Linear space

Good morning ,

One of the exercises I encountered is this one;

Let V be the subset of  $$\mathbb{R}^4$$  spanned by the vectors

$$v_1 = \begin{pmatrix} 1 \\ -2\\ 0\\ 3 \end{pmatrix}$$ , $$v_2 = \begin{pmatrix} 2 \\ 3 \\ 0 \\ -1 \\ \end{pmatrix}$$ , $$v_3 = \begin{pmatrix} 2 \\ -1 \\ 2 \\ 1 \\ \end{pmatrix}$$

Prove that V is a linear space over  $$\mathbb{R}$$

So I build an answer based on some other answer I found and was hoping someone would be willing to give some feedback on it. I usually have trouble composing proofs and I'm not sure whether I'm writing pure gibberish or whether this is actually correct.

Proof:

$$\begin{array}{lcl} \mbox{Fix arbitrary p,q,r } \in V \mbox{ and } \lambda,\mu \in \mathbb{R} \\ \mbox{Then } l = \alpha_{1l}v_1+\alpha_{2l}v_2 +\alpha_{3l}v_3 \mbox{ for some } \alpha_{1l},\alpha_{2l},\alpha_{3l} \in \mathbb{R}, l \in [p,q,r].\\ \mbox{Let us now verify the ten properties of linear spaces over } \mathbb{R}. \\ \mbox{1. } p+q = \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 + \alpha_{1q}v_1+\alpha_{2q}v_2 +\alpha_{3q}v_3\\ = (\alpha_{1p}+\alpha_{1q})v_1+(\alpha_{2p}+\alpha_{2q})v_2 +(\alpha_{3p}+\alpha_{3q})v_3 \in \mathbb{R}\\ \end{array}$$

$$\begin{array}{lcl} \mbox{2. } p+q = \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 + \alpha_{1q}v_1+\alpha_{2q}v_2 +\alpha_{3q}v_3\\ = \alpha_{1q}v_1+\alpha_{2q}v_2 +\alpha_{3q}v_3 + \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3\\ = q+p \\ \mbox{3. } p+(q+r) = \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 + (\alpha_{1q}+\alpha_{1r})v_1+(\alpha_{2q}+\alpha_{2r})v_2 +(\alpha_{3q}+\alpha_{3r})v_3 \\ = (\alpha_{1p}+\alpha_{1q})v_1+(\alpha_{2p}+\alpha_{2q})v_2 +(\alpha_{3p}+\alpha_{3q})v_3 + \alpha_{1r}v_1+\alpha_{2r}v_2 +\alpha_{3r}v_3 \\ \end{array}$$

$$\begin{array}{lcl} \mbox{4. Let 0 be the zero element of V. Then } 0 = 0v_1+0v_2+0v_3 \in V \mbox{ and } 0+v = v \mbox{ for all } v \in V\\ \mbox{5. Let 0 be the zero element of V and fix an arbitrary } p = \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 \mbox{. Let } z:= -\alpha_{1p}v_1-\alpha_{2p}v_2 -\alpha_{3p}v_3 = -p \in V \\ \mbox{Then: }p+z = (\alpha_{1p}-\alpha_{1q})v_1+(\alpha_{2p}-\alpha_{2q})v_2 +(\alpha_{3p}-\alpha_{3q})v_3 = 0\\ \end{array}$$

$$\begin{array}{lcl} \mbox{6. } \lambda(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = (\lambda \alpha_{1p})v_1+ (\lambda \alpha_{2p})v_2 + (\lambda \alpha_{3p})v_3 \in V \\ \mbox{7. } \lambda(p+q)= \lambda(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 + \alpha_{1q}v_1+\alpha_{2q}v_2 +\alpha_{3q}v_3) = \lambda(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) + \lambda(\alpha_{1q}v_1+\alpha_{2q}v_2 +\alpha_{3q}v_3) = \lambda p + \lambda q\\ \end{array}$$

$$\begin{array}{lcl} \mbox{8. } (\lambda + \mu)p = (\lambda + \mu)(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \lambda(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) + \mu(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \lambda p + \mu p\\ \mbox{9. }\lambda( \mu p) = \lambda ( \mu (\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \lambda \mu (\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \mu \lambda (\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \mu (\lambda (\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3)) = \mu (\lambda p)\\ \end{array}$$

$$\mbox{10. } 1 \times p = 1(\alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3) = \alpha_{1p}v_1+\alpha_{2p}v_2 +\alpha_{3p}v_3 = p$$

$$\begin{array}{lcl}\mbox{ Since V follows all ten conditions of a linear space over } \mathbb{R} \mbox{ , V is a linear space over } \mathbb{R}\\ \mbox{q.e.d.} \end{array}$$

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