Sketch the unit ball $B(0, 1)$ in $\mathbb{R}^2$ that has the norm:
$||(x,y)||=\sqrt{2x^2+3y^2}$
Im ok at sketching unit balls normally but the norm given has stumped me in making progress with this question so any help will be appreciated.
Sketch the unit ball $B(0, 1)$ in $\mathbb{R}^2$ equipped with the norm $||(x,y)||=\sqrt{2x^2+3y^2}$
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metric-spaces
norm
normed-spaces
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3The boundary of the ball is the curve $2 x^2 + 3 y^2 = 1$. This is an ellipse... – 2017-02-20
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0If $\Vert\cdot\Vert_e$ is the euclidean norm, then the map $\phi:(\mathbb{R}^2,\Vert\cdot\Vert_e)\to(\mathbb{R}^2,\Vert\cdot\Vert)$, $\phi(x,y)=(x/\sqrt{2},y/\sqrt{3})$, is a linear isometry, so it maps unit balls to unit balls. – 2017-02-20
2 Answers
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The boundary of that unit ball would be the points $x,y$ so that $$\sqrt{2x^2+3y^2}=1 \iff 2x^2+3y^2=1$$
As they've already pointed out in the comments, that's the equation of an ellipse. It's easier to sketch it if you give some easy values to $x$ and $y$. For example, $x=0 \Rightarrow y=\frac{\pm1}{\sqrt3}$, $y=0 \Rightarrow x=\frac{\pm1}{\sqrt2}$...
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It's an ellipse passing through $||v_i||$ for $v_i\in\big\{(1,0),(0,1),(-1,0),(0,-1)\big\}$.
Note: For generating such pictures by yourself you can use Geogebra and enter 1 = sqrt(2x^2 + 3y^2) in the bar at the bottom of the window.
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0[Wolfram](https://www.wolframalpha.com/input/?i=1%3Dsqrt(2x%5E2%2B3y%5E2)) is also good for these kind of tasks. – 2017-02-20