Distribution of $\mathbf{v}^{\top} \Sigma^{- 1} \mathbf{v}$, when $\mathbf{v}$ is a multivariate normal with covariance $\Sigma$? What is the distribution of the quadratic form $\mathbf{v}^{\top} \Sigma^{-1} \mathbf{v}$, when $\mathbf{v}$ is a multivariate normal with covariance $\Sigma$ and zero means?
I suspect this is related to the Chi-square_distribution. In general there is no simple closed form expression for the distribution of a quadratic form of this type. But here the quadratic matrix is constructed with the inverse covariance matrix, which I hope makes the result simple.
 A: In a univariate settings $x \sim N(\mu,\sigma^2)$, standard normal $z \sim N(0,1)$ is obtained by centering (subtracting mean $\mu$) and scaling (dividing by standard deviation $\sigma$): $z = \frac{(x-\mu)}{\sigma}$.
The chi-square distribution is obtained from the sum of the squares of these standard normal variates, $\sum\limits_{i=1}^{k} z_i^2 \sim \chi(k)$.
Factoring $\Sigma$ using its eigenvector matrix $U$ and eigenvalue diagonal matrix $\Lambda$,  $$\Sigma = U \Lambda^2 U^\mathsf{T}$$  and
$$\Sigma^{-1} = U \Lambda^{-2} U^\mathsf{T}.$$
In a multivariate setting, $v_{n \times 1} \sim N(\mu_{n \times 1},\Sigma_{n \times n})$, standard vector normal $\mathbf{z}_{n \times 1} \sim N(\mathbf{0}_{n \times 1},\mathbf{I}_{n})$ is obtained by centering $-\mu$, rotating $U^\mathsf{T}$, and scaling $\Lambda^{-1}$:
$$\mathbf{z} = \Lambda^{-1} U^\mathsf{T}(\mathbf{v}-\mathbf{\mu}) $$
Define $$\Sigma^{-\frac{1}{2}} = \Lambda^{-1} U^\mathsf{T}$$.$$\mathbf{z} = \Sigma^{-\frac{1}{2}} (\mathbf{v}-\mathbf{\mu}) $$
Here $\mathbf{\mu} = 0$, so standard normal vector $\mathbf{z}$ is obtained by scaling and rotating normal vector $\mathbf{v}$, $\mathbf{z} = \Sigma^{-\frac{1}{2}} \mathbf{v}$.
Rewriting $$\mathbf{v}^\mathsf{T} \Sigma^{-1} \mathbf{v} = (\Sigma^{-\frac{1}{2}} \mathbf{v})^\mathsf{T} \Sigma^{-\frac{1}{2}} \mathbf{v} = \mathbf{z}^\mathsf{T} \mathbf{z}$$ it is clear $\mathbf{v}^\mathsf{T} \Sigma^{-1} \mathbf{v}$ is the sum of the squared elements of $\mathbf{z}$.
Therefore, if $\mathbf{v}$ is an n-length vector, $\mathbf{v} \sim N(\mathbf{0,\Sigma}),$ $$\mathbf{v}^\mathsf{T} \Sigma^{-1} \mathbf{v} \sim \chi(n)\quad.$$
