Showing boundedness of a set defined by equality

Question

I am trying to show that the set given by:

$$S = \{\mathbf x \in \Bbb R^2 \mid x^2 + 3xy + 3y^2 = 3\}$$

is bounded.

I am able to show that this is true whenever $(x,y) \in \Bbb R^2$ is such that:

$x>0$ and $y>0$; or
$x<0$ and $y<0$.

However, I having some trouble showing that this for the other two cases where $x$ and $y$ have opposite signs.

Am I on the right track so far?

Thanks in advance.

score 2 · Answer 1 · answered Jan 08 '16 at 14:56

2

$$x^2+3xy+3y^2=\left(\frac{9}{10}x+\frac{5}{3}y\right)^2+\frac{19}{100}x^2+\frac{2}{9}y^2$$ shows, that $x$ and $y$ must be bounded to get $3$.

answered Jan 08 '16 at 14:56

How did you come up with this equality ? – Gabriel Romon Jan 08 '16 at 16:44
I thought about the binomial theorem to write the expression as a sum of quadratic terms. Obviously there are many other ways to do this when you think about the binomial theorem. – Jan 08 '16 at 16:55
By "binomial theorem", you mean the identity $(a+b)^2=a^2+2ab+b^2$ ? – Gabriel Romon Jan 08 '16 at 16:58
Yes, I thought of that. – Jan 08 '16 at 17:00
thanks guys for your help I'm studying optimization and a question I was doing had this domain which needed to be compact for Weierstrass' theorem. – bluemackillop Jan 09 '16 at 08:10

Michael Albanese · Answer 2 · 2016-01-13T12:00:48.283

I think you're going to have difficulty trying to establish the boundedness of $S$ using this argument. The reason why is that the Cartesian coordinates are not well-suited to describe $S$. Below I have given another method to establish the boundedness of $S$.

Note that $x^2 + 3xy + 3y^2 = {\bf x}^TA{\bf x}$ where

$$A = \begin{bmatrix} 1 & \frac{3}{2}\\ \frac{3}{2} & 3\end{bmatrix}.$$

As $A$ is symmetric, it can be diagonalised. That is, there is an orthogonal matrix $P$ such that $P^TAP$ is a diagonal matrix $\operatorname{diag}(\lambda_1, \lambda_2)$ where $\lambda_1$, $\lambda_2$ are the eigenvalues of $A$. If we introduce new variables ${\bf \hat{x}} = (\hat{x}, \hat{y})$ given by ${\bf\hat{x}} = P^{-1}{\bf x}$, then as ${\bf x} = P{\bf\hat{x}}$ we have

$${\bf x}^TA{\bf x} = {\bf \hat{x}}^TP^TAP{\bf\hat{x}} = {\bf\hat{x}}^T\operatorname{diag}(\lambda_1, \lambda_2){\bf\hat{x}} = \lambda_1\hat{x}^2 + \lambda_2\hat{y}^2.$$

In these new coordinates, we see that $S$ is simply a conic.

If $\lambda_1, \lambda_2 > 0$, then $S$ is an ellipse.
If $\lambda_1 > 0$, $\lambda_2 < 0$ (or vice versa), then $S$ is a hyperbola.
If $\lambda_1 = 0$, $\lambda_2 > 0$ (or vice versa), then $S$ is the union of two lines.
If $\lambda_1, \lambda_2 \leq 0$, $S$ is empty.

As we explicitly have the matrix $A$, we can calculate $\lambda_1$ and $\lambda_2$. However, we just need to know the signs of the eigenvalues, which is a bit easier to determine. As $\det A = \lambda_1\lambda_2 = \frac{3}{4} > 0$ and $\operatorname{tr} A = \lambda_1 + \lambda_2 = 4 > 0$, both $\lambda_1$ and $\lambda_2$ are positive. Therefore, $S$ is an ellipse, and hence bounded.

Note, one can ensure that $P$ has determinant $1$, and hence $P$ is a rotation matrix. The change of variables ${\bf\hat{x}} = P^{-1}{\bf x}$ rotate the $x$ and $y$ axes in such a way that the expression $x^2 + 3xy + 3y^2$ is put in standard form. These coordinates are the ones which we can consider to be well-suited to describe $S$. This is the standard procedure for putting a quadric into standard form.

Showing boundedness of a set defined by equality

2 Answers2