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Are the vectors $\left[\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right]$, $\left[\begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ 0 \end{matrix}\right]$ and $\left[\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\right]$ linearly dependent?

I think yes, because if we multiply the second vector by $-2$ and add it to the 3rd, we obtain the first. If I am correct, is it also necessary to reduce the matrix to its echelon form? Why?

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    $\begingroup$ When a set of vectors contains the vector $\vec{0}$ it is always linearly dependent. If the set is $\{\vec{0},\vec{v}_2,\cdots, \vec{v}_n\}$ then $\vec{0}=0\vec{v}_2+\cdots 0\vec{v}_n.$ $\endgroup$ – mfl Mar 19 '15 at 16:09
  • $\begingroup$ Your argument is correct, but as has been observed there is a "simpler" reason. $\endgroup$ – André Nicolas Mar 19 '15 at 16:10
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They are linear dependent most obviously because there is a zero vector in it. Since

$$c_1 \vec{0} + 0 \vec{v}_1 + 0 \vec{v}_2=0$$

for any $c_1$. By definition they have to be linearly dependent. You don't need to reduce them to row echelon form.

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Your reasoning is headed in the right direction; indeed there is a linear combination of the second and third vectors that yields zero. this, in and of itself, tells you the vectors are dependent, no matter what the first vector is.

Moreover, any collection of vectors that includes the zero vector is automatically linearly dependent, can you see why?

In other words, these vectors fail to be independent in two different ways.

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The 2nd and the 3rd vector are linearly dependent, whatever the first one may be. This is because the 2nd vector is a scalar multiple of the third one, and vice versa.

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