# Can a vector space have multiple spanning sets?

Maybe this is obvious, but can a vector space have multiple spanning sets or is there only a single spanning set for every vector space?

Thanks

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Of course there are multiple sets.

Even for just one-dimensional vector spaces, at least over fields with more than two elements, every non-zero scalar is a spanning set.

If $\{v_1,\ldots,v_k\}$ spans $V$ then $\{v_1+v_2,v_2,\ldots,v_k\}$ is a different spanning set, and one can replace more vectors by others, or by more complicated expressions.

In fact if $V$ is spanned by $v_1,\ldots, v_k$ and $v\in V$ is a vector such that $v=\sum_{i=1}^k\alpha_i v_i$, and $\alpha_i\neq 0$ then we can show that $\{v_1,\ldots,v_{i-1},v,v_{i+1},\ldots,v_k\}$ is a spanning set.

Furthermore! If we only require spanning the space, and not being linearly independent then there are even more spanning sets than that. Anything which contains a basis, really.

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OK yeah that makes sense. I think I was just confused about the definition of a spanning set in general but I'm kind of figuring it out. – user58437 Jan 19 '13 at 21:48

If $S$ spans $V$, so does $\{\alpha v:v\in S\}$ for every non-zero scalar $\alpha$. Unless $V$ is over the two-element field, this automatically gives you more than one spanning set. More trivially, both $V$ and $V\setminus\{\vec 0\}$ span $V$.

Less trivially, if $\dim V\ge 2$, let $u$ and $v$ be linearly independent vectors in $V$. Extend $\{u,v\}$ to a basis $B$ for $V$ containing $u$ and $v$. Then $B\cup\{u+v\}$ is a spanning set different from $B$.

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What if it is a vector space over a finite field with two elements only? – Git Gud Jan 19 '13 at 21:35
@GitGud: If it’s one-dimensional, it’s essentially the field, and the spanning sets are $V$ and $V\setminus\{\vec 0\}$. If its dimension is at least $2$, see the addition to my answer. – Brian M. Scott Jan 19 '13 at 21:39
Even then it has more that one spanning set if its dimension is greater than one, since if $\,\{v\,,\,u\}\,$ is a spanning set , then so is the set $\,\{v+u\,,\,u\}\,$ (proof?) – DonAntonio Jan 19 '13 at 21:40
Right. Thank you, both. – Git Gud Jan 19 '13 at 21:42
@GitGud: You’re welcome. – Brian M. Scott Jan 19 '13 at 21:42

If you have any spanning set $S$ that is proper in the vector space $V$, $S+\{v\}$ is still a spanning set for any $v\in V\setminus S$.

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Nice and neat answer. Short one. +1 – Babak S. Jan 27 '13 at 14:16

If $V$ is a vector space then I think that we can say that: $$\{B \cup X / X \subset V \, \text{and} \, B \,\text{a basis of} \, V \}$$ is the set of all apanning sets of $V$.

In auther words we obtains a spanning set of $V$ bay taking a basis of $V$ or by adding elments to a basis of $V$.

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