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If the vectors $\vec w_1$, $\vec w_2$, $\vec w_3$ are linearly independent, are vectors $\vec w_1$, $\vec w_1 + \vec w_2$, $\vec w_1 + \vec w_2 + \vec w_3$ linearly independent as well? Explain

At first I thought the second set of vectors would not be linearly independent, since you are adding the components of one vector(s) onto another. However when I try to write one vector as a scalar multiple of the others, I can't seem to find any ways to do so. I tried to do the guess and check method to find situations where the vectors are linearly dependent but failed. Can anyone please put me on the right direction for explaining whether or not the new set of vectors is linearly independent or not? Thank you very much!

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  • $\begingroup$ Have you tried proving that they are independent? $\endgroup$ – Eric Wofsey Oct 10 '15 at 21:20
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Hint:

Start by using the definition. Consider the vector equation $$x(\vec w_1)+y(\vec w_1 + \vec w_2)+z(\vec w_1 + \vec w_2 + \vec w_3)=\vec 0.$$ If this equation has only the trivial solution ($x,y,z=0$) then the vectors are linearly independent. This equation can be rewritten as $$(x+y+z)\vec w_1 + (y+z) \vec w_2 + z \vec w_3 = \vec 0.$$

You are given that $\vec w_1, \vec w_2, \vec w_3$ are linearly independent. What does that tell you about the coefficients in the vector equation we obtained?

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Let \begin{align} \vec v_1 & = \vec w_1 \\ \vec v_2 & = \vec w_1 + \vec w_2 \\ \vec v_3 & = \vec w_1 + \vec w_2 + \vec w_3 \end{align} Showing that $\{ \vec v_1, \vec v_2, \vec v_3 \}$ is linearly independent is equivalent to showing that it spans the same subspace as $\{ \vec w_1, \vec w_2, \vec w_3 \}$, which is trivial because \begin{align} \vec w_1 & = \vec v_1 \\ \vec w_2 & = \vec v_2 - \vec v_1 \\ \vec w_3 & = \vec v_3 - \vec v_2. \end{align}

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