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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.

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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.

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Is everything correct and complete?

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  • 2
    $\begingroup$ Everything is ok. $\endgroup$ – Alberto Saracco Nov 2 '19 at 10:19
  • $\begingroup$ 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 $\endgroup$ – Mary Star Nov 2 '19 at 10:23
  • $\begingroup$ @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. $\endgroup$ – lonza leggiera Nov 2 '19 at 10:29
  • $\begingroup$ 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. $\endgroup$ – Arthur Nov 2 '19 at 10:45
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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.

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