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Let $\,S\,$ be the set of vectors in $\,\mathbb R^3$ that are of the form $\;\begin{bmatrix} a \\ b \\ a-2b \\ \end{bmatrix} \;$ where $a$ and $b$ are real numbers.

Prove $\,S\,$ is a subspace of $\,\mathbb R^3$.


So I know in order to show that a set of vectors is a subspace the $0$ vector must be an element of the set, vector addition must be closed, and scalar multiplication must be closed.

I started off showing the $0$ vector is an element of the set by having: $\;0\cdot a+0\cdot b+0\cdot \left(a-2b\right)=0\,$ which it does so that part is proven.

Now how do I show the closure? For addition I wrote:

\begin{align} a &= c_1a+c_2b+c_3\left(a-2b\right) \\ b &= d_1a+d_2b+d_3\left(a-2b\right) \end{align}

and then combined $\,a+b\,$ to be $\,\left(c_1+d_1\right)a + \left(c_2+d_2\right)b + \left(c_3+d_3\right)\left(a-2b\right)\,$ and since these are linear combinations of one another it is closed under addition? Is this correct? And how do I show closed under scalar multiplication?

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    $\begingroup$ You haven't understood what the set $S$ is. For example, put a=1, b=2. Then (1,2,-3) is an element of the set. So, what do you have to put a and b to in order to get the zero vector (0,0,0)? $\endgroup$ Commented Mar 15, 2016 at 14:16
  • $\begingroup$ For that case I would need (0,0,2-2(1)) $\endgroup$
    – Lil
    Commented Mar 15, 2016 at 14:17
  • $\begingroup$ Well, that's (0,0,0). But what are a and b here? $\endgroup$ Commented Mar 15, 2016 at 14:19
  • $\begingroup$ for the first two spaces I want a and b just to be 0 because no operation is done but or the bottom row I want a to be positive 2 and b to be 1? $\endgroup$
    – Lil
    Commented Mar 15, 2016 at 14:22
  • $\begingroup$ You mean entries - not spaces. The problem is: you cannot change a and b once you have chosen them. $\endgroup$ Commented Mar 15, 2016 at 14:23

2 Answers 2

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Let $\vec{v}_1, \vec{v}_2$ be two arbitrary vectors of this form:

$$ \vec{v}_1=(a_1,b_1,a_1-2b_1)^T,\\ \vec{v}_2=(a_2,b_2,a_2-2b_2)^T. $$

What we need to show is that their arbitrary linear combination $x\vec{v}_1+y\vec{v}_1$ is again of such form: $$ x\vec{v}_1+y\vec{v}_1=(xa_1+ya_2,xb_1+yb_2,x(a_1-2b_1)+y(a_2-2b_2))^T=(xa_1+ya_2,xb_1+yb_2,xa_1+ya_2-2(xb_1+yb_2))^T $$

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A vector belonging to $\mathbb R^3$ is an elements of $S$ iff you can write it as: $$v_{a,b}=a(1,0,1)^T+b(0,1,-2)^T$$ Note that for $\lambda\in\mathbb R$: $$\lambda v_{a,b}=v_{\lambda a,\lambda b}\in S$$ and: $$v_{a_1,b_1}+v_{a_2,b_2}=v_{a_1+a_2,b_1+b_2}\in S$$

So $S$ is closed under scalair multiplication and addition.

That is enough for it to be a subspace of $\mathbb R^3$.

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