$\chi^2$ distribution straight from a joint distribution?

Question

If $(X,Y)$ are normally distributed with mean $0$ and variance matrix $\Sigma$, then why is $$(X,Y)\Sigma^{-1}(X,Y)^T$$ a $\chi^2$ distribution with $2$ degrees of freedom?

Answer 1

If $Z = \Sigma^{-1/2} (X,Y)^T$, then $Z$ is normally distributed with mean 0 and covariance matrix $$ \mathbb E[Z Z^T] = \Sigma^{-1/2} \mathbb E[(X,Y)^T (X,Y)] \Sigma^{-1/2} = \Sigma^{-1/2} \Sigma \Sigma^{-1/2} = I$$ i.e. its two entries $Z_1$ and $Z_2$ are independent standard normal random variables. Now $(X,Y) \Sigma^{-1} (X,Y)^T = Z^T Z = Z_1^2 + Z_2^2$ is the sum of squares of two independent standard normal random variables. This is one characterization of the $\chi^2$ distribution with $2$ degrees of freedom.