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Is this set a linear subspace of $\mathbb{R}^{2}$? $U=\left\{\begin{pmatrix} 0\\ 0 \end{pmatrix}\right\}$

I think $U \subseteq \mathbb{R}^{2}$ because we have only zero and this are included in $\mathbb{R}^{2}$.

We can clearly also see that $U \neq \emptyset$ which is good.

Now we need to show that there exists $U_{1},U_{2} \in U$ and $\lambda \in \mathbb{R}$ such that

  • $U_{1}+U_{2} \in U$

  • $\lambda \cdot U_{1} \in U$

The second will not work because if we take a $\lambda \neq 0$ we will have something else than $\begin{pmatrix} 0\\ 0 \end{pmatrix}$ so this won't work because it's not in the set $U$

So I say that $U$ isn't linear subspace of $\mathbb{R}^{2}$

Is it correct or did I do it wrong again? : /

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    $\begingroup$ You are mis-using quantifiers. The conditions to be a linear subspace are that for all $U_1,U_2 \in U$ and $\lambda \in \mathbb{R}$, we have $U_1 + U_2 \in U$ and $\lambda \cdot U_1 \in U$. $\endgroup$ – Lee Mosher Dec 3 '16 at 14:33
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Is it correct or did I do it wrong again? : /

You did it wrong:

Now we need to show that there exists $U_{1},U_{2} \in U$ and $\lambda \in \mathbb{R}$ such that

Not EXISTS, you have to show that for all $U_{1},U_{2} \in U$ and $\lambda \in \mathbb{R}$

The second will not work because if we take a $\lambda \neq 0$ we will have something else than $\begin{pmatrix} 0\\ 0 \end{pmatrix}$

Will we? For me for all $\lambda \in \Bbb R$ it holds $$\lambda \begin{pmatrix} 0\\ 0 \end{pmatrix} = \begin{pmatrix} \lambda\cdot 0\\ \lambda\cdot 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix} \in U$$

So everthing is fine… and yes, it's a linear subspace, the smallest one…

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  • $\begingroup$ Oh thank you now I see it. How would you do the other with $U_{1}+U_{2} \in U$? Maybe this: $$\begin{pmatrix} 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$? $\endgroup$ – cnmesr Dec 3 '16 at 14:14
  • $\begingroup$ Indeed, that is how it's done. $\endgroup$ – David Wheeler Dec 3 '16 at 14:29

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