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My linear algebra notes state the following lemma: If $(v_1, ...,v_m)$ is linearly dependent in $V$ and $v_1 \neq 0$ then there exists $j \in \{2,...,m\}$ such that $v_j \in span(v_1,...,v_{j-1})$ where $(...)$ denotes an ordered list.

But if at least one $v_i$ is $\neq 0$ then the list can be reordered and the lemma applied. Is $v_1 \neq 0$ just another way of saying $v_i \neq 0$ for at least one $i$?

newb
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3 Answers3

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Your book is correct, but silly. It should not have excluded $v_1=0$, but allowed $j=1$ instead. By convention $\operatorname{span}()=\{0\}$ (it is important that the span of every set of vectors is a subspace, so the empty set should give the null subspace), and $v_1=0$ (which all by itself makes the set linearly dependent) would not be an exception, because one can then take $j=1$ (and if there is no linear dependence among the remaining vectors, one has to take $j=1$). So it is just another case of unfounded fear of the void.

The result states (or should) that given an ordered sequence of linearly dependent vectors, there is always one of them that is in the span of set of vectors preceding it. This is always true. Indeed, you can always take the this vector to be the first one to make the sequence-up-to-there linearly dependent. The empty sequence is always independent, and a sequence with one vector is linearly dependent only if that vector is zero, in which case it is in the (zero-dimensional) span of the (empty) set of preceding vectors. If the first is not zero but after some independent vectors a zero vector comes along, then it also in the span of the set of preceding vectors, but that span now has positive dimension. Of course a nonzero vector in the same span, in place of that zero vector, would also have made the sequence dependent; this is in fact the more common case.

Marc van Leeuwen
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  • But why is it important that the vectors are ordered? Why not state and prove the following lemma instead: If $v_1,...,v_n$ are linearly dependent then there is $k $ with the property that $v_k$ can be expressed as linear combination of the remaining vectors and removing $v_k$ leaves the linear span unchanged? – newb Aug 07 '13 at 16:27
  • @newb: That is a perfectly valid but weaker statement. There are cases where expressing a vector using later vectors in the list is not useful enough. For instance in the incomplete basis theorem, from a list $I$ of independent vectors followed by a list $S$ of vectors that (by themselves) span the vector space, construct a basis containing$~I$. Only throwing out vectors dependent on the previous vectors will ensure that elements of $I$ remain, though they have linear dependencies; using your lemma would not ensure this. – Marc van Leeuwen Aug 07 '13 at 16:49
  • But couldn't I prove the incomplete basis theorem using my lemma as follows: Start with the sets $D$ empty and $S$ containing the incomplete basis. Then while $S$ does not contain $dim(V)$ many linearly independent vectors proceed as follows: pick $v \in V \setminus (D \cup S)$. If ${v}$ union $S$ is linearly independent add $v$ to $S$. If not add $v$ to the set $D$ ("discarded vectors"). Now argue why this algorithm terminates and produces a basis for $V$. – newb Aug 08 '13 at 07:44
  • @newb: That's all right (if you start adding the independent vectors $I$ they will not be discarded), but doesn't really use your lemma. Instead you use that if $I$ is independent but $I\cup{v}$ is dependent, then necessarily $v\in span(I)$, which is precisely the basic ingredient of "my" (or better your books) lemma. I might add your lemma is fairly trivial: a linear dependence relation by definition involves a nonzero coefficient, and by singling out the term containing it and dividing by the coefficient one gets the expression. While not hard either, the other lemma needs some more work. – Marc van Leeuwen Aug 08 '13 at 07:58
  • Thank you very much, I understand it now. In particular, the proof and the dependence lemma here can be altered so as to allow $v_1=0$ and $j=1$. The proof remains the same (except deleting the mention of $v_1 \neq 0$. – newb Aug 08 '13 at 08:15
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$v_1\ne0$ simply means that $v_1\ne 0$. All other $v_j$ may be zero. Note that $v_1$ does play a special role insofar as it is the only of the given vectors that is definitely in all $\operatorname{span}(v_1,\ldots, v_{j-1})$ of the claim. Note that the claim as stated fails if $v_1=0$ and $v_2,\ldots, v_m$ are linearly independant.

You can apply the theorem to different permutations of the $v_i$. But note that the claim differs if you permute the vectors! What you get is that some vector is a linear combination of the others (and that holds even if all vectors are zero). The theorem at hand is concerned with a given ordered seqeuence of vectors and want a vector to be a linear combination of preceeding vectors.

Hagen von Eitzen
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You're essentially correct. But it's another way of saying $v_i\neq 0$ for some $v_{i}$, not all $v_{i}$.

Alex Lapanowski
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