Dimension of subspace this is the question:
What is the dimension of the subspace of $\mathbb{R}^4$ spanned (generated) by the vectors
\begin{array}{ccc}
1 & 2 & 1 & 0 \\
0 & 0 & 0 & 0 \\
-1 & -1 & 0 & 1 \\
0&1&1&1 
\end{array} 
I ended up with 
\begin{array}{ccc}
1 & 2 & 1 & 0 \\
0 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 
\end{array} 
then:
\begin{array}{ccc}
1 & 2 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 
\end{array} 
I think i am supposed to find the value of each variable but with only one line with 3 variables, how can i determine if it is linearly independent or dependent? that would then lead me to the answer regarding the dimension of the subspace i believe.
thanks in advance   
 A: The second matrix you obtained is correct - it is row equivalent to the first matrix, but the third is incorrect - only one of the rows with three 1's should become a zero row...if you do not understand why then it is a good idea to just study the row operations again.
Anyway, so that would leave you with a matrix with two nonzero rows that is row equivalent to the first matrix, and the two remaining row vectors are linearly independent, and hence the dimension of the subspace spanned by the given vectors is two.
So the reason why this so involves quite a bit of theory you have to work through, but in summary: 1> The dimension of the subspace spanned by the given vectors is the amount of vectors in a basis for such a subspace. 2> If you have linearly independent vectors spanning a subspace, then those vectors form a basis for the subspace. 3> The row space / column space of a matrix is spanned by the rows / columns of the matrix. 4> The dimension of the row space / column space of a matrix is known as its rank. 5> Row equivalent / Column equivalent / Equivalent matrices have the same rank.
So really the method you have followed to determine the dimension of the subspace involves all of the above. I hope that helps...
