Vector version of the proof that $\overline{X}$ and $S^2$ are independent

Solution 1:

For $P\equiv \iota(\iota^{\top}\iota)^{-1}\iota$, it's symmetric and we have $P^{2}=\iota(\iota^{\top}\iota)^{-1}\iota=P$, meaning it's idempotent. Using this result, we deduce $(I-P)(I-P)=I-P-P+P^{2}=I-P$, thus $I-P$ is also idempotent.

Therefore, $$S^{2}=(n-1)^{-1}X^{\top}\underbrace{(I-P)}X=(n-1)^{-1}X^{\top}\underbrace{(I-P)(I-P)}X=(n-1)^{-1}\bigl((I-P)X\bigr)^{\top}\bigl((I-P)X\bigr),$$ hence a function of $(I-P)X$.

Update for your question: Note that $(I-P)X$ and $\iota^{\top} X$ are all normal distributed, hence uncorrelated$\iff$independent. $$ \mathrm{Cov}\bigl((I-P)X,\iota^{\top}X\bigr)=\mathbb{E}[(I-P)X(\iota^{\top}X)]-\mathbb{E}[(I-P)X]\mathbb{E}[\iota^{\top} X]=0-(I-P)\mu(\iota^{\top}\mu)=0.$$