Basis in Linear Algebra I am taking an introductory linear algebra course, and I am stuck on this problem:
Explain why the set $W= \{(a,b,c)\ |\  a+b+c=0\}$ is a subspace of $\mathbb R^3$. After, find a basis for the subspace $W$. 
I don't really know how to start it, so if you guys could answer my question, I would greatly appreciate it. 
thanks
 A: Prove the basic conditions a subset must fulfill to be a subspace: (a) check that the zero vector is in $\;W\;$ , (b) show that if two vectors are in $\;W\;$ so is their sum, and (c) if a vector $\;w\in W\; $ and $\;r\;$ is any scalar, then also $\;rw\in W\;$ .
As for a basis: observe that $\;a+b=0=c\;$ leaves us with only one free choice, for example $\;b\;$ . Once $\;b\;$ is chosen we already have no more choices as it must be $\;a=-b\;,\;\;c=0\;$ . This tells us that the dimension is $\;1\; $ (Remember: the ammount of free choices), so now only take your favourite number, which is $\;1\;$  (it can't be zero as then you get the zero vector, which can't be part of any basis. Why?), and $\;\{(1,-1,0)\}\;$ is a basis of $\;W\;$ .
A: The subspace is the null space of the matrix 
\begin{bmatrix}
1&1&1\\
\end{bmatrix}
and hence is a $2$ dimensional subspace by the rank nullity theorem. One can check that $(1,-1,0)^T$ and $(0,1,-1)^T$ are two linearly independent vectors in the null space so that they form a basis for $W$.
A: By what you write we have that $c=0$ which means we've already lost 1 dimension, then we have $a+b=0$ which means it's a line only, a line is definitionally a subspace of the 3 dimensional space.
A: Just go by the definition of subspace: a subspace is a subset of the space that is also a vector space. You can easily prove that any linear combination of vectors $(a, b, c)$ for which $a+b + c = 0$, also satisfies this condition. Therefore, it is indeed a subspace.
I can elaborate on this if the rest of the proof is not clear to you.
