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Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.

The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\textbf{x}\in\mathbb R^n: f(\textbf{x})=0\}$ is closed. I would need to have a hint for the boundedness of $V$. Thanks!

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4 Answers 4

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Note that $(a-b)^2 \ge 0$, so $ab \le \frac{a^2+b^2}{2}$. Thus, since $x^2 - xy + y^2 - 1 = 0$ we have $x^2 + y^2 = 1 + xy \le 1 + \frac{x^2+y^2}{2}$ and so $\frac{1}{2} (x^2 + y^2) \le 1$ and finally $x^2 + y^2 \le 2$.

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The equation $x^2 - x y + y^2 = 1$ can be rewritten as follows

$$\begin{bmatrix} x\\ y\end{bmatrix}^{\top} \begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$

Computing the spectral decomposition,

$$\begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} = \frac 12 \begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 0.5 & 0\\ 0 & 1.5\end{bmatrix} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}$$

we then obtain the following

$$\frac{1}{2} \left( \frac{x + y}{\sqrt 2} \right)^2 + \frac{3}{2} \left( \frac{x - y}{\sqrt 2} \right)^2 = 1$$

Thus, the set is an ellipse, which is closed and bounded. The ellipse's semimajor and semiminor axes are $\sqrt 2 \approx 1.414$ and $\sqrt{\frac 23} \approx 0.816$, respectively. Plotting the ellipse,

enter image description here

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Changing coordinates to $z=x-y/2$ proves this curve is an ellipse.

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  • $\begingroup$ Are you sure??? $\endgroup$ Commented Apr 18, 2017 at 11:42
  • $\begingroup$ @RodrigodeAzevedo Since $z^2+\frac{3y^2}{4}=1$, neither $y$ nor $z$ nor $x$ can be large. The result is a bounded conic section. $\endgroup$
    – J.G.
    Commented Apr 18, 2017 at 13:09
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Show that:

{ $(x,y) \in \mathbb{R}: x^2 -xy +y^2 = 1$} is a bounded.set.

Consider polar coordinates.

$x = r cos (\phi) ; y= r sin (\phi).$.

We get :

$r^2 - r^2 cos (\phi) sin(\phi) = 1$ ;

$r^2 (1 - (1/2) sin(2\phi)) = 1 $;

Note : $-1 \le sin(2\phi) \le 1$.

Hence:

$r^2 \le 2 $, now back to $ x,y$:

$x^2 + y^2 \le 2$.

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