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I am trying to solve the following problem:

Let $ S = \{\ (1,1,0), (5,2,3)\ \} $

Let $ V = \{ \ (x,y,z)\ |\ x - y -z = 0 \} $ be a subset of $ \mathbb{R}^3 $.

Please show that $ \operatorname{span}(S) = V $

We know that in order for a span to be $\mathbb{R}^n$, it needs to contain $n$ vectors. However, there are only 2 vectors inside $ S $ so how is this possible? I am aware that there is a given condition in $V$ but how does that contribute to the solution? Could someone please advise me?

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    Hint: The condition in $V$ makes it a plane. So 2 base vectors suffices to span it. – 2017-02-14
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    More hint: Refer this http://math.stackexchange.com/questions/892076/vectors-spanning-a-plane – 2017-02-14
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    Off the cuff solution. The two vectors are linearly independent. Take their cross product, and take the dot product with the plane. It will be a zero. This proves that the plane it generates will be parallel to the plane given. However both the planes pass through zero. Hence both are planes are the same. And the two vectors span the given plane. – 2017-02-14

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First, recall that the dimension of a vector space is the number of vectors in a basis for it. Be careful to distinguish between the dimension of the total space that you’re working in from that of its subspaces. The former, $\mathbb R^3$, is indeed three-dimensional, but its subspaces can be of any dimension from $0$ to $3$.

So, for instance, even though the elements of $S$ are coordinate triples, the dimension of $S$ is at most two, because it’s generated by only two vectors. It’s exactly two if the vectors are linearly independent, in which case they form a basis for $S$.

On to the problem at hand. Start by plugging the two vectors that generate $S$ into the defining equation for $V$: $$\begin{align}1-1-0&=0 \\ 5-2-3&=0.\end{align}$$ Both of the vectors thus satisfy this equation, and so any linear combination of them will also satisfy the equation (try it!). This shows that $S\subseteq V$.

To prove equality, you need to show either that $\dim S=\dim V$ or that $V\subseteq S$. For the former, you might start be finding out whether or not the two vectors that generate $S$ are linearly independent (for a pair of vectors that’s really easy). That’ll tell you if $\dim S$ is $1$ or $2$, after which you’ll need to work out the dimension of $V$ and compare them. I’ll leave the details for you to work out.

If you want to try to prove $V\subseteq S$ instead, one approach is to first characterize the elements of $V$ in a different way. The defining equation of $V$ is $x-y-z=0$, so $V$ clearly consists of all elements of $\mathbb R^3$ of the form $(y+z,y,z)$. (This can also help you figure out the dimension of $V$, by the way.) Now, can you express such a vector as a linear combination of $(1,1,0)$ and $(5,2,3)$? In other words, solve $$(y+z,y,z)=\alpha(1,1,0)+\beta(5,2,3)=(\alpha+5\beta,\alpha+2\beta,3\beta)$$ for the scalars $\alpha$ and $\beta$. This will show that $(y+z,y,z)\in S$, and so $V\subseteq S$. Again, I leave working out the details to you.

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Note that $V$ has two dimensions, so if you have two vectors linearly independent in it, the span of them will generate the vector space.

As $V$ is defined by $x-y-z=0$ then $x=y+z$, making $y=1$ we have $v_1 = (1,1,0)$ and making $z=1$ we have $v_2 = (1,0,1)$, consisting of a base for $V$.

The first element of $S$ is $v_1$, and the second is an element of $V$ because $(5,2,3) = 2 v_1 + 3 v_2$. Note that $(1,1,0)$ and $(5,2,3$) are linearly indepedent, so they span $V$.

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    Why does $V $ have 2 dimensions? Aren't the vectors in the set in $ \mathbb{R}^3 $ space? – 2017-02-14
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    The constraint $x-y-z=0$ forms a hyperplane in $R^3$. You can see it as a homogeneous system of equations, with 3 variables and 1 equation. – 2017-02-14
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    @LanceHAOH The *parent* space $\mathbb R^3$ is indeed three-dimensional, but the subspace $V$ is only two-dimensional. – 2017-02-14
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Let's call the two vectors that span $S$ $a$ and $b$. You can rewrite the set S as $$S = \left\{x\in\mathbb{R}^3|\,x=c_1a+c_2b;\,c_1,c_2\in\mathbb{R}\right\},$$ which is the definition of the span. Insert this into the condition $$x-y-z=0$$ which defines the set $V$. You will see that this condition is satisfied for any $c_1, c_2 \in \mathbb{R}$. Thus $S \subset V$. Conversely, let $d = (d_1,d_2,d_3)^T$ be a vector, which is in $V$. Thus we have $d_1=d_2+d_3$. Any such vector can be written as $$d=\begin{pmatrix}d_2\\d_2\\0\end{pmatrix}+\begin{pmatrix}d_3\\0\\d_3\end{pmatrix}=(d_2-\frac{2}{3}d_3)\begin{pmatrix}1\\1\\0\end{pmatrix}+\frac{1}{3}d_3\begin{pmatrix}5\\2\\3\end{pmatrix}.$$ As both coefficients $(d_2-\frac{2}{3}d_3)$, $\frac{1}{3}d_3$ are in $\mathbb{R}$ we conclude $V \subset S$ and therefore $$S=V.$$

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If $(x,y,z)\in \mathrm{span}(S)$, then by definition $$(x,y,z)=\lambda(1,1,0)+\mu(5,2,3)=(\lambda+5\mu,\lambda+2\mu,3\mu)$$ for some $\lambda,\mu\in\mathbb R$. So $x-y-z=\lambda+5\mu-(\lambda+2\mu)-3\mu = 0$. Therefore $(x,y,z)\in V$.

Conversely, if $(x,y,z)\in V$, then $x=y+z$. Put $\lambda=y-\frac23 z, \mu=\frac13z$. Then $$\lambda(1,1,0)+\mu(5,2,3)=(\lambda+5\mu,\lambda+2\mu,3\mu)=(y+z,y,z)=(x,y,z)$$

Therefore $(x,y,z)\in\mathrm{span}(S)$.