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I'm watching Linear Independence and Subspaces lecture, and the prof proves that in any linear dependent space there's a vector which is a linear combination of finite number of other vectors.

He looks at $\alpha_1*x_1 + ...+ \alpha_n*x_n = \theta$, finds the largest index k of a non-zero coefficient $\alpha_k$, and shows that we can express $x_k$ as a l.c of the previous vectors like so: $-\alpha_1/\alpha_k*x1 + ... + -\alpha_{k-1}/\alpha_k*x_{k-1} = x_k$

HOWEVER, it might be that $\alpha_k$ is the ONLY one which isn't zero! so $\alpha_1 .. \alpha_{k-1}$ are all zero! how do we deal with that???

related to Vector in a linearly dependent set is a linear combination of other vectors in that set?, but here I'm after just one vector, not every vector in the set.

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If $\alpha_k$ is the only nonzero coefficient, we have that $x_k=0$, thus any set containing it is linearly dependent.

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  • $\begingroup$ do you mean that if xk is the zero vector, we can represent it as a trivial linear combination of other vectors, with all coefficients zero? does that count as a legit linear combination? $\endgroup$ – ihadanny Jan 10 '16 at 7:15
  • $\begingroup$ Yes, it does, a linear combination is just a sum of vectors multiplied by scalars, whichever those scalars are. $\endgroup$ – YoTengoUnLCD Jan 10 '16 at 14:52
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There is no problem if only one $\alpha$ is nonzero. It is still a linear combination of the other vectors. For a concrete example, consider $(1,0), (0,1), (2,0)$. These are three vectors in a two-dimensional space, and so are certainly linearly dependent. The linear dependence relation here is that $(2,0) = 2 (1,0) + 0 (0,1)$.

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  • $\begingroup$ thanks @mixedmath - I wasn't clear on my question, tried to rephrase. I'm not worried that one of the coefficients in the linear combination would be 0, I'm worried that they all would be zero! $\endgroup$ – ihadanny Jan 9 '16 at 18:47
  • $\begingroup$ @ihadanny I've addressed that in my answer. $\endgroup$ – YoTengoUnLCD Jan 9 '16 at 18:52

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