A represents the quadratic form $q(x, y) = x^2 + 4xy + y^2 $; for what values of $(x, y) \in \mathbb{R}^2$ is $ q(x, y) =0$?

Question

A represents the quadratic form $q(x, y) = x^2 + 4xy + y^2$ ; for what values of $(x, y) \in \mathbb{R}^2 $ is $ q(x, y) = 0$ ?

Anyone here to help me, I would really appreciate

Answer 1

Note that $x^2+4xy+y^2=(x+2y)^2-3y^2$. So the solutions are the lines $x=\sqrt3y-2y$ and $x=-\sqrt3y-2y$

Answer 2

If you set the coordinate vector $r = [x, y]^T $ , then the given quadratic form equation is expressible as follows

$ r^T Q r = 0 $

where

$Q = \begin{bmatrix} 1 && 2 \\ 2 && 1 \end{bmatrix} $

$Q$ is symmetric, so it is diagonalizable. To diagonalize it, compute

$ \theta = \dfrac{1}{2} \tan^{-1} \big( \dfrac{ 2 Q_{12}}{Q_{11} - Q_{22}} \big) = \dfrac{\pi}{4} $

The entries of the diagonal matrix are

$ D_{11} = \dfrac{1}{2} \left( Q_{11} + Q_{22} \right) + \dfrac{1}{2} \left(Q_{11} - Q_{22}\right) \cos(2 \theta) + Q_{12} \sin(2 \theta) = 3 $

$ D_{22} = \dfrac{1}{2} \left( Q_{11} + Q_{22} \right) - \dfrac{1}{2} \left(Q_{11} - Q_{22}\right) \cos(2 \theta) - Q_{12} \sin(2 \theta) = -1 $

Now, with this diagonalization, the above quadratic form equation becomes,

$ r^T R D R^T r = 0 $

Let $ z = R^T r $, then $ z^T D z= 0 $ , which means $ 3 z_1^2 - z_2^2 = 0 $, which has two solutions,

First solution: $(z_1, z_2) = t (1, \sqrt{3}) , t \in \mathbb{R} $

Second solution: $(z_1, z_2) = t (1, -\sqrt{3}) , t \in \mathbb{R} $

Now, recalling that $ r = R z $, and that

$ R = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} $

Then, the above two solution in $z$ gives us the following corresponding solutions for $r$

First solution: $r = (x, y) = t \left(\dfrac{1}{\sqrt{2}}\right) ( 1 - \sqrt{3} , 1 + \sqrt{3} ) $

Second solution: $r = (x, y) = t \left(\dfrac{1}{\sqrt{2}}\right) (1 + \sqrt{3}, 1 - \sqrt{3} ) $

Stated differently, the above two solutions are straight lines passing through the origin with slopes $ m_1 = \dfrac{ 1 + \sqrt{3} }{ 1- \sqrt{3} } = -2 - \sqrt{3} $ for the first one and $ m_2 = \dfrac{ 1 - \sqrt{3} }{1 + \sqrt{3}} = - 2 + \sqrt{3} $ for the second one.

That is, the solutions are

$ y = (-2 - \sqrt{3} ) x $

and

$ y = (-2 + \sqrt{3} ) x $

Verification:

Plug the first solution into $ x^2 + 4 xy + y^2 $ you get

$ x^2 + 4 x (-2 - \sqrt{3}) x + (-2 - \sqrt{3} )^2 x^2 = x^2 ( 1 - 8 - 4 \sqrt{3} + 4 + 3 + 4 \sqrt{3} ) = 0 $

And similarly for the second solution.

This also verifies the solutions obtained by @Guillerminho77 in a much much simpler way.