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I have the following result:

The set $\{\mathbf{v}_1,...,\mathbf{v}_n\}$ is linearly dependent if and only if some $\mathbf{v}_i$ is a linear combination of its predecessors. That is, $\mathbf{v}_i \in$ span$\{\mathbf{v}_1,...,\mathbf{v}_{i-1}\}$ for some $i \in \{1,...,n\}$

I was curious what happens in the case where $\mathbf{v}_1$ is the zero vector and $\mathbf{v}_2,...,\mathbf{v}_n$ are linearly independent. This set is linearly dependent and I think you can say the second sentence in the above statement is true since the predecessors of $\mathbf{v}_1$ is the empty set and the span of the empty set is the zero vector. But I'm unsure about the first sentence. Can you have a linear combination of the empty set?

Bernard
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Henry Brown
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  • When you wrote “some $\mathbf v_1$”, I suppose that you meant “some $\mathbf v_i$”. – José Carlos Santos Dec 23 '20 at 08:44
  • The textbook you are using may (?) exclude the $0$ vector from being included. Otherwise, the book may include something about all possible permutations. – Clayton Dec 23 '20 at 08:45
  • @JoséCarlosSantos Also I think that. I think is a typo. –  Dec 23 '20 at 08:45
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    As given, the result is false. Either once must not include the zero vector as the beginning one, or else the wording mustbe changed. For example, for any non-zero vector $;v;$ in some linear spacve, the set $;{0,,v};$ is l.d., yet there is no vector which is a linear comb. of the preceeding one...unles, of course, one could take, for the first vector, the empty set....then, under this agreement, the result would be true. – DonAntonio Dec 23 '20 at 08:46
  • @Jose Carlos Santos, Yes, it was some v_i rather than v_1 – Henry Brown Dec 23 '20 at 08:58
  • @Clayton, the result is stated as given. The proof seems to ignore the case I've described (And DonAntonio describes). – Henry Brown Dec 23 '20 at 09:04

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