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I am given the following matrix $A$ and I need to find a nullspace of this matrix.

$$A = \begin{pmatrix} 2&1&4&-1 \\ 1&1&1&1 \\ 1&0&3&-2 \\ -3&-2&-5&0 \end{pmatrix}$$

I have found a row reduced form of this matrix, which is: $$A' = \begin{pmatrix} 1&0&3&-2 \\ 0&1&-2&3 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}$$ And then I used the formula $A'x=0$, which gave me: $$A'x = \begin{pmatrix} 1&0&3&-2 \\ 0&1&-2&3 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ Hence I obtained the following system of linear equations: $$\begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases}$$ So I just said that $x_3=\alpha$, $x_4=\beta$ and the nullspace is: $$nullspace(A)=\{2\beta-3\alpha,2\alpha-3\beta,\alpha,\beta) \ | \ \alpha,\beta \in \mathbb{R}\}$$

Is my thinking correct? Thank you guys!

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  • $\begingroup$ your method seems fine to me $\endgroup$ Commented May 10, 2015 at 17:02
  • $\begingroup$ I forgot about one thing. How to find a basis of the nullspace? $\endgroup$
    – marco11
    Commented May 10, 2015 at 17:12

2 Answers 2

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Since $x_1=2x_4-3x_3$ and $x_2=2x_3-3x_4\Rightarrow$

if $(x_1,x_2,x_3,x_4)\in$ nullspace($A$): $$(x_1,x_2,x_3,x_4)=(2x_4-3x_3,2x_3-3x_4,x_3,x_4)=x_3(-3,2,1,0)+x_4(2,-3,0,1)$$ So Nullspace$(A)=\langle (-3,2,1,0),(2,-3,0,1) \rangle$.

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  • $\begingroup$ I am very sorry, I wrote one index wrongly in the first equation of the system of equations. Now it is corrected. $\endgroup$
    – marco11
    Commented May 10, 2015 at 17:03
  • $\begingroup$ I edited answer, can you take a look and rate me later. $\endgroup$
    – L F
    Commented May 10, 2015 at 17:13
  • $\begingroup$ Thank you! Could you also suggest how to find a basis of this nullspace? $\endgroup$
    – marco11
    Commented May 10, 2015 at 17:15
  • $\begingroup$ read the last two lines, this is the way to find a basis when you have a system of equations, also I wrote the basis $(-3,2,1,0),(2,-3,0,1)$, you have to check that they're li. if not, just take one of them because the other will be multiple of the first. (If it result helpfull you can upvote my answer and select it as correct answer in the left side of the asnwer.) $\endgroup$
    – L F
    Commented May 10, 2015 at 17:18
  • $\begingroup$ Thank you very much. I really appreciate your help. $\endgroup$
    – marco11
    Commented May 10, 2015 at 17:27
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Everything is right. But after finding the equations \begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases} This $x_1$ and $x_2$ are pivot variables and $x_3$ and $x_4$ are free variables. The number of non zero rows is the rank of the matrix, in our case $2$ and hence nullity of the matrix is $2$. since $dim(W) = Rank + Nullity$. Now since $nullity =2$, the usual basis is $\{(1,0),(0,1)\}$. Hence Substitute $x_3=1$ and $x_4=0$ in two equations, we will get $x_1=-3$ and $x_2 =2$ and we get a point $(-3,2,1,0)$ and again substitute $x_3=0$ and $x_4=1$ in two equations, we will get $x_1 = 2$ and $x_2 =-3$ and we will get another point $(2,-3,0,1)$. These two points $(-3,2,1,0)$ and $(2,-3,0,1)$ serves as a basis for the nullspace of the matrix.

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  • $\begingroup$ That's brilliant! $\endgroup$
    – marco11
    Commented May 10, 2015 at 17:56

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