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Let $W$ be the set of all $x=(x_1,x_2,x_3,x_4,x_5)$ in $\mathbb{R}^5$ which satisfy

$2x_1-x_2+\frac{4}{3}x_3-x_4=0$

$x_1+\frac{2}{3}x_3-x_5=0$

$9x_1-3x_2+6x_3-3x_4-3x_5=0$.

Find a finite set of vectors which spans $W$?

This question is from linear algebra Hoffman and K. Guys please help in this question I don't know what to do in this question.

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    $\begingroup$ This is a linear system. Do you know how to use Gaussian elimination to put it in Reduced Row Echelon Form? That's a standard way to solve such a problem. $\endgroup$ – Julien Apr 5 '13 at 2:07
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Hint:

As Julien mentions, write it in matrix form and do Row-Reduced-Echelon-Form (RREF).

$\displaystyle \begin{bmatrix}2 & -1 & \frac{4}{3} & -1 & 0\\1 & 0 & \frac{2}{3} & 0 & -1 \\ 9 & -3 & 6 & -3 & -3\end{bmatrix}$

The RREF produces:

$\displaystyle \begin{bmatrix}1 & 0 & 2/3 & 0 & -1\\0 & 1 & 0 & 1 & -2\\0 & 0 & 0 & 0 & 0 \end{bmatrix}$

Do you know how to proceed?

Update

From the RREF, you now know that the system has a solution and is thus consistent.

That is, you can solve the RREF system (you know how - do it, but don't need to for this problem).

This means that those two vectors span (know which ones) $W$ in $\mathbb{R}^5$.

Recommendation: It is important for you to review what span means as that might be confusing you (I am guessing), but you should you understand this concept.

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  • $\begingroup$ Ok i understood upto this.Then how will i get finite set of vectors which span W $\endgroup$ – David Jones Apr 5 '13 at 6:42
  • $\begingroup$ @DavidJones: see update. $\endgroup$ – Amzoti Apr 5 '13 at 12:33
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    $\begingroup$ Nice work, and nice extra help with the update! +1 $\endgroup$ – Namaste Apr 6 '13 at 0:27

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