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:**

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

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

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

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

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

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

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

Reinout

reinout-g
Jun 8, 2014

#1**+5 **

Don't worry Reinout. Rosala will be along soon. she is bound to have some feedback for you.

Melody
Jun 8, 2014

#3**+14 **

i knew it reinout , i knew it ! u wont be able to work without my "all the best " ! lol! i dont know why reinout , i am starting to think that do u have einsteins dna in ur blood or what !lol!i just cant understand ur maths , when u write something i dont get it , is it the ques or the answer !lol!by the way "all the best " for this one too!

rosala
Jun 8, 2014

#4**+5 **

Good sniff out, rosala.....reinout is * INDEED* a distant cousin of the famous Dr. Einstein...

But, I suspect that it's all relative...........

CPhill
Jun 8, 2014

#6**+5 **

Haha, imagine how brilliant the people are that can actually answer my questions!

p.s. Nice joke CPhill, I almost missed it

reinout-g
Jun 8, 2014