$$X_{i}:=\text{ith sample which was extracted from}~~\mathcal{N}(\mu,\sigma^2)\tag{1}$$
$$i\in\mathbb{N}_{n}^{*}~~\Leftrightarrow~~i\in\{1,2,\ldots,n\}\tag{2}$$
I assume these random variables are mutually independent.
$$Z_{i}:={X_{i}-\mu\over\sigma}~~\leftarrow~~\text{standardization}\tag{3}$$
$$\therefore~~Z_{i}\sim\mathcal N(0,1)\tag{4}$$
$$\sum_{i=1}^{n}Z_{i}^2=\sum_{i=1}^{n}\left({X_{i}-\mu\over\sigma}\right)^2~~\sim~~\chi_{n}^2\tag{5}$$
$$\bar{X}:={1\over n}\sum_{i=1}^{n}X_{i}\tag{6}$$
I've been confused to understand why the following can be said.
$$\sum_{i=1}^{n}\left({X_{i}-\bar X\over\sigma}\right)^2\sim\chi_{n-1}^2\tag{7}$$
The book says about it with following statements.
$$\sum_{i=1}^{n}\left({X_{i}-\bar X\over\sigma}\right)=0~\text{is held hence,}\tag{8}$$
$${X_{i}-\bar X\over\sigma}~\text{is linearly independent and,}\tag{9}$$
$$\text{number of degrees of freedom is substracted by 1 compared to}~\mu\tag{10}$$
I can't even derive eqn8 in the first place.
I need your idea.
ADD
I derived the eqn8. It was too trivial.
My this post is relavant with the following statements in the book.
$$\text{population}\sim\mathcal N(\mu,\sigma^2)\tag{11}$$
$$X_{1},\ldots,X_{n}:=\text{randomly extracted samples from the population}\tag{12}$$
$$U:={\bar X-\mu\over\sqrt{{S^2\over n}}}\tag{13}$$
$$U\sim\mathcal t_{n-1}\tag{14}$$
$$\left(\bar X={1\over n}\sum_{i=1}^{n}X_{i}~~,~~S^2={1\over n-1}\sum_{i=1}^{n}\left(X_{i}-\bar X\right)^2\right)\tag{15}$$
