Distributions – Proving $\sum_{i=1}^n(X_i-\overline X_n)^2-\sum_{i=1}^m(X_i-\overline X_m)^2 \sim \chi^2_{n-m}$

Suppose $$X_1,X_2,\ldots,X_n$$ are i.i.d $$N(0,1)$$ random variables. For $$2\le m, let $$S_m^2=\sum_{i=1}^m(X_i-\overline X_m)^2$$ and $$S_n^2=\sum_{i=1}^n(X_i-\overline X_n)^2$$ where $$\overline X_m=\frac1m\sum_{i=1}^m X_i$$ and $$\overline X_n=\frac1n\sum_{i=1}^n X_i$$. I am trying to prove that $$T=S_n^2-S_m^2 \sim \chi^2_{n-m}$$.

My idea is to write $$T$$ as a quadratic form $$X^TAX$$ where $$X=(X_1,\ldots,X_n)^T$$ and $$A$$ is a symmetric matrix of order $$n$$. Then $$T$$ would have a $$\chi^2$$ distribution if and only if $$A$$ is idempotent, the degrees of freedom of $$T$$ being the rank of $$A$$ (or the trace of $$A$$ since $$A$$ is idempotent).

Now $$S_n^2=X^TA_1X$$ where $$A_1=I_n-\frac1n \mathbf1_n\mathbf1_n^T$$ and $$\mathbf1_n$$ is a vector of all ones.

If $$Y=(X_1,\ldots,X_m)^T$$, then similarly, $$S_m^2=Y^TA_2Y$$ with $$A_2=I_m-\frac1m \mathbf1_m\mathbf1_m^T$$.

So I think $$T=X^TA_1X-Y^TA_2Y=X^TAX\,,$$

where $$A=A_1-\begin{pmatrix}A_2 & O_{m\times \overline{n-m}} \\ O_{\overline{n-m}\times m} & O_{n-m}\end{pmatrix}$$

I can show that $$A_1$$ and $$A_2$$ are idempotent, but verifying $$A$$ is idempotent is somewhat cumbersome. Is there any easier way out? Alternatively, is $$S_n^2-S_m^2$$ independent of $$S_m^2$$? I understand this would solve the problem since $$S_n^2 \sim \chi^2_{n-1}$$ and $$S_m^2 \sim \chi^2_{m-1}$$. Again, according to a theorem on quadratic forms, I just need to show $$T$$ is non-negative definite. Then from $$S_n^2 \sim \chi^2_{n-1}$$ and $$S_m^2 \sim \chi^2_{m-1}$$, it would follow that $$T\sim \chi^2_{(n-1)-(m-1)}$$. Any suggestions are welcome.

This is geometry.

There's not much to prove, actually, because you already know a lot.

1. From $$X_1, \ldots, X_m$$ there exist $$m-1$$ orthonormal linear combinations $$U_1, \ldots, U_{m-1}$$ that have iid standard Normal distributions independent of $$\bar X_m,$$ for which $$S_m^2 = U_1^2 + U_2^2 + \cdots + U_{m-1}^2.$$ (This is the standard variance decomposition associated with the mean.)

2. $$X_{m+1}, \ldots, X_n$$ are independent of $$X_1,\ldots X_m$$ and therefore $$(\bar X_m, U_1, U_2, \ldots, U_{m-1}, X_{m+1}, X_{m+1}, \ldots, X_n)$$ are independent.

3. Because $$n\bar X_n = m\bar X_m + (X_{m+1} + \cdots + X_n),$$ $$\bar X_n$$ is independent of $$U_1, \ldots, U_{m-1}.$$

4. Independence among Normal variables is equivalent to orthogonal linear combinations. Linear algebra tells us the $$m-1$$ linear combinations corresponding to $$U_1, \ldots, U_{m-1}$$ can be extended to an orthonormal basis $$U_1, \ldots, U_{m-1}, U_m, \ldots, U_{n-1}$$ of the space orthogonal to $$\bar X_n.$$ (This is a standard, important theorem. If you haven't seen it, prove it by induction--it's simple.)

5. Similarly (exactly as in $$(1)$$), there exist orthonormal $$V_1, \ldots, V_{n-1}$$ that are independent of $$\bar X_n$$ and for which $$S_n^2 = V_1^2 + \cdots + V_{n-1}^2.$$

6. Since $$(U_1, \ldots, U_{n-1})$$ and $$(V_1, \ldots, V_{n-1})$$ are both orthonormal bases for the space orthogonal to $$\bar X_n,$$ their sums of squares are equal: $$S_n^2 = U_1^2 + \cdots + U_{n-1}^2.$$

7. Subtracting, we find $$S_n^2 - S_m^2 = U_{m}^2 + U_{m+1}^2 + \cdots + U_{n-1}^2$$ is the sum of $$n-m$$ orthogonal standard Normal variables, whence (by definition) it has a $$\chi^2(n-m)$$ distribution, QED.