Constraints on Vector Space and Dimensions How can I use the rank-nullity theorem to show that if $$V =\left\{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}: a+b+c = 0, a+2b+3c =0\right\}$$ then $\dim($V$) = 2?$
I don't see how the theorem ban be applied to this problem.
 A: Note: I assume you meant $a+b+c=0$ instead of $x+y+z=0$, and same for the other equation.
Write a matrix with the first two rows [1, 1, 1, 0] and [1, 2, 3, 0].  Then complete the matrix to a nonsingular matrix.  The remaining two rows will give you the basis of the solution set to this equation.
The rank-nullity theorem tells you that you started with 4 unknowns, and thus your Dim = 4.  Also, the 2 constraints you have tell you which vectors will go to zero, and thus determine your nullspace.  Then rk + NS = Dim translates to rk + 2 = 4, so the rank of your solution set will be 4 - 2 = 2.
A: Consider the matrix
$$A = \left( \begin{array}{cccc}
1 & 1 & 1 & 0 \\
1 & 2 & 3 & 0 \\
1 & 1 & 1 & 0 \\
1 & 2 & 3 & 0
\end{array} \right).
$$
This matrix has rank $2$, note that the bottom two rows are linearly dependent (they are repeats of the top two rows).  Moreover, any vector in $V$ is in the null space, by your definition of $V$.  Therefore, the rank-nullity theorem says that dim$(V) = 4 - \operatorname{rank}(A) = 4 - 2 = 2$.
