Sample variance of a normal distribution If my sample variance of a random sample from a $\mathop{\mathcal N}\left(\mu,\sigma^2\right)$ distribution  is
$$S_2^2=\frac{1}{n}\sum_{i=1}^n{\left(X_i-\bar{X}\right)}^2,$$
then
$$Q \mathrel {:=}\frac{\sum_{i=1}^n{\left(X_i-\bar{X}\right)^2}}{\sigma^2}\frac{n}{n} = \dfrac{nS_2^2}{\sigma^2}.$$
Does this mean that $Q$ has a $\chi^2$-distribution with $n$ degrees of freedom?
 A: With $\mathbf X \mathrel{:=} \left(X_1, \ldots, X_n\right)^\top$ we have $\mathbf X \sim \mathop{\mathcal N}\left(\mu  \mathbf 1_n, \sigma^2  I_n\right)$ and
$$
\sum_{i=1}^n \left(X_i - \bar X\right)^2 = \left(C_n \mathbf X\right)^\top \left(C_n \mathbf X\right) = \mathbf X^\top C_n \mathbf X, 
$$
where $C_n$ denotes the centering matrix of order $n$.
Idempotency and symmetry of $C_n$ allow for the decomposition $C_n = AA^\top$ with $A^\top A = I_{\mathop{\mathrm{rk}}\left(C_n\right)}$ and $A \in \mathbb R^{n \times \mathop{\mathrm{rk}}\left(C_n\right)}$.
With $\mathop{\mathrm{rk}}\left(C_n\right) = \mathop{\mathrm{tr}}\left(C_n\right) = n-1$ this gives
$$
\frac{1}{\sigma}A^\top \mathbf X \sim \mathop{\mathcal N}\left(\frac{\mu}{\sigma} A^\top \mathbf 1_n, I_{n-1}\right)
$$
and hence
$$
Q = \sum_{i=1}^n \frac{\left(X_i - \bar X\right)^2}{\sigma^2}
= \frac{1}{{\sigma^2}} \mathbf X^\top C_n \mathbf X
= \left\|\frac{1}{\sigma}A^\top \mathbf X\right\|_2^2 \sim \mathop{\chi_{n-1}^2}\left(\lambda\right),
$$
where $\lambda = \left\|\frac{\mu}{\sigma} A^\top \mathbf 1_n\right\|_2^2$ denotes the non-centrality parameter of the non-central chi-squared distribution with $n-1$ degrees of freedom.
Finally, $\left\|\frac{\mu}{\sigma} A^\top \mathbf 1_n\right\|_2^2 = \frac{\mu^2}{\sigma^2}\mathbf 1_n^\top C_n \mathbf 1_n = 0$ yields $Q \sim \mathop{\chi_{n-1}^2}$, i.e., $Q$ follows a (central) chi-squared distribution with $n-1$ degrees of freedom.
