Distribution of $\mathbf{v}^{\top} \Sigma^{- 1} \mathbf{v}$, when $\mathbf{v}$ is a multivariate normal with covariance $\Sigma$? [duplicate]

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.

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, $$\mathbf{v} \sim N(\mathbf{0,\Sigma}),$$ $$\mathbf{v}^\mathsf{T} \Sigma^{-1} \mathbf{v} \sim \chi(n)\quad.$$