How is the basis of this subspace the answer below?
I know for a basis, there are two conditions:
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The set is linearly independent.
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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 $2$.
Best Answer
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.