I'm trying to convince myself that the following two definitions of the signature of a symmetric matrix are the same. The first one says it is the triple $(p,n,z)$ where $p$ is the number of $1$'s, $n$ is the number of $-1$'s, and $z$ is the number of $0$'s in the normal form of the symmetric matrix. The other definition says it is the triple $(p,n,z)$ where $p$, $n$, and $z$ are the number of positive, negative, and zero eigenvalues of the matrix, respectively.
My first guess was that this is obvious, since the diagonal elements of the matrix $P^tAP$, where $A$ is a given symmetric matrix, which is the normal form of $A$, are the eigenvalues. But now I doubt whether it is true, because this is certainly true for $P^{-1}AP$, however, the source I use does not claim that the matrix $P$ which reduces $A$ to the normal form must be orthogonal.
This same issue comes up over and over and over and over and over again. For symmetric matrices over the reals, the equivalence relation that preserves all the signs, as in Sylvester's Law of Inertia, is congruence. If $A$ and $B$ are real symmetric, it means there is an invertible matrix $Q$ such that $$ Q^T A Q = B. $$
Give me an example symmetric matrix $H$ with integer coefficients, I will put here how to solve $P^T HP = D$ diagonal. For this problem, all integers, we can arrange that $P$ has all rational entries, which means $D$ does as well, and $\det P = \pm 1.$ The diagonal entries of $D$ will not be the eigenvalues (which could very well be irrational)
Here is an example with really, really ugly eigenvalues: $$ H = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$ $$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 11 }{ 8 } & 1 & 0 & 0 \\ \frac{ 4 }{ 25 } & - \frac{ 12 }{ 25 } & 1 & 0 \\ - \frac{ 271 }{ 143 } & \frac{ 241 }{ 143 } & - \frac{ 228 }{ 143 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 271 }{ 143 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 241 }{ 143 } \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) $$
Having reached this diagonal matrix $D,$ we can continue to get only $0,1,-1$ on the diagonal by taking $R^T P^T H P R,$ where
$$ R = R^T = \left( \begin{array}{rrrr} \frac{1}{\sqrt8} & 0 & 0 & 0 \\ 0 & \frac{ \sqrt 8 }{ 5 } & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ \sqrt {143} } & 0 \\ 0 & 0 & 0 & \frac{ \sqrt {143} }{ \sqrt {569} } \\ \end{array} \right) $$
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$$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 8 & 0 & 4 & 3 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 4 & - \frac{ 3 }{ 2 } & 7 & 12 \\ 3 & \frac{ 23 }{ 8 } & 12 & 17 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 3 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 0 & - \frac{ 3 }{ 2 } & 5 & \frac{ 21 }{ 2 } \\ 3 & \frac{ 23 }{ 8 } & \frac{ 21 }{ 2 } & 17 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & - \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 0 & - \frac{ 3 }{ 2 } & 5 & \frac{ 21 }{ 2 } \\ 0 & \frac{ 23 }{ 8 } & \frac{ 21 }{ 2 } & \frac{ 127 }{ 8 } \\ \end{array} \right) $$
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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 3 }{ 8 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & \frac{ 23 }{ 8 } \\ 0 & 0 & \frac{ 143 }{ 25 } & \frac{ 228 }{ 25 } \\ 0 & \frac{ 23 }{ 8 } & \frac{ 228 }{ 25 } & \frac{ 127 }{ 8 } \\ \end{array} \right) $$
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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 41 }{ 25 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & \frac{ 228 }{ 25 } \\ 0 & 0 & \frac{ 228 }{ 25 } & \frac{ 463 }{ 25 } \\ \end{array} \right) $$
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$$ E_{6} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{6} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 271 }{ 143 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 241 }{ 143 } \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{6} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{6} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 11 }{ 8 } & 1 & 0 & 0 \\ \frac{ 4 }{ 25 } & - \frac{ 12 }{ 25 } & 1 & 0 \\ - \frac{ 271 }{ 143 } & \frac{ 241 }{ 143 } & - \frac{ 228 }{ 143 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 271 }{ 143 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 241 }{ 143 } \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 11 }{ 8 } & 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 12 }{ 25 } & 1 & 0 \\ \frac{ 3 }{ 8 } & - \frac{ 23 }{ 25 } & \frac{ 228 }{ 143 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$
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