Distribution of $(X_1-\mu)^T\Sigma^{-1}(X_1-\mu)$ is chi-squared?

Question

If $(X_i)_{i=1}^{20}\sim N_6(\mu,\Sigma),$ then find the distribution of

$$ (X_1-\mu)^T\Sigma^{-1}(X_1-\mu)$$

The solution is $\chi^2_6,$ but could someone show why? I only know that the sum of standard normal variables is itself chi-squared but I'm not sure how to approach this.

Answer 1

The (finite-dimensional version of the) spectral theorem has a quick corollary that says a positive-definite symmetric real matrix $\Sigma$ has a positive-definite symmetric real square root $\Sigma^{1/2},$ and the inverse of $\Sigma^{1/2}$ is the positive-definite symmetric real square root of $\Sigma^{-1},$ and we denote it $\Sigma^{-1/2}.$ So $$ \Sigma^{-1/2} (X_1-\mu) \sim \operatorname N_6(0, I_6). $$ Therefore the distribution of the sum of squares of the components is $\chi^2_6.$

This, however, does not work if $\Sigma$ is singular. But if $\Sigma$ is non-negative-definite and symmetric and has real entries then $(X_1-\mu)^T \Sigma^{-1} (X_1-\mu) \sim \chi^2_{\operatorname{rank}\Sigma}$ (where $\text{“ } \Sigma^{-1}\text{ ''}$ denotes a sort of generalized inverse, the details of whose nature are omitted here).

So a question is: are you familiar with the spectral theorem?