reference for linear algebra books that teach reverse Hermite method for symmetric matrices

I think I have the energy today to fill in the details of this png image of a calculation

from this question: Finding $P$ such that $P^TAP$ is a diagonal matrix

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 1 & 1 \\ 0 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & - 2 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & \frac{ 1 }{ 2 } & - 1 \\ 2 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & - 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) $$

$$ H = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) $$ $$ D_0 = H $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 0 & 4 \\ 2 & 4 & 4 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 2 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 4 \\ 0 & 4 & 0 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 4 & - 2 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & - 1 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 4 \\ 0 & 4 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrr} 1 & - 4 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & - 2 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 1 & 1 \\ 0 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & - 2 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & \frac{ 1 }{ 2 } & - 1 \\ 2 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & - 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 4 & 8 \\ 2 & 8 & 4 \\ \end{array} \right) $$

I actually just read this in Shilov's Linear Algebra (Dover edition) while reviewing for my prelims. He handles this at the beginning of chapter 7; he states it as a theorem about finding a canonical basis for quadratic forms, but since those are the same as symmetric bilinear forms in characteristic$\neq 2$, and since the matrix of a bilinear form transforms as $A\mapsto P^t AP$, that is exactly the theorem you're looking for.

You can find a description of a very similar method in "Schaum's Outline of Linear Algebra", by Lipschutz and Lipson.

In the first edition it's introduced in exercise 12.9 (page 270). In the fifth edition, it's introduced as Algorithm 12.1 (page 361); you can find it in this answer.

After some more research, I found another similar algorithm in "Schaum's Outline of Matrix Operations", by Bronson, on page 145 (Chapter 16).