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Let $\ u_1, u_2, u_3 $ be linearly independent vectors in $\ V$. and

$\ v_1 = u_1 + u_2+ u_3, \ \ v_2 = u_1 - u_2 + u_3 , \ v_3 = -u_1 + 3u_2 - u_3 $

Does (1) $\operatorname{span}\{ u_1,u_2,u_3 \} =\operatorname{span}\{ v_1 , v_2 , v_3 \} $ ?

It is clear that $\operatorname{span}\{ v_1, v_2, v_3 \}\subseteq\operatorname{span}\{ u_1,u_2,u_3 \} $ So need to prove that $\operatorname{span}\{u_1,u_2,u_3 \} \subseteq\operatorname{span}\{ v_1, v_2, v_3 \} $

So I chose random vector $\ w_1 \in\operatorname{span}\{u_1,u_2,u_3\} $ , $\ w_1 = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3$ and then trying to express it using $\ v_1, v_2, v_3 $

Now trying to solve it using the equations I only get

$\ u_2 = \frac{v_2 + v_3}{2} $ and can't express $\ u_1, u_3 $ using the v's. So I guess it means those vectors are linearly dependent and therefore equation (1) is wrong?

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Simply note that

$$u_2=\frac{v_1+v_3}4=\frac{v_1-v_2}2$$

therefore $v_1,v_2,v_3$ are not linearly independent and

$$\ Sp \{ u_1,u_2,u_3 \} \neq Sp\{ v_1 , v_2 , v_3 \}$$

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  • $\begingroup$ Right.. Thanks!! $\endgroup$ – bm1125 Aug 28 '18 at 9:05
  • $\begingroup$ @bm1125 You are wecome! Bye $\endgroup$ – gimusi Aug 28 '18 at 9:06
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Clearly, $v_1-2v_2=v_3$. Therefore, yes, the vectors are linearly dependent and so they don't span the whole space.

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Note that $$\det \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ -1 & 3 & -1 \\ \end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace

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