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Let $Q \in \mathbb{R}^{8 \times 4}, x \in \mathbb{R}^4$

$Qx = \begin{bmatrix} 2 & 1 & 1 & 0\\ 0 & 1 & 0 & 1\\ -2 & 0 & -1 & 1\\ 0 & -2 & 0 & -2\\ 2 & -2 & 1 & -3\\ 0 & 4 & 0 & 4\\ -2 & 6 & -1 & 7\\ 0 & -8 & 0 & -8 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{bmatrix}$

Matrix $Q$ has rank=2 , and $rank(Q)+dim(Ker(Q))=4$

If I solve the system (with Gaussian elimination), I get the solution $S=\{(\alpha,0,2\alpha,0)|\alpha \in \mathbb{R} \}$

Is the solution wrong? if $dim(Ker(Q))=2$ , then the solutinon should have $\alpha$ and $\beta$ ?

EDIT: It was an error in the Gaussian elimination

$S=\{(\alpha,\beta,-2\alpha,-\beta)|\alpha,\beta \in \mathbb{R} \}$

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    $\begingroup$ You should have $(1,0,-2,0)$ and $(0,1,-1,-1)$ as a base to the nullspace. $\endgroup$ Jun 4, 2013 at 17:38
  • $\begingroup$ Thank you! how did you find the base? $\endgroup$
    – Cristi
    Jun 4, 2013 at 17:42
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    $\begingroup$ If $\operatorname{rk} Q = 2$, then there are two linearly independent rows. Pick two linearly independent rows and figure out their null space. $\endgroup$
    – copper.hat
    Jun 4, 2013 at 17:45
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    $\begingroup$ @Cristi: you made a mistake in your gaussian elimination I think, your method was correct. $\endgroup$ Jun 4, 2013 at 17:46
  • $\begingroup$ @Cristi You may answer your own question and accept it (i.e. click the "tick" symbol on the left of the answer), so that it no longer appears in the "unanswered" queue. This helps the community. $\endgroup$
    – user1551
    Jun 5, 2013 at 1:50

1 Answer 1

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Solution is $S=\{(\alpha,\beta,-2\alpha,-\beta)|\alpha,\beta \in \mathbb{R} \}$

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    $\begingroup$ No! The solution is $\{(\alpha,\beta,-2\alpha-\beta,-\beta) : \alpha,\beta\in\mathbb R\}$ $\endgroup$ Jun 5, 2013 at 7:58

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