Describe an equation geometrically

Question

Finishing the last few stuff left for my end-term semester exams on Linear Algebra II, I bumped across a collection of identical exercises, posting one below :

Describe geometrically, giving as much information as possible (axis, focis, distances etc.), the geometrical space of the points $(x,y)\in \mathbb R^2$ which satisfy the equation :

$13x^2 + 7y^2 - 6\sqrt{3} xy = 4$

Now, I know this is an ellipse (examined it through Mathematica/Wolfram). Also by sheer interexchange it's obvious that it's an ellipse but my question is how do you describe it by a form of mathematical approach ? One attempt that came across my mind is going through the normal form of the quadratic form with respect to scalar.

Is this a correct approach ? Can someone give a hint on where to start from ? Any help would be appreciated !

Answer 1

Yes, this is a correct approach. More precisely, consider the symmetric matrix

$$ A = \begin{pmatrix} 13 & -3\sqrt{3} \\ -3\sqrt{3} & 7 \end{pmatrix} $$

whose associated quadratic form is

$$ q_A(\vec{v}) = \vec{v}^T A \vec{v} = 13x^2 - 6\sqrt{3}xy + 7y^2, \,\,\, \vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}. $$

You want to find the geometric shape of the level set $q_A(\vec{v}) = 4$. The matrix $A$ is orthogonally similar to the diagonal matrix $D = \operatorname{diag}(4,16)$ which means that you can find an orthogonal matrix $O$ so that $A = O D O^T$. Now, the level set $q_D(\vec{w}) = 4$ is an ellipse (whose equation is $4x^2 + 16y^2 = 4 \iff x^2 + (2y)^2 = 1$) whose diameters are of length $2,1$. Since

$$ q_A(\vec{v}) = q_D(O^T\vec{v}) = q_D(\vec{w}) = 4, \,\,\, \vec{w} = O^T\vec{v} $$

the level set $q_A(\vec{v}) = 4$ is obtained by rotating the ellipse $x^2 + (2y)^2 = 1$ around the origin using $O$. The axes of $q_A(\vec{v}) = 4$ are the eigenvectors of the matrix $A$ / columns of $O$.

Answer 2

In quadratic form, this is $v^T A v=4$ where $v=(x,y)^T$ and $A=\begin{pmatrix}13 &-3\sqrt{3}\\ -3\sqrt{3} &7\end{pmatrix}$. This has determinant $64$ and trace $20$, which by inspection implies eigenvalues $4,16$. Consequently there exist eigenvectors $v_1,v_2$ such that $Av_1=4v_2$ and $Av_2=16v_2$; their construction is left to the interested reader.

Since $A$ is real and symmetric, these provide an orthogonal basis; for convenience, I'll assume they're both normalized to 1. Therefore we may write $v=c_1 v_1+c_2 v_2$ for some $c_1,c_2$, and then the above quadratic form becomes

$$v^T A v=(c_1 v_1+c_2 v_2)^T A(c_1 v_1+c_2 v_2)=(c_1 v_1^T+c_2 v_2^T)(4c_1v_1+16c_2 v_2) = (2c_1)^2+(4c_2)^2.$$ Now, $(2c_1)^2+(4c_2)^2=4$ is an ellipse in the coordinates $(c_1,c_2)$; the semi-major and semi-minor axes are $2$ and $1$ respectively. But the change of variables $(c_1,c_2)=(v_1^T,v_2^T) v$ corresponds directly to a rotation of the $xy$-axes. So the figure is still an ellipse with these axes.