# Showing a set of Vectors is a basis for a Vector Space V

Consider the Vector Space $V=\{(x,y,z) \vert 2x-y+z=0\}$ and Let $S=\{(0,1,1),(1,2,0)\}$ and $T=\{(1,1,-1),(1,0,-2)\}$

1) Show that both $S$ and $T$ are bases for $V$

I represented $V$ explicitly: $V=\{a(\frac{1}{2},1,0)+b(-\frac{1}{2},0,1)\vert a,b\in \mathbb R\} = \text{span}\{(\frac{1}{2},1,0),(-\frac{1}{2},0,1)\}$

I know that in order for $S$ or $T$ to be a basis for $V$, it must satisfy 2 conditions:

• The vectors in $S$ and $T$ must be Linearly Independent to each other. (This is trivial because the 2 vectors in each other are definitely not scalar multiples of each other.)
• $S$ spans $V$, $T$ spans $V$ (This is the problem, I am not sure on how to go ahead with this, especially with the sets I am dealing with contain only some vectors.)

I know how to show that a set of vectors is a basis for $\mathbb R^n$ but I am not sure how to go about checking point 2. Any help here is appreciated!

Since $V$ is spanned by two linearly independent vectors, $V$ is $2$-dimensional. Therefore, any two linearly independent vectors of $V$ are a basis of $V$.

• As in all vectors in $V$ can be represented as a linear combination of the vectors in $S$? The same goes for $T$? Any idea on how do I check if a set spans another set in general? – Derp Oct 19 '17 at 15:38
• @Derp Yes, yes, and no, respectively. – José Carlos Santos Oct 19 '17 at 15:51
• Many thanks for the help! – Derp Oct 19 '17 at 15:52

I think the explicit expression is not hard to do. If

$$V = span \{ (\frac{1}{2},1,0) , (-\frac{1}{2},0,1) \} = span \{v_1,v_2\}$$

and

$$S = span \{ (0,1,1), (1,2,0)\} = span \{ s_1,s_2 \}$$ just note that $v_1 = \frac{1}{2}s_2$ and $v_2=\frac{1}{2}(2s_1-s_2)$, so any linear combinations of $v_1$ and $v_2$ can be write as linear combinations of $s_1$ and $s_2$. And do the same for $T$.

• Oh okay, so suppose $S$ spans $V$, it basically means that I just need to check if all the vectors in $V$ can be written as a linear combination of $S$? Sorry just wanted to confirm... – Derp Oct 19 '17 at 15:49
• Yes. Thats the definition of span. – Sou Oct 19 '17 at 16:13