My question asks me to consider a matrix A

$\displaystyle \left( \begin{array}{cccc} 1 & 2 & 2 & 3 \\ 2 & 5 & 4 & 8 \\ -1 & -3 & -2 & -5 \\ 0 & 2 & 0 & 4 \end{array}\right)$

Then find a basis for the nullspace of A, and hence then dimension of the nullspace;

which I find to be $\displaystyle \{ \left( \begin{array}{c} -2 \\ 0 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right) \}$.

Since there are two basis vectors the dimension is 2, right?

My problem then is "Using the rank-nullity theorem or otherwise, determine the dimension of the subspace of R^{4} spanned by the four columns of A".

R-N states that dimension null = #cols - rank...

So dimension null =2 , #cols = 4, so I'm guessing rank = 2...

I'm confused by the mention of the "dimension of the subspace of R^{4}" though.

Is the answer they're looking for simply 2? And how ought I to phrase this to explain my answer, rather than merely subtracting one number from another?

Thanks in advance