The book of Linear Algebra and Its Applications by Peter Lax talks about Law of Inertia for positive, negative zero eigenvalues of a linear map. But the book does not present a practice application. For example
$$M = \left(\begin{array}{cccc} 0 & 5 & 1 & 0\\ 5 & 0 & 5 & 0\\ 1 & 5 & 0 & 5\\ 0 & 0 & 5 & 0 \end{array}\right)$$
I know that $M$ has two positive eigenvalues and two negative eigenvalues, but how to use the Law of Inertia for get this? I mean determine the sign without explicitly calculating
Does require some sort of calculation, either completing the square (Lagrange's method) or doing the same thing backwards. Here, $PQ=QP=I.$ We see that the diagonal elements of $D$ are two positive and two negative numbers. Numerically, the eigenvalues, which we are ignoring, are $$ -7.672402397924414971791413120, \; \; -3.561373056202417532539267706, \; \; 2.671382004610319545357997188, \; \; 8.562393449516512958972683639$$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 1 & - \frac{ 1 }{ 5 } & 1 & 0 \\ - \frac{ 5 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 5 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 1 & - \frac{ 5 }{ 2 } \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 5 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & \frac{ 25 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 3 }{ 5 } & - \frac{ 4 }{ 5 } & 1 & 0 \\ 0 & 0 & - \frac{ 5 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & \frac{ 25 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 5 } & 0 \\ - 1 & 1 & - \frac{ 4 }{ 5 } & 0 \\ 0 & 0 & 1 & - \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 10 & 5 & 6 & 0 \\ 5 & 0 & 5 & 0 \\ 6 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 10 & 0 & 6 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 2 & 0 \\ 6 & 2 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & - \frac{ 3 }{ 5 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } & 0 \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 5 } & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 2 & 0 \\ 0 & 2 & - \frac{ 18 }{ 5 } & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & \frac{ 4 }{ 5 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 1 & 0 \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 5 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 5 } & 0 \\ - 1 & 1 & - \frac{ 4 }{ 5 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$
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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 1 & - \frac{ 5 }{ 2 } \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 5 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 5 } & 0 \\ - 1 & 1 & - \frac{ 4 }{ 5 } & 0 \\ 0 & 0 & 1 & - \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & \frac{ 25 }{ 2 } \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 1 & - \frac{ 1 }{ 5 } & 1 & 0 \\ - \frac{ 5 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 5 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 1 & - \frac{ 5 }{ 2 } \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 5 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & \frac{ 25 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 3 }{ 5 } & - \frac{ 4 }{ 5 } & 1 & 0 \\ 0 & 0 & - \frac{ 5 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 10 & 0 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & \frac{ 25 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 5 } & 0 \\ - 1 & 1 & - \frac{ 4 }{ 5 } & 0 \\ 0 & 0 & 1 & - \frac{ 5 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 5 & 1 & 0 \\ 5 & 0 & 5 & 0 \\ 1 & 5 & 0 & 5 \\ 0 & 0 & 5 & 0 \\ \end{array} \right) $$