We know that to prove vectors $v_1,v_2,v_3$ linearly dependent we must find scalers $x_1,x_2,x_3$ not all equal to $0$ such that $x_1v_1+x_2v_2+x_3v_3=0$. But the doubt I have is on the other alternative condition,which says if one of them can be expressed as a linear combination of other $2$,then they are linearly dependent. So let's assume the $x_1,x_2,x_3$ found earlier satisfy $x_1=x_2=0$ with $x_3$ non zero,then we can write $v_3=\frac{x_1}{-x_3}v_1+\frac{x_2}{-x_3}v_2=0v_1+0v_2$. Here we see that $v_3$ is not a linear combination of $v_1,v_2$ since coefficients of $v_1,v_2$ are zero hence linearly independent. But by first condition they are linearly dependent due to not all coefficients equal to $0$. Isn't it contradictory? Where is the mistake I am making?
3 Answers
The mistake that you made is that you said "$v_3$ is not a linear combination of $v_1,v_2$...". Note that $v_3 = 0$ and a zero vector is always a linear combination of $v_1, v_2$ as you already wrote: $0=v_3 = 0v_1+0v_2$. And this implies that $v_1,v_2,v_3$ form a linearly dependent set of vectors because one of them is the zero vector.
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The definition of "the vector $x_0$ is a linear combination of $x_1,\cdots,x_n$" is that, exists scalars $a_1,\cdots, a_n$ in the corresponding field s.t. \begin{equation} x_0=a_1x_1+\cdots+a_nx_n, \end{equation} and we do not require $a_i\ne 0$ for all $i$.
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In your case, $v_3$ is the zero vector, so by convention (or definition), it must be linearly dependent with other vectors.
In fact your definition of linearly dependence is equivalent to your definition of linearly independent. Suppose a set of vector is not linearly independent, we must find some $a_1,\cdots,a_n$ such that they are not all zero, and $$a_1v_1+\cdots+a_nv_n=0$$ Since they are not all zero, we can certainly find some candidates $a_i\ne0$, then $$v_i=\dfrac{a_1}{a_i}v_1+\cdots+\dfrac{a_n}{a_i}v_n$$
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