Are these sets subspaces?

In summary, the given sets are subsets of R2 and R3, and we want to determine if they are subspaces. The first set, U1, is a circle with radius 2, and is not a subspace because it does not satisfy the necessary condition that the origin must be contained. The second set, U2, is a line through the origin and is therefore a subspace. The third set, U3, is the xz-plane with y=0 and is a subspace because it contains the origin and satisfies the necessary conditions for a subspace in R3. The fourth set, U4, is a plane parallel to the xz-plane with y=1 and is not a subspace since it
  • #1
mathmari
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Hey! :eek:

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)

    $$$$
  • $U_2$ :

    [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)

    $$$$
  • This is the $xz$ plane with $y=0$.

    This is a subspace, since the zero vector is contained, or not? (Wondering)

    $$$$
  • This is the $xz$ plane with $y=1$.

    This is not a subspace since the zero vector is not contained. Is that correct? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

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:

  • $U_1$ :



    This is a subspace, isn't it? But how can we explain that from the graph? (Wondering)

    $$$$
  • $U_2$ :



    Since this line goes through the origin, it is a subspace, or not? (Wondering)

    $$$$
  • This is the $xz$ plane with $y=0$.

    This is a subspace, since the zero vector is contained, or not? (Wondering)

    $$$$
  • This is the $xz$ plane with $y=1$.

    This is not a subspace since the zero vector is not contained. Is that correct? (Wondering)
Your graph for the first one looks odd because your graph does not have the same size units on the two axes. The graph should be a circle, not an ellipse. As for "subspace", are you clear on what a subspace is? If u and v are in the space then u+ v and au, for a any number, must be in the space.

For the first one, u= (2, 0) is in the space but 2(2, 0)= (4, 0) is not.

For the second one we can write u= (2a, -a) and v= (2b, -b). Then u+ v= (2a+ 2b, -a- b)= (2(a+ b), -(a+ b)) so is also in the set. And for any number, x, x(2a, -a)= (2(xa), -(xa)) so is in the set. Yes, the origin, (0, 0)= (2(0), -0) is in the set. Geometrically a subspace of R2 is a line through the origin. The first example was not a line.

Yes, the third is the xz-plane. In R3, any subspace is either a line containing the origin or a plane containing the origin. This is a plane containing the origin. A proof that this is a subspace would be to take two such vectors as u= (a, 0, b) and v= (c, 0, d). Then u+ v= (a+c, 0, b+ d) which is in the set and, for x any number, xu= (xa, 0, xb), in the set.

Strictly speaking the last one is NOT "the xz-plane with y= 1". The "xz-plane" always has y= 0. This is a plane parallel to the xz-plane. You are right that this is not a subspace because it does not contain the origin. More specifically, u= (a, 1, b) is in the set but 2u= (2a, 2, 2b) is not.

You seem to be under that impression that a set is a subspace as long as it contains the origin. That is "necessary" condition, but is not "sufficient". A subspace of R2 must be a line containing the origin and a subspace of R3 must be a line or plane containing the origin.
 

1. Are all subsets of a vector space considered subspaces?

Yes, any subset of a vector space that satisfies the three properties of a subspace (non-empty, closed under addition, and closed under scalar multiplication) is considered a subspace.

2. Can a subspace contain the zero vector?

Yes, all subspaces must contain the zero vector since it is closed under scalar multiplication and addition.

3. How do I prove that a set is a subspace?

To prove that a set is a subspace, you must show that it satisfies the three properties of a subspace: non-empty, closed under addition, and closed under scalar multiplication. You can do this by showing that the set contains the zero vector, is closed under addition, and is closed under scalar multiplication.

4. Can a subspace have more than one basis?

Yes, a subspace can have multiple bases. Bases are not unique, so a subspace can have different sets of vectors that span the same subspace.

5. Is the intersection of two subspaces always a subspace?

Yes, the intersection of two subspaces is always a subspace. This is because the intersection will contain the zero vector, is closed under addition, and is closed under scalar multiplication.

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