Why is Hotelling's $T^2 \sim \chi^2_p$ for large $n$? I'm interested in some proof (simple if possible) as to why Hotelling's $T^2$ is chi-squared distributed for large n. I understand and can show that the Mahalanobis Distance is in fact chi-squared distributed (as bellow), but I have a little bit of trouble showing it should be the same case for the Hotelling's $T^2$ case since there is the component $n$ and I'm not sure what to do with it.
Hotelling's $T^2$: $n(\bar{\boldsymbol{X}} - \boldsymbol{\mu})^T\boldsymbol{S}^{-1}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})$
I know that for large $n$ we can assume $\boldsymbol{S}^{-1} \approx \boldsymbol{\Sigma}^{-1}$, and that $\boldsymbol{\Sigma}^{-1} = \boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}$, so far large $n$ we can update the Hotelling's $T^2$ formula to:
$n(\bar{\boldsymbol{X}} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-1}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})$
and expand it to
$n(\bar{\boldsymbol{X}} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})$
Mahalanobis Distance proof:
$$
\begin{align}
D &= (\boldsymbol{X} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-1}(\boldsymbol{X} - \boldsymbol{\mu}) \\
&= (\boldsymbol{X} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}(\boldsymbol{X} - \boldsymbol{\mu}) \\
&= \big(\boldsymbol{\Sigma}^{-\frac{1}{2}}(\boldsymbol{X} - \boldsymbol{\mu})\big)^T\big(\boldsymbol{\Sigma}^{-\frac{1}{2}}(\boldsymbol{X} - \boldsymbol{\mu})\big)\\
&= \boldsymbol{Y}^T\boldsymbol{Y} \\
&= ||\boldsymbol{Y}||^2\\
&= \sum \limits_{k=1}^lY_k^2 \\
D &\sim \chi_k^2
\end{align}
$$
I know also that $\frac{n-p}{(n-1)p}T^2 \sim F_{p, n-p}$ and that an F distribution with large n and low p is approximately $\chi_p^2$ distributed, but when trying to connect this information to write a proof I end up being lost and frustrated.
 A: Turns out the answer to this was rather simple. We can say then that Hotelling's $T^2$ is in fact the t-test formula squared.
\begin{align}
t &= \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}} \\
t^2 &= \bigg( \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}} \bigg)^2 \\
&= \bigg( \frac{n^{\frac{1}{2}}(\bar{X}-\mu)}{\sigma} \bigg)\bigg( \frac{n^{\frac{1}{2}}(\bar{X}-\mu)}{\sigma} \bigg)
\end{align}
Now, we can see this formula under the light of a multivariate case, and that we are likely dealing with sample statistics rather than population ones. However, as $n \rightarrow \infty$ the sample variance-covariance matrix $\boldsymbol{S}$ gets closer to the population variance-covariance matrix $\boldsymbol{\Sigma}$.
As we know that the square root matrix of $\boldsymbol{\Sigma}$ is also symmetric we have that $\boldsymbol{\Sigma^{-1}} = \boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}$. Then we have
\begin{align}
T^2 &= n(\bar{\boldsymbol{X}} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-1}(\bar{\boldsymbol{X}} - \boldsymbol{\mu}) \\
&= n^{\frac{1}{2}}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})^T\boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}n^{\frac{1}{2}}(\bar{\boldsymbol{X}} - \boldsymbol{\mu}) \\
&= \big(n^{\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})\big)^T\big(n^{\frac{1}{2}}\boldsymbol{\Sigma}^{-\frac{1}{2}}(\bar{\boldsymbol{X}} - \boldsymbol{\mu})\big)\\
&= \boldsymbol{Y}^T\boldsymbol{Y} \\
&= ||\boldsymbol{Y}||^2\\
&= \sum \limits_{k=1}^lY_k^2 \\
&\sim \chi_k^2
\end{align}
