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 $2$.

  • 1
    $\begingroup$ $$\begin{bmatrix}s-2t\\s+t \\ 3t\end{bmatrix} = s\mathbf{v}_1 + t\mathbf{v}_2$$so the parametrisation and the two given basis vectors span the same space. $\endgroup$ – Arthur May 9 '15 at 23:27
  • $\begingroup$ Who do you mean by a "pivot?" $\endgroup$ – 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 $\{a_1 v_1 +a_2 v_2 + \cdots + a_n v_n : v_i\in S, a_i \in K\}$.

With matrices and whatnot this can be rephrased as you indicated in terms of how many pivots there are. It would be an excellent exercise to work this out!

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 $H$.


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