# Basis of a subspace (What does it mean for vectors to span a subspace?)

How is the basis of this subspace the answer below? I know for a basis, there are two conditions:

1. The set is linearly independent.

2. The set spans H.

I thought in order for the vectors to span H, there has to be a pivot in each row, but there are three rows and only two pivots. I know that the set is linearly independent, but I don't understand how the set spans H.

Here is the subspace:

$$\left\{\begin{bmatrix} s-2t \\ s+t \\ 3t \end{bmatrix} : s,t \in \mathbb{R} \right\}$$

Here is the solution (finding the basis):

This subspace is $H=\mathrm{Span}\{\mathbf{v}_1,\mathbf{v}_2\}$, where $\mathbf{v}_1=\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $\mathbf{v}_2=\begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}$. Since $\mathbf{v}_1$ and $\mathbf{v}_2$ are not multiples of each other, $\{\mathbf{v}_1,\mathbf{v}_2\}$ is linearly independent and thus is a basis for $H$. Hence the dimension of $H$ is $2$.

• $$\begin{bmatrix}s-2t\\s+t \\ 3t\end{bmatrix} = s\mathbf{v}_1 + t\mathbf{v}_2$$so the parametrisation and the two given basis vectors span the same space. – Arthur May 9 '15 at 23:27
• Who do you mean by a "pivot?" – Thomas Andrews May 9 '15 at 23:30

The space has dimension $d$ at most $2$, since it is the image of a linear mapping from $\mathbb{R}^2$ to $\mathbb{R}^3$: $$\ell\colon (s,t)\mapsto (s-2t,s+t,3t)$$
Thus, any set of $2$ independent vectors has to be a basis, since a basis is a set of $d$ linearly independent vectors.
Span means that your set of vectors generates the whole space. More rigorously, if $V$ is a vector space over a field $K$, the span of a set of vectors $S$, $\operatorname{Span}(S)$, is defined to be $\{a_1 v_1 +a_2 v_2 + \cdots + a_n v_n : v_i\in S, a_i \in K\}$.
In terms of your particular example anyway, you want to show that every vector $v$ in $H$ can be written as $v=a_1v_1+a_2v_2$ for $a_1,a_2\in \Bbb{R}$. But of course by definition of $H$, if $v\in H$, $$v=\pmatrix{s-2t \\ s+t\\3t}=s\pmatrix{1\\1\\0}+t\pmatrix{-2\\1\\3}=sv_1+tv_2.$$
Therefore $v_1$ and $v_2$ span $H$.