Show $\beta$ is a basis for subspace $V$

I'm working on problems in the section about dimensions of subspaces associated with a matrix.

The problem states to use the results of the section to show that $\beta$ is a basis for the subspace $V$.

$\beta = {[0, 1, 4], [4, -7, 0]}$

$V =$ {$[-s+t, 2s-t, s+3t]$$\epsilon$$\mathbb{R}^3$: r, s, and t are scalars}

How would I solve a problem like this ?

• The question is kind of hard to answer since we don't know what "the results of the section" are. – Fryie Nov 2 '15 at 1:31

Hint: the two vectors are linearly independent and belong to $V$; moreover $V$ is not the whole of $\mathbb{R}^3$.
If you want to use matrices and linear maps, you can consider the linear map $$f\colon\mathbb{R}^2\to\mathbb{R}^3,\qquad f([s,t])=[-s+t, 2s-t, s+3t]$$ Compute its associate matrix and see whether the two given vectors belong to the image. Then you're basically done.
If $V$ has dimension $n$ and $v_1,...,v_n$ are $n$ linearly independent vectors in $V$, then they are a basis of $V$. You can start working from there.