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Consider $$ A =\left( \begin{matrix} 1 & -1 & 0 & -2 \\ 0 & 0 & 1 & -1 \\ \end{matrix} \right) $$ and find a basis and the dimension of $S(A,0)$, where $S(A,0)$ is the subspace of all the solutions $\mathbf{x}\in\mathbb{R}^n$ to the linear equations $A\mathbf{x}=\mathbf{0}$.

We need to solve: $$w-x-2z=0$$ $$y-z=0$$ So for any $\lambda,\mu\in\mathbb{R}$ we get $w=\lambda$, $x=\lambda - 2\mu$, $y=\mu$, $z=\mu$.

Does this mean that $S(A,0)$ is spanned by two vectors? For example;

  • $\lambda = 1, \mu = 0$ gives the vector $(1,1,0,0)$
  • $\lambda = 0, \mu = 1$ gives the vector $(0,-2,1,1)$

So then the dimension of $S(A,0)$ would be 2?

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  • $\begingroup$ All looks fine to me. $\endgroup$ – David Nov 26 '14 at 0:22
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It's worth mentioning that one can easily check what the dimension of $S$ (usually called the null space) is going to be.

For an $m \times n$ matrix $A$, the rank-nullity theorem states that $\operatorname{Rank}(A) + \operatorname{Null}(A) = n$.

In this case, $n = 4$, and $\operatorname{Rank}(A) = 2$ since it is already in echelon form.

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