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Use the rank-nullity theorem to show that if $V =\begin{bmatrix}a\\b\\c\\d\end{bmatrix}: x+y+z = 0, x+2y+3z =0$ that $\dim($V$) = 2.$

I verified that the dimension is 2 by finding a basis from the RREF of the system of constraints, but I'm not sure what role the rank-nullity theorem plays here.

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Hint Your space is the Nullspace of $$A=\begin{bmatrix} 1& 1 &1 & 0 \\ 1 & 2 &3 &0 \end{bmatrix}$$ [the zeros correspond to the fourth variable which is missing from the equation].

Now the two rows of you matrix are not proportional, hence linearly independent. Thus $\operatorname{rank}(A)=2$.

What does the rank-nullity theorem say?

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  • $\begingroup$ So null($A$) = 0 and rank($A$) = 2? I don't see why null($A$) has to be zero though. $\endgroup$ – 100001 Mar 4 '16 at 2:21
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    $\begingroup$ @MathLogic The rank-nullity theorem says that the rank plus the nullity is the number of COLUMNS of A ;) $\endgroup$ – N. S. Mar 4 '16 at 2:48
  • $\begingroup$ So null($V$) = 2. But I still don't see how that shows that dim($V$) = 2? $\endgroup$ – 100001 Mar 4 '16 at 2:55
  • $\begingroup$ $W$ is the Null space of the matrix ;) $\endgroup$ – N. S. Mar 4 '16 at 3:34
  • $\begingroup$ I'm sorry, I don't understand. Could you expound upon your explanations a bit more? I can't figure out what you are trying to get at. $\endgroup$ – 100001 Mar 4 '16 at 3:41

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