Proof of Sylvester's law of inertia by comparing subspaces

Proof

We show that the number of positive entries is the same, by finding two subspaces that would have to meet.

1. Let $q(x) = x^T A x$ for $x \in \R^n$. Let $d_1, \dots, d_n$ be the diagonal entries of $D$, and $d'_1, \dots, d'_n$ those of $D'$.
2. For $y \in \R^n$, the transpose of a product gives $q(Py) = y^T P^T A P y = y^T D y = \sum_i d_i y_i^2$.
3. In the same way, $q(P' z) = \sum_j d'_j z_j^2$ for $z \in \R^n$.
4. Let $p$ be the number of $i$ with $d_i > 0$, and $p'$ the number of $j$ with $d'_j > 0$. Let $I = \set{i : d_i > 0}$ and $J = \set{j : d'_j \le 0}$.
5. Let $S$ be the span of the columns $P e_i$ with $i \in I$. Let $T$ be the span of the columns $P' e_j$ with $j \in J$.
6. The columns of an invertible matrix $P$ are linearly independent: if $Pc = 0$, then $c = P^{-1} P c = 0$. The same holds for $P'$.
7. By step 6 and the definition of dimension, $\dim S = p$ and $\dim T = n - p'$.
8. Let $x \in S$ with $x \ne 0$. Then $x = Py$, where $y \ne 0$ and $y_i = 0$ for $i \notin I$. By step 2, $q(x) = \sum_{i \in I} d_i y_i^2 > 0$.
9. Let $x \in T$. Then $x = P' z$, where $z_j = 0$ for $j \notin J$. By step 3, $q(x) = \sum_{j \in J} d'_j z_j^2 \le 0$.
10. By steps 8 and 9, $S \cap T = \set{0}$, so $\dim(S \cap T) = 0$.
11. By Grassmann's formula, $\dim(S + T) = p + (n - p')$.
12. The subspace $S + T$ lies in $\R^n$, which has dimension $n$. By the dimension of a subspace, $p + n - p' \le n$, so $p \le p'$.
13. Exchanging the roles of $(P, D)$ and $(P', D')$ gives $p' \le p$. So $p = p'$.
14. The matrix $-A$ is symmetric, with $P^T (-A) P = -D$ and $P'^T (-A) P' = -D'$. The positive entries of $-D$ are the negative entries of $D$.
15. Applying step 13 to $-A$, the matrices $D$ and $D'$ have the same number of negative diagonal entries.
16. The number of zero diagonal entries is $n$ minus the numbers of positive and negative entries. By steps 13 and 15, it is also the same for $D$ and $D'$.