in \mathbb R^n$ and we want to see what the distribution of the projection of $X$ onto the subspace given by $u$ is. This is
$$
u(u^Tu)^{-1}u^TX = (u^TX) u
$$
so $u^TX$ is the coordinate in this one dimensional subspace and $(u^TX)u$ is the corresponding element of $\mathbb R^n$.
This is a linear transformation of a Gaussian so
$$
u^TX \sim \mathcal N(u^T\mu, u^T\Sigma u) 
$$
which is just a univariate Gaussian now with variance $\sigma^2 = u^T\Sigma u$. The optimizers of $u \mapsto u^T \Sigma u$ are given by the eigenvectors of $\Sigma$, and in particular the smallest variance of $u^T X$ will be for an eigenvector of $\lambda_\min(\Sigma)$ while the largest variance will be for an eigenvector of $\lambda_\max(\Sigma)$. For the mean, we have
$$
u^T\mu = \|\mu\|\cos\theta
$$
so the mean will contract or stay the same depending on the angle between $u$ and $\mu$.

(Updating to address the comments)$\newcommand{\E}{\operatorname{E}}$
The result that $\text{Var}(u^TX) = u^T\Sigma u$ follows from the definition of the variance and only requires that $\mu$ and $\Sigma$ are finite and well-defined:
$$
\text{Var}(u^TX) = \E[(u^TX - u^T\mu)^2] = \E[(u^TX - u^T\mu)(u^TX - u^T\mu)^T] \\
= u^T\E[(X - \mu)(X - \mu)^T]u = u^T\Sigma u.
$$
Normality is not required.
