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When given an equation of the form $$Ax^2+Bxy+Cy^2 + Dx + Ey + F$$ where $B \not= 0$ and it is not a degenerate conic, then you can use $\Delta = B^2 -4AC $ to see what type of conic it is, and then (after a lengthy process) you are able to find an equation with $B = 0$, where your $x$ and $y$ are parameterized into $x'$and $y'$, where $x',y'$ are a set axis with each one being either perpendicular to your axis of symmetry of parallel to it.

While I know how to do this process by way of lengthy algebraic manipulation, I was wondering if there was a way to do this with linear algebra? I know that we can express the standard form equation as $$x^{\top}Ax + B^{\top}x + F$$ where $x = \left[ \begin{array}{c} x \\y\end{array}\right]$, $A = \left[ \begin{array}{cc} A & B/2\\ B/2 & C\end{array}\right]$, and $B = \left[ \begin{array}{c} D \\E\end{array}\right]$. I don't really know where to go from there, though.

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  • $\begingroup$ $x^{\top}Ax$ is a quadratic form that can be reduced to a canonical form (a sum of squares) by changing a basis. $\endgroup$
    – user35603
    Apr 20, 2014 at 8:25
  • $\begingroup$ @user35603 How would I find my new basis vectors? $\endgroup$ Apr 20, 2014 at 8:57
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    $\begingroup$ See this pdf. $\endgroup$ Apr 20, 2014 at 10:24
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    $\begingroup$ @JuanSebastianLozanoMuñoz: Or this pdf. $\endgroup$
    – user35603
    Apr 20, 2014 at 10:39

1 Answer 1

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The matrix $A$ is symmetric, and assuming the coefficients $a,b,\ldots,f$ are real or complex, $A$ is then orthogonally (unitarily) similar to a real diagonal matrix (I will just assume we're working with real numbers from here on). So you have $A=P^TDP$. Now let $X=PX'$ where $X'=(x',y')^T$ is your new coordinate system. So then we have \begin{equation} X^TAX=(PX')^TA(PX')=X'^T(P^TAP)X'=X'^TDX'=\lambda_1x'^2+\lambda_2y'^2, \end{equation} which is what we want. So if we then take $B^TPX'=(d',e')(x',y')^T=d'x'+e'y'$the full equation then becomes: \begin{equation}X'^TDX'+B^TPX'+f=\lambda_1x'^2+\lambda_2y'^2+d'x'+e'y'+f.\end{equation}

So, in a nutshell, you just have to do orthogonal diagonalization on $A$, and then do the substitution as above and multiply out. The matrix $P$ is then a rotation transformation matrix (if det$(P)=-1$ then you can just interchange the columns of $P$ and $D$ so that the new matrix $P'$ then has determinant 1, ensuring it is a rotation and not a reflection about a line through the origin)

(The source for most of this: Linear Algebra (4th ed) by Friedberg et al, p.388-392)

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