Sylvester's law of inertia

Suppose $Q$ is diagonalized in two bases $\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ and $\{\mathbf{f}_1,\dots,\mathbf{f}_n\}$, with positive, negative, and zero counts $(n_+, n_-, n_0)$ in the first basis and $(p_+, p_-, p_0)$ in the second. Order each basis so the positive coefficients come first, then the negative ones, then the zeros.

Suppose toward a contradiction that $n_+ > p_+$. Let $V_+ = \mathrm{span}(\mathbf{e}_1,\dots,\mathbf{e}_{n_+})$, a subspace of dimension $n_+$ on which $Q(\mathbf{x}) > 0$ for every nonzero $\mathbf{x} \in V_+$ (since only the positive squares in the $\mathbf{e}$-expansion are active on $V_+$).

Similarly, let $W_- = \mathrm{span}(\mathbf{f}_{p_+ + 1},\dots,\mathbf{f}_n)$, a subspace of dimension $n - p_+$ on which $Q(\mathbf{x}) \le 0$ for every $\mathbf{x} \in W_-$ (since only the negative and zero squares in the $\mathbf{f}$-expansion are active on $W_-$).

Now count dimensions inside $\mathbb{R}^n$: since $n_+ > p_+$, we have $\dim V_+ + \dim W_- = n_+ + (n - p_+) > n$. By the dimension formula for subspaces, two subspaces whose dimensions add up to more than $n$ cannot have trivial intersection; hence there exists a nonzero vector $\mathbf{v} \in V_+ \cap W_-$.

Because $\mathbf{v} \in V_+$ and $\mathbf{v} \neq \mathbf{0}$, we must have $Q(\mathbf{v}) > 0$; because $\mathbf{v} \in W_-$, we must simultaneously have $Q(\mathbf{v}) \le 0$, an outright contradiction. Therefore $n_+ \le p_+$, and by symmetry of the two bases $p_+ \le n_+$, so $n_+ = p_+$. Applying the exact same argument to $-Q$ gives $n_- = p_-$, and finally $n_0 = n - n_+ - n_- = p_0$.