# The Unapologetic Mathematician

## Spanning sets

Today I just want to point out a dual proposition to the one I refined last week. At that time we stated that any linearly independent set can be expanded to a basis. This followed from the fact that a basis is a maximal linearly independent set. But a basis is also a minimal spanning set, and that will lead to our flipped result.

First, a basis must be a spanning set, and if we remove any vector it can no longer be a spanning set. For if the remaining vectors spanned, then the removed vector could be written as a finite linear combination of them, and this would contradict the linear independence of the basis.

On the other hand, if a spanning set $\left\{e_i\right\}$ is minimal it must be linearly independent. For if it were not, we could write out a nontrivial finite linear combination $0=\sum r^ie_i$

For some index — say $i_0$ — we have $r^{i_0}\neq0$. Then we can rearrange this to write $e_{i_0}=\frac{1}{r^{i_0}}\sum\limits_{i\neq i_0}r^ie_i$

and so we may omit the vector $e_{i_0}$ from the set and the remaining vectors still span. This contradicts the assumed minimality of the original spanning set.

So a basis is a minimal spanning set. This leads us to the proposition that any spanning set may be narrowed to a basis. And the proof is again the same technique we used to show that every vector space has a basis. It’s just that this time we flip the whole thing over. Now we consider the set of subsets of our spanning set which span the vector space, and we use Zorn’s lemma to show that this must contain a minimal spanning set. This will then be a basis contained in our original spanning set.

Of course, as usual, in the finite-dimensional case we don’t need Zorn’s lemma, so the squeamish can relax in that case.

June 30, 2008 - Posted by | Algebra, Linear Algebra

## 5 Comments »

1. […] Any vector in the image of can be written as for some vector . If we pick a basis latex V\$, then we can write . Thus the vectors span the image of . And thus they contain a basis for the image. […]

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2. I am afraid, we cannot “flip the whole thing over” and use Zorn’s lemma for the spanning subsets, since we cannot assert that every chain is bounded from below. Indeed, the chain of the sets {n,n+1,…} (n=1,2,…) consists of spanning subsets of R, but does not posess a lower bound (its intersection is empty). Comment by Alexander E. Gutman | September 11, 2009 | Reply

3. That seems to be right. Luckily, the finite-dimensional case still holds fine, and that’s all that we need for most of the sequel. Comment by John Armstrong | September 11, 2009 | Reply

4. […] let’s say is some finite collection of vectors which span (it doesn’t matter if they’re linearly independent or not). Let be a linear […]

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5. […] so we don’t really want zero to be in our collection . We also may as well assume that spans , because it certainly spans some subspace of and anything that happens off of this subspace is […]

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