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Suppose that we have the system:

$$-x_1 + x_2 + x_3 + x_4 = 0$$ $$x_2 + x_4 + x_5 = 0$$ $$x_1-x_3+x_5 = 0$$

Give a basis for the solutionspace (i.e. the Nullspace) of this system of equations and express the basis in terms of its basis elements.

My attempt at a solution: (please tell me if I'm going wrong somewhere)

So first of all I used row reduction to simplify this system into 1 equation. I'll just tell you the steps where $R_1, R_2, R_3$ are Row $1, 2, 3$.

$$-R_1$$ $$R_3-R_1$$ $$R_2-R_3, R_3 - R_2$$

The system simplifies to this equation: $$x_1 -x_2-x_3-x_4+ 0x_5= 0$$ $$=>x_1 -x_2-x_3-x_4 = 0$$

Let $C_1, C_2, C_3, C_4, C_5$ denote Column $1,2,3,4,5.$

$$\langle C_1,C_2,C_3,C_4,C_5 \rangle \subseteq \mathbb R^3 \text { is the column space}$$

$$\langle R_1,R_2,R_3 \rangle \subseteq \mathbb R^5 \text { is the row space}$$

Let $S$ denote the nullspace

$S= $ {$(x_1,x_2,x_3,x_4,x_5) \in \mathbb R^5: x_1 = x_2+x_3+x_4$ }

$$(x_2+x_3+x_4, x_2, x_3, x_4, x_5) = x_1(0,0,0,0,0) + x_2(1,1,0,0,0) + x_3(1,0,1,0,0) + x_4(1,0,0,1,0) +x_5(0,0,0,0,0)$$

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2 Answers 2

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Row reduction on the matrix goes as follows: \begin{align} \begin{bmatrix} -1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & -1 & 0 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & -1 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & -1 & 0 & 1 \end{bmatrix} &&R_1\gets -R_1\\[6px] &\to \begin{bmatrix} 1 & -1 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \end{bmatrix} &&R_3\gets R_3-R_1\\[6px] &\to \begin{bmatrix} 1 & -1 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} &&R_3\gets R_3-R_2\\[6px] &\to \begin{bmatrix} 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} &&R_1\gets R_1+R_2\\[6px] \end{align} The free variables are $x_3$, $x_4$ and $x_5$, with $$ \begin{cases} x_1=x_3-x_5\\ x_2=-x_4-x_5 \end{cases} $$ so a basis of the null space is given by $$ \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} -1 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ Since the first two columns are the dominant ones (those which correspond to the nonfree variables), a basis for the column space is $$ \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} $$ Note that the first two columns are easily seen to be linearly independent, so the rank of the matrix is at least $2$; therefore you can't have four free variables as you'd get with your (wrong) row reduction, because $2+4=6$ would violate the rank-nullity theorem.

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  • $\begingroup$ Wait so aren't we supposed to say that $x_5 = x_3 - x_1$ then $x_2 = -x_4 - (x_3 - x_1)$ so $x_2 = x_1-x_3-x_4$ and $(x_1, x_1-x_3-x_4, x_3,x_4, x_5) = x_1(1,1,0,0,0) + x_2(0,0,0,0,0) + x_3(0,-1,1,0,0) + x_4(1,-1,0,1,0) + x_5(0,0,0,0,1)$. I don't really understand what is being asked here $\endgroup$ Apr 6, 2015 at 11:18
  • $\begingroup$ I know that's basically what I did in the first place but how can it be wrong $\endgroup$ Apr 6, 2015 at 11:19
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    $\begingroup$ @StephanCasey A basis can be chosen in several different ways; I prefer a more systematic procedure. However, your way of solving the system is wrong. It's true that you can use $x_5$ and $x_2$ as nonfree variables, but what you write is eventually wrong, because you don't use $x_5=x_3-x_1$, so the four nonzero vectors you show are not a basis of the null space. $\endgroup$
    – egreg
    Apr 6, 2015 at 11:22
  • $\begingroup$ I don't really see how you obtained your basis of the null space. I understand what you mean about the approach though $\endgroup$ Apr 6, 2015 at 11:29
  • $\begingroup$ Oh I see. You go $(x_3-x_5,-x_4-x_5, x_3, x_4, x_5) = x_1(0,0,0,0,0) + x_2(0,0,0,0,0) +x_3(1,0,1,0,0)+x_4(0,-1,0,1,0) + x_5(-1,-1,0,0,-1) = x_3(1,0,1,0,0)+x_4(0,-1,0,1,0) + x_5(-1,-1,0,0,-1)$ $\endgroup$ Apr 6, 2015 at 11:36
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In your row reduction process, you can not do

$$R_2-R_3, R_3-R_2$$

one after another. After $R_2-R_3$, $R_2$ is eliminated. So you end up with

$$x_1-x_2-x_3-x_4=0\\x_2+x_4+x_5=0$$

So your null space can be represented by $x_3,x_4,x_5$.

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