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  "body": "\n## Proof\n\nWe show that the number of positive entries is the same, by finding two subspaces that would have to meet.\n\n1. 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 $D'$.\n2. For $y \\in \\R^n$, the [[transpose of a product|09GK5PS824]] gives $q(Py) = y^T P^T A P y = y^T D y = \\sum_i d_i y_i^2$.\n3. In the same way, $q(P' z) = \\sum_j d'_j z_j^2$ for $z \\in \\R^n$.\n4. Let $p$ be the number of $i$ with $d_i \u003e 0$, and $p'$ the number of $j$ with $d'_j \u003e 0$. Let $I = \\set{i : d_i \u003e 0}$ and $J = \\set{j : d'_j \\le 0}$.\n5. Let $S$ be the [[span|0HR573CHNM]] of the columns $P e_i$ with $i \\in I$. Let $T$ be the span of the columns $P' e_j$ with $j \\in J$.\n6. The columns of an [[invertible|0TWTS88TWN]] matrix $P$ are [[linearly independent|0EVHJFTG9Z]]: if $Pc = 0$, then $c = P^{-1} P c = 0$. The same holds for $P'$.\n7. By step 6 and the definition of [[dimension|05G2G0P224]], $\\dim S = p$ and $\\dim T = n - p'$.\n8. 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, $q(x) = \\sum_{i \\in I} d_i y_i^2 \u003e 0$.\n9. Let $x \\in T$. Then $x = P' z$, where $z_j = 0$ for $j \\notin J$. By step 3, $q(x) = \\sum_{j \\in J} d'_j z_j^2 \\le 0$.\n10. By steps 8 and 9, $S \\cap T = \\set{0}$, so $\\dim(S \\cap T) = 0$.\n11. By [[Grassmann's formula|06HTJNCMPA]], $\\dim(S + T) = p + (n - p')$.\n12. The subspace $S + T$ lies in $\\R^n$, which has [[dimension $n$|0QK7FPE6VE]]. By [[the dimension of a subspace|0QW084460E]], $p + n - p' \\le n$, so $p \\le p'$.\n13. Exchanging the roles of $(P, D)$ and $(P', D')$ gives $p' \\le p$. So $p = p'$.\n14. The matrix $-A$ is [[symmetric|0N28QN4CYA]], with $P^T (-A) P = -D$ and $P'^T (-A) P' = -D'$. The positive entries of $-D$ are the negative entries of $D$.\n15. Applying step 13 to $-A$, the matrices $D$ and $D'$ have the same number of negative diagonal entries.\n16. 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 $D'$.\n\n## Idea\n\nThe positive entries of $D$ give a subspace of dimension $p$ on which the form is positive. The other entries of $D'$ give a subspace of dimension $n - p'$ on which it is not positive. These subspaces meet only in $0$, so their dimensions add to at most $n$, which forces $p \\le p'$.\n",
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