In a problem sheet for my linear algebra course I was asked to find the standard form of the quadratic form $Q=8x_1^2 + 2x_2^2 +3x_3^2 +8x_2x_3$.
Following the steps in the lecture notes I arrived at the expression $Q=8y_1^2 + \frac{1}{2}((\sqrt{65}+5)y_2^2 + (5-\sqrt{65})y_3^2)$.
I am not sure whether the co-efficients are correct as the solution has not been given yet but my question is what benefit does re writing Q in this form give, and how does it link with the rest of linear algebra?
Edit:
let $X= \begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix}$,
$Q=X^T M X$,
Where $M$ is the matrix of coefficients,$ \begin{pmatrix} 8&0&0\\0&2&4\\0&4&3\end{pmatrix}$.
$X$ and $Y$ are related by $X=PY$. Where $P$ is the eigen vector matrix of M.
If you drop any requirement about orthogonality, you can solve $Q^T D Q = H$ with all rational numbers, $D$ diagonal, and $\det Q = 1.$ The equivalence relation is called congruence, so some books call this process congruence diagonalization. It is simply repeated "completing the square" done in reverse order.
In turn this shows what you get with Sylvester Inertia.
$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 3 \\ \end{array} \right) $$
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 3 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 3 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 5 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 5 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & 2 & 4 \\ 0 & 4 & 3 \\ \end{array} \right) $$