Thread: Dimension of a subspace of R^4

1. Dimension of a subspace of R^4

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?

Yes, you've gotten the answer already. $\displaystyle R^4$ merely refers to the 4 dimensional vector space with each vector component being in the reals. Each one of your rows or columns is a point in that space.