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I'm taking about the

$$\begin{pmatrix} A &B/2 \\ B/2 &C \end{pmatrix}$$

matrix for conic $Ax^2 + Bxy+Cy^2 + Dx+\dots =0$.

I don't get what the kind of transformation that matrix is doing and why I get the princple axis from its eigenvectors.

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  • $\begingroup$ oh crap :| that's what i meant $\endgroup$ – Vrisk Oct 26 '17 at 8:55
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Let's translate the origin to get rid of the linear terms, so that the curve is given implicitly by the equation $$p^TMp + c = 0$$ where $M$ is your matrix and $p = (x,y)$. Let's expand $p$ in the eigenvectors of $M$: $$p = \alpha_1 v_1 + \alpha_2v_2.$$ Then $$\alpha_1^2 \lambda_1 + \alpha_2^2 \lambda_2 + c = 0.$$

Notice

1) Solutions are invariant under negating the sign of either $\alpha_1$ or $\alpha_2$. This means the curve is symmetric with respect to both the $v_1$ and $v_2$ axes. On its own this may already answer your question, depending on how you define the principal axes. But we can go deeper:

2) If $\lambda_1$ and $\lambda_2$ have the same sign, WLOG we can assume both are positive and that $\lambda_1 < \lambda_2$. Then the curve is an ellipse and the closest point on the curve to the origin is at $\pm\sqrt{c/\lambda_2}v_2$. The farthest point is at $\pm\sqrt{c/\lambda_1}v_1$. Hence $v_1$ and $v_2$ are the major and minor axes of the ellipse.

3) If they have opposite sign, we can assume $c>0$ and $\lambda_1 < 0 < \lambda_2$. Then the curve is a hyperbola and the closest point to the origin is at $\pm\sqrt{c/\lambda_1}v_1$. You can go arbitrarily far in the $v_2$ direction without ever hitting the curve.

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  • $\begingroup$ hey what did you do with p when you expanded it? it's a vector (x, y) right? $\endgroup$ – Vrisk Oct 26 '17 at 9:28
  • $\begingroup$ sorry to bother but I'm having difficulty figuring out how you got that quadratic in alpha, can you please provide some pointers? $\endgroup$ – Vrisk Oct 27 '17 at 18:27
  • $\begingroup$ @Vrisk Do you know about eigenvectors and eigenvalues? In particular, use the fact that for symmetric matrices $M$ the eigenvectors are orthogonal: $v_1 \cdot v_2 = 0$ and in particular $v_1^TMv_2 = v_2^TMv_1 = 0$. $\endgroup$ – user7530 Oct 27 '17 at 18:39

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