- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
We have the following subsets:
\begin{align*}&U_1:=\left \{\begin{pmatrix}x \\ y\end{pmatrix} \mid x^2+y^2\leq 4\right \} \subseteq \mathbb{R}^2\\ &U_2:=\left \{\begin{pmatrix}2a \\ -a\end{pmatrix} \mid a\in \mathbb{R}\right \} \subseteq \mathbb{R}^2 \\ &U_3:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=0\right \}\subseteq \mathbb{R}^3 \\ &U_4:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=1\right \}\subseteq \mathbb{R}^3\end{align*}
I want to sketch these sets and check in that way if these are subspaces. We have the following graphs:
We have the following subsets:
\begin{align*}&U_1:=\left \{\begin{pmatrix}x \\ y\end{pmatrix} \mid x^2+y^2\leq 4\right \} \subseteq \mathbb{R}^2\\ &U_2:=\left \{\begin{pmatrix}2a \\ -a\end{pmatrix} \mid a\in \mathbb{R}\right \} \subseteq \mathbb{R}^2 \\ &U_3:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=0\right \}\subseteq \mathbb{R}^3 \\ &U_4:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=1\right \}\subseteq \mathbb{R}^3\end{align*}
I want to sketch these sets and check in that way if these are subspaces. We have the following graphs:
- $U_1$ :
[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-5,"ymin":-5.5,"xmax":5,"ymax":5.5}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"x^{2\\ }+y^2\\ \\le\\ 4"}]}}[/DESMOS]
This is a subspace, isn't it? But how can we explain that from the graph? (Wondering)
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- $U_2$ :
[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-10,"ymin":-5.5,"xmax":10,"ymax":5.5}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"x=-2y"}]}}[/DESMOS]
Since this line goes through the origin, it is a subspace, or not? (Wondering)
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- This is the $xz$ plane with $y=0$.
This is a subspace, since the zero vector is contained, or not? (Wondering)
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- This is the $xz$ plane with $y=1$.
This is not a subspace since the zero vector is not contained. Is that correct? (Wondering)