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Prove that if $S=\{v_1, v_2, v_3\}$ is a linearly dependent set of vectors in a vector space $V$, and $v_4$ is any vector in $V$ that is not in $S$, then $\{v_1, v_2, v_3, v_4\}$ is also linearly dependent.

If the vectors in $S$ are linearly dependent, each vector in $S$ can be written as linear combinations of the other vectors. If add a vector $v_4$, and multiply this vector by a scalar $k=0$, then the vectors $\{v_1,v_2,v_3\}$ can still be written as linear combinations of each other.

Is this a proof?

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A set of vectors is linearly dependent if and only if there is a nontrivial linear combination of the vectors in the set which is zero. Nontrivial here means that there is at least one nonzero coefficient in the linear combination.

It should be immediately obvious then that you can append ANY set of vectors, finite or infinite, even uncountable, to a linearly dependent set and the result is still a linearly dependent set. That's because once you have a nontrivial linear combination summing to zero, you can throw any additional vectors into that sum with all additional coefficients identically zero; this means that (1) the sum is still zero, and (2) the linear combination is still nontrivial.

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Linear dependence does not say that all vectors in the set can be written as a non-trivial linear combination of the other vectors in the set. Consider $\{ (1, 0, 0), (2, 0, 0), (0, 1, 0) \}$. Notice the first two vectors are linearly dependent, which implies the set is linearly dependent. However, $(0, 1, 0)$ is independent of the other two vectors.

So if $S$ is linearly dependent, let $v_{1} \in S$ be a dependent vector such that $v_{1} = \sum_{i=2}^{n} a_{i} v_{i}$. So if we add $v_{n+1}$ to $S$, we can still write $v_{1} = \sum_{i=2}^{n} a_{i} v_{i} + 0v_{n+1}$. Hence, $S \cup \{v_{n+1}\}$ is still linearly dependent. So your idea is right, but you should finesse the verbage some to make it more exacting.

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  • $\begingroup$ Can the Plus/Minus Theorem be used here? $\endgroup$ – Al Jebr Apr 23 '14 at 2:54
  • $\begingroup$ So, we can add a vector and multiply that vector by $0$ to say it is still linearly dependent? $\endgroup$ – Al Jebr Apr 23 '14 at 2:58
  • $\begingroup$ To show linear dependence of the set $S$, we pick a vector $v_{1}$ and exhibit a non-trivial linear combination of vectors $v_{2}, ..., v_{n}$ that form $v_{1}$. Taking a new vector $v_{n+1}$ does not change the fact that we have such a non-trivial linear combination. Hence, linear dependence holds. Note that the set of vectors is said to be linearly (in)dependent, not the linear combination of vectors. $\endgroup$ – ml0105 Apr 23 '14 at 3:59
  • $\begingroup$ I did not understand your last sentence. Can you rephrase? $\endgroup$ – Al Jebr Apr 23 '14 at 4:18
  • $\begingroup$ As an example, let $S = \{ (1, 0, 0), (0, 1, 0), (0, 0, 1)\}$. We talk about $S$ (or the vectors in $S$) being linearly (in)dependent. Linear combinations formed from vectors in $S$ are used to test for independence, but the linear combinations themselves are not (in)dependent. Note that in this case, the vectors in $S$ are linearly independent. $\endgroup$ – ml0105 Apr 23 '14 at 4:21

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