Find a basis for this matrix I have a matrix that only contain variables and zeros, like this:
$$ 
\begin{bmatrix} 
0 & -a & -b \\
a & 0 & -c \\
b & c & 0 \\
\end{bmatrix} $$
I usually would find the basis for this by row reduction and then take the columns with leading ones as basis, but how do I do when there is just variables?
 A: The variables are also just numbers, so this isn't much different from doing what you described. (And the process you described finds a basis for the rowspace, which is the interpretation I'll use for my solution. I'm also assuming the matrix has enties in a field.)
The only complication is that the size of the basis may change depending on how many variables are zero. In all cases here, it happens that the basis of the rowspace will be either empty or will have two elements.
First of all, the determinant is $0$, and so the matrix can't ever have more than two linearly independent rows.
If $a=b=c=0$, then the basis is empty.
If $a$ isn't zero, then $(0,-a,-b)$ and $(a,0,-c)$ are linearly independent and form a basis.
If $a=0$ but on of $b$ or $c$ isn't, then it's obvious that $(b,c,0)$ and $(0,0,x)$ for a basis, where $x$ is whichever one of $b,c$ that is not zero.

It seems also that the question is not very clear on what "basis" we are looking for. Another plausible interpretation is this:

Find a basis for the vector space of matrices $\left\{\begin{bmatrix} 
0 & -a & -b \\
a & 0 & -c \\
b & c & 0 \\
\end{bmatrix} \mid a,b,c\in F  \right\}$

That is fairly easy to do by inspection:
$\begin{bmatrix} 
0 & -a & -b \\
a & 0 & -c \\
b & c & 0 \\
\end{bmatrix}=
\begin{bmatrix} 
0 & -a & 0 \\
a & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}+\begin{bmatrix} 
0 & 0 & -b \\
0 & 0 & 0 \\
b & 0 & 0 \\
\end{bmatrix}+\begin{bmatrix} 
0 & 0 & 0 \\
0 & 0 & -c \\
0 & c & 0 \\
\end{bmatrix}\\
=
a\begin{bmatrix} 
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}+b\begin{bmatrix} 
0 & 0 & -1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{bmatrix}+c\begin{bmatrix} 
0 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0 \\
\end{bmatrix}$
So the last three matrices are a good choice of basis.
