An Exercise of noncentral $\chi^2$ distribution. Let $Y_1,\ldots,Y_n$ be independent random variables with $Y_k$ distributed as $N\sim(a_k,\sigma^2)$, and $\bar Y=\sum_{k=1}^{n}\frac{Y_k}{n}$ denote the sample mean, $S^2$ denotes the sample variance.
Then show that 
$$\frac{(n-1)S^2}{\sigma^2}=\frac{1}{\sigma^2}\sum_{k=1}^{n}(Y_k-\bar Y)^2$$
is distributed as noncentral $\chi^2$ with $(n-1)$ degrees of freedom and noncentrality parameter $\lambda=\sum_{k=1}^n\frac{(a_k-\bar a)^2}{\sigma^2}$, where $\bar a=\sum_{k=1}^{n}\frac{a_k}{n}$.
I have only the idea of Moment Generating Function(MGF) Technique to prove this type of proof based on transformation. But this is almost impossible to me use the MGF technique for the given exercise.
How can i prove it easily?
 A: Since the $Y$s are independent and the distributions are normal and the variances are all equal, the distribution is spherically symmetric, so the distribution of sum of squares will still be the same if we look at a rotated coordinate system.
Let $P$ be the orthogonal projection of the column vector $(Y_1,\ldots, Y_n)^T$ onto the one-dimensional space $x_1=\cdots=x_n$, and let $Q=I-P$ be the complementary orthogonal projection onto the $(n-1)$-dimensional space $x_1+\cdots+x_n=0$.  Then
$$
Q\begin{bmatrix} Y_1 \\  \vdots \\ Y_n \end{bmatrix} = \begin{bmatrix} Y_1-\bar Y \\  \vdots \\ Y_n-\bar Y \end{bmatrix} \text{ and }Q\begin{bmatrix} a_1 \\  \vdots \\ a_n \end{bmatrix} = \begin{bmatrix} a_1-\bar a \\  \vdots \\ a_n-\bar a \end{bmatrix}.
$$
If we express everything in a rotated coordinate system with one axis in the direction of the space $x_1=\cdots=x_n$ and the others orthogonal to that, this is expressed as
$$
Q\begin{bmatrix} U_1 \\ U_2 \\  \vdots \\ U_n \end{bmatrix} = \begin{bmatrix} 0 \\ U_2 \\  \vdots \\ U_n \end{bmatrix} \text{ and }Q\begin{bmatrix} b_1 \\ b_2 \\  \vdots \\ b_n \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\  \vdots \\ b_n \end{bmatrix}.
$$
Then
$$
\frac{1}{\sigma^2}(U_2^2+\cdots +U_n^2) \sim \chi^2_{n-1,b_2^2+\cdots+b_n^2}.
$$
But
$$
(Y_1-\bar Y)^2 + \cdots+(Y_n-\bar Y)^2 = U_2^2+\cdots+U_n^2
$$
and
$$
(a_1-\bar a)^2+\cdots+(a_n-\bar a)^2 = b_2^2+\cdots+b_n^2.
$$
