# Bases of subspaces

I need some help with this homework question.

How do I determine whether the set of $\{(1,-1,0),(0,1,-1)\}$ forms a basis for the subspace of $\mathbb{R}^3$ consisting of all $(x,y,z)$ such that $x+y+z=0$?

I know that for a vector space to form a basis, it has to span $V$ and be linearly independent.

Do correct me if I'm wrong in any areas

Check if your two vectors are linearly independent (that is if they are not collinear). Then check that both vectors belong to your subset. If you can answer positively to both questions then you are done since your subspace has dimension 2 (therefore any linearly independent collection of vectors of $V$ of exactly two vectors is a basis).

• that's definitely not true, you are glossing over the fact that the vectors need to span the subspace. – TSF Apr 17 '18 at 13:56
• @TonyS.F. Except that you'd have 2 linearly independent vectors in a sub space of dimension 2 so they would span the whole sub space. – Bill O'Haran Apr 17 '18 at 13:58
• To check if they are linearly independent, can I use gaussian elimination to reduce it to row echelon form? – Jenny Apr 17 '18 at 13:58
• @Jenny Yes yoou can! – Must Apr 17 '18 at 14:00
• Can I use the same method to check that it spans V as well? – Jenny Apr 17 '18 at 14:03

If you answer yes to the following, then the full answer is yes! Let $V$ be the vector space defined by $V = \{(x,y,z) \mid x+y+z=0\}$

1) Are the vectors you are checking as a basis linearly independent?

2) Are the vectors you are checking as a basis in $V$?

3) If (1) and (2) are both yes, then the final question to answer is: is the dimension of $V$ equal to the number of vectors you are checking as a basis?

Here a hint: a generic vector in the subspace you consider is $$(x,-x-z,z) = x(1,-1,0)+z(0,-1,1).$$

• How did you get this : (x,y,−x−y)? – Jenny Apr 17 '18 at 14:02
• I see... another question is to check if they are linearly independent, I can use gaussian elimination to reduce it to row echelon form, however to check if it spans, can I use the gaussian elimination method as well? – Jenny Apr 17 '18 at 14:05
• I edited the answer. It is better to write $y=−x−z$, so that $(1,−1,0), (0,-1,1)$ are generators of any vector in your subspace. Since they are just two vectors, to prove they are linearly independent it is enough to see they are not proportional. Note: writing $(x,y,z) = x(1,-1,0)+z(0,-1,1)$ already shows that your set of vectors generates the subspace. – Gibbs Apr 17 '18 at 14:29

This set subspace is the space of all vectors of the form,

$$v = \begin{pmatrix}a\\ b\\ -a-b\end{pmatrix}$$

which is obviously a 2 dimensional space. Now, we need to be able to write $v$ as a linear combination of our proposed basis vectors.