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When expressing a vector in a linearly dependent set as a linear combination of other vectors in the set, will there always be a unique solution or is it possible that there could be an infinite amount ways to express that vector as a linear combination of the others? (Excluding the case of a $(0, 0)$ vector)

I'm just wondering because my book asked us to show whether a set was dependent, and in the process we formed the equation $-\dfrac{2}{7}t(3, 4) + \dfrac{8}{7}t(-1, 1) + t(2,0) = (0,0)$ , which has an infinite amount of solutions. The problem then asked us to solve for one of the vectors, which gave the equation $(2, 0) = \dfrac{2}{7}(3, 4) - \dfrac{8}{7}(-1, 1)$. I noticed that the $t$'s canceled out and there was a unique solution for expressing $(2,0)$ as a linear combination, so I wondered if this will always be the case.

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  • $\begingroup$ Not generally. Consider the vector x=1 as your basis. Then the vector y=5*x had only one expression for that given basis. This is in the vector space of real numbers $\endgroup$ – Michael Stachowsky Dec 9 '16 at 4:09
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If the set of remaining vectors is itself linearly dependent, then there will be some combination thereof adding up to $0$. Let for example $u=a, v=-a, w=2a$, then $w=2 u = 3 u + v = \cdots$

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The set $$\{(1,1)\ , \ (2,2)\} $$ is linearly dependent. There is only one way of expresing $(2,2)$ as a linear combination of $(1,1)$.

The set $$\{(1,1),(2,2),(3,3)\}$$ is linearly dependent. There are infinitely many ways of expressing $(2,2)$ as a linear combination of $(1,1)$ and $(3,3)$.

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