# Are the sets subspaces?

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 check if these are subspaces.



I have done the following:

• $$U_1$$ :

The set is non-empty, $$(0,0)^T\in U_1$$.

Let $$(x_1, y_1)^T, (x_2, y_2)^T\in U_1$$. That means that $$x_1^2+y_1^2\leq 4$$ and $$x_2^2+y_2^2\leq 4$$. For the sum $$(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$$ we have \begin{align*}(x_1+x_2)^2+(y_1+y_2)^2&=x_1^2+2x_1x_2+x_2^2+y_1^2+2y_1y_2+y_2^2 \\ & =\left (x_1^2+y_1^2\right )+\left (x_2^2+y_2^2\right )+2x_1x_2+2y_1y_2 \\ & \leq 4+4+2x_1x_2+2y_1y_2\end{align*} This is not necessarily less than $$4$$ and so this is not a subspace. Is that correct?

• $$U_2$$ :

The set is non-empty, $$(0,0)^T\in U_2$$.

Let $$(x_1, y_1)^T, (x_2, y_2)^T\in U_2$$. That means that $$y_1=-2x_1$$ and $$y_2=-2x_2$$. For the sum $$(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$$ we have $$y_1+y_2=-2x_1-2x_2=-2(x_1+x_2)$$ and that means that the sum is also in $$U_2$$.

Let $$r\in \mathbb{R}$$ and $$(x_1, y_1)^T\in U_2$$. That means that $$y_1=-2x_1$$. For the scalar multiplication $$r(x_1, y_1)^T=(rx_1, ry_1)^T=(rx_1, ry_1)^T$$ we have $$ry_1=r\cdot (-2x_1)=-2(rx_1)$$ and that means that the result is also in $$U_2$$.

Therefore $$U_2$$ is a subspace.

• $$U_3$$ :

The set is non-empty, $$(0,0,0)^T\in U_3$$.

Let $$(x_1, y_1, z_1)^T, (x_2, y_2, z_2)^T\in U_3$$. That means that $$y_1=y_2=0$$. For the sum $$(x_1, y_1, z_1)^T+ (x_2, y_2, z_2)^T=(x_1+x_2, y_1+y_2, z_1+z_2)^T$$ we have $$y_1+y_2=0+0=0$$ and that means that the sum is also in $$U_3$$.

Let $$r\in \mathbb{R}$$ and $$(x_1, y_1, z_1)^T\in U_3$$. That means that $$y_1=0$$. For the scalar multiplication $$r(x_1, y_1, z_1)^T=(rx_1, ry_1, rz_1)^T$$ we have $$ry_1=r\cdot 0=0$$ and that means that the result is also in $$U_3$$.

Therefore $$U_3$$ is a subspace.

• $$U_4$$ :

Since the zero vector is nott included, $$U_4$$ is not a subspace.



Is everything correct and complete?

• Everything is ok. – Alberto Saracco Nov 2 '19 at 10:19
• Great!! Thank you!! :-) If we want to to check the same but instead of a proof as above, we would use their graphs, how could we do that? For example at $U_1$ which is a circle and the inner part of the circle, how do we see that this is not a subspace? @AlbertoSaracco – Mary Star Nov 2 '19 at 10:23
• @Mary Star Subspaces of a vector space are either just the single point $\ (0,0,\dots,0)\$, or (complete) lines, planes or hyperplanes passing through that point. – lonza leggiera Nov 2 '19 at 10:29
• I would take two concrete vectors in $U_1$ whose sum is not in $U_1$. Like $(2,0)$ and $(0,2)$. It's easier to both write and read than algebraic manipulations on arbitrary entries. – Arthur Nov 2 '19 at 10:45

## 1 Answer

For every one want to know:

$$U_1: \qquad$$ There is no bounded subspace but $$\{0\}$$

$$U_2 = \langle \binom{2}{-1} \rangle: \qquad$$ Multiples of any vector produce a subspace of degree $$1$$.

$$U_2' = \langle v_i \rangle_i: \qquad$$ linear combinations of any set of vectors produce a subspace of degree equals to the max number of independent vectors in generating set.

$$U_{3,4} = y=y_0: \qquad$$ Any linear equation produces a subspace of degree $$n-1$$ where your space is of degree $$n$$.

$$U'_{3,4} : \qquad$$ A system of linear equations produce a subspace of degree $$n-k$$ where you have $$k$$ equations.