Nullspace Dimension I have come across a problem which is not consistent with what I know about subspaces and the Fundamental Theorem of Linear Algebra. Consider a matrix  
$$\begin{bmatrix}1&2&3&5&7&8\\1&1&4&6&8&9\end{bmatrix}.$$ 
Its RRE form is then:
$$\begin{bmatrix}1&0&5&7&9&10\\0&1&-1&-1&-1&-1\end{bmatrix}.$$
The first two columns are the pivot columns, the rest is just a linear combination of the two leading columns. So far so good. We then say that the dimension of the nullspace is just the number of free variables (in this case $4$). But is not a dimension of a subspace the number of the basis vectors in the subspace? How can we say that the dimension is $4$ when in this example there can be only $2$ basis vectors for the nullspace? So the dimension of the nullspace is $2$, the number of pivot columns is also $2$ - but this does not add up to $6$.

Oh, nevermind - I was looking at the columns of the matrix rather than looking at the solution vectors. It then becomes quite clear, thanks.
