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Are these sets subspaces?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
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)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.