I am trying to understand how to classify quadratic forms and I have picked the following example: trying to classify $2x^2+2y^2+5xy-4x+y-6=0$.
What I have done:
I have considered the part $2x^2+2y^2+5xy=0$ with associated matrix $ A=\begin{bmatrix} 2 & \frac{9}{2} \\ \frac{5}{2} & 2 \end{bmatrix}$ with eigenvalues $\lambda_1=\frac{9}{2},\ \lambda_2 =-\frac{1}{2}$ and corresponding eigenvectors $\vec{v}_{\lambda_1}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\vec{v}_{\lambda_1}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix}$. This allows us to diagonalize $A$ and write the quadratic in the form $\frac{9}{2}(x')^2-\frac{1}{2}(y')^2=0$ in the new basis, which is a pair of intersecting lines.
My question is: how do I deal with the terms $-4x+y-6$?
Thank you very much.
UPDATE (SOLVED): this is the equation of a conic with a mixed term $xy$ which implies it has been rotated by an angle $\tan (2\theta)=\frac{5}{2-2}=\infty\Rightarrow 2\theta =\frac{\pi}{2}\Rightarrow \theta =\frac{\pi}{4}\Rightarrow$ we can eliminate the mixed term with the rotation $$x=X\cos\left(\frac{\pi}{4}\right)-Y\sin\left(\frac{\pi}{4}\right)=\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}, y=X\sin\left(\frac{\pi}{4}\right)+Y\cos\left(\frac{\pi}{4}\right)=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}$$ so the equation $2x^2+2y^2+5xy-4x+y-6=0$ becomes \begin{align*} &2\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\right)^2+5\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\right)\left(\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\right)+2\left(\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\right)^2-4\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\right)\\ &+\left(\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\right)-6=\frac{9}{2}X^2-\frac{1}{2}Y^2-\frac{3}{\sqrt{2}}X+\frac{5}{\sqrt{2}}Y-6=0 \end{align*} and by completing the square we find $\left(\frac{3}{\sqrt{2}}X-\frac{1}{2}\right)^2-\left(\frac{1}{\sqrt{2}}Y-\frac{5}{2}\right)^2=0\Rightarrow$ the conic represents the pair of intersecting lines $Y=3X+2\sqrt{2}, Y =-3X+3\sqrt{2}$ in the new reference frame $(X,Y)$ and by using $X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}, Y=-\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$ we get the equation of the lines in the original reference frame $(x,y)$: $\fbox{$y=-2-2x$ and $y=\frac{3}{2}-\frac{x}{2}$}$.
Summary: your expression is
$$ (x+2y-3)(2x+y+2) .$$
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$$ P^T H P = D $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 5 }{ 4 } & 1 & 0 \\ - \frac{ 7 }{ 3 } & \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & - \frac{ 7 }{ 3 } \\ 0 & 1 & \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$
$$ Q^T D Q = H $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 5 }{ 4 } & 1 & 0 \\ - 1 & - \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & - 1 \\ 0 & 1 & - \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
The expression $Q^T DQ = H ,$ where $H$ is the Hessian matrix of second partials, tells us that twice your quadratic expression is $$ 4 (x+ \frac{5}{4}y - 1)^2 - \frac{9}{4} (y- \frac{8}{3})^2 $$ where the third row has a coefficient $0.$ This is because of the $0$ diagonal element in $D.$ The result is that, being evidently a difference of squares, twice your expression factors, and the zero set is two lines.
Wrote it out, twice your expression is $$ (2x+4y-6)(2x+y+2) ,$$ so the expression not doubled is $$ (x+2y-3)(2x+y+2) .$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 4 & 0 & - 4 \\ 0 & - \frac{ 9 }{ 4 } & 6 \\ - 4 & 6 & - 12 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 6 \\ 0 & 6 & - 16 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & - \frac{ 7 }{ 3 } \\ 0 & 1 & \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & - 1 \\ 0 & 1 & - \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
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$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 5 }{ 4 } & 1 & 0 \\ - \frac{ 7 }{ 3 } & \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & - \frac{ 7 }{ 3 } \\ 0 & 1 & \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$
$$ Q^T D Q = H $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 5 }{ 4 } & 1 & 0 \\ - 1 & - \frac{ 8 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - \frac{ 9 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & - 1 \\ 0 & 1 & - \frac{ 8 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 5 & - 4 \\ 5 & 4 & 1 \\ - 4 & 1 & - 12 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$