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To Prove:

If $S=[{v_1,v_2,...,v_k}]$ is a subset of vector space $V$. Then $span(S)$ is the smallest subspace of $V$ containing set $S$.

I know that $L[S]$ is a subspace of $V$. But in most arguments for proving the $L[S]$ is the smallest subspace containing $S$ , I find that if $W$ is another subspace of $V$ containing $S$ then, proving $S \subset W$ means $S$ is the smallest. I couldn't understand that if $S \subset W$ proves that $L[S]$ is the smallest containing $S$. Please elaborate.

Jyotishraj Thoudam
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  • See the related question https://math.stackexchange.com/questions/2142109/if-s-is-a-subset-of-a-vector-space-v-then-spans-equals-the-intersection?rq=1 from the handy list at the right. – amd Aug 17 '17 at 19:13

2 Answers2

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Theorem: Let $S$ be a subset of a vector space $V$. Then $\text{span}(S)$ is the smallest vector subspace of $V$ which contains $S$.

What kind of set $T$ would disprove the theorem? It would have to be (1) a vector subspace which (2) contains every vector in $S$, and which (3) is smaller than span(S) so that $T\subsetneq\text{span}(S)$.

Let's suppose $T$ has the first two qualifications. We'll show that it can't have the third qualification; that's enough to establish the proof. In particular, suppose:

  1. $T$ is not just a set; it's a vector subspace. This means that $T$ is closed under linear combinations. (every linear combination of vectors in $T$ also belongs to $T$.)

  2. $T$ contains every member of $S$. ($T \supseteq S$).

By the first two properties, $T$ contains every vector in $S$ and $T$ is closed under linear combinations, so it follows that $T$ contains every linear combination of vectors in $S$. But $\text{span}(S)$ is by definition the set of all linear combinations of vectors in $S$. Therefore $T\supseteq \text{span}(S)$. Therefore $T$ is at least as large as $\text{span}(S)$, so the theorem is true.

user326210
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Here is your statement written slightly different. Maybe that will help.

Theorem: If $W\subset V$ is a subspace, such that $v_1,…,v_k∈W,$ then $\text{span}(v_1, …,v_k)\subset W$. Proof: Since $v_1,…,v_k∈W$ and $W$ is a subspace all linear combinations $$α_1v_1 + … + α_kv_k∈W$$ Since $\text{span}(v_1,…,v_k)$ contains (only) these linear combination it follows $\text{span}(v_1,…,v_k)\subset W$.

Written in words that theorem states, that any subspace $W$, that exists and contains $v_1,…,v_n$, also contains $\text{span}(v_1,…,v_k)$. Hence $W$ is larger (or equal) to $\text{span}(v_1,…,v_k)$, since it contains $\text{span}(v_1,…,v_k)$ and could contain some more elements.

And since any other subspace $W$ is larger (or equal) it follows that $\text{span}(v_1,…,v_k)$ the smallest subspace.

That is like saying: Any number in $ℕ∪\{0\}$ is larger (or equal) to $0$, hence $0$ is the smallest number in $ℕ∪\{0\}$.

Or even closer to the original problem: Any subset $M⊂(ℕ∪\{0\})$ with $0∈M$ is larger or equal to $\{0\}$, hence $\{0\}$ has to be the smallest subset of $ℕ∪\{0\}$ that contains $0$.

P. Siehr
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  • $L[S]$ is just span of S – Jyotishraj Thoudam Aug 17 '17 at 08:01
  • Ok, thanks. Never used that notation before. I edited my answer a bit, to be more clear. At least I hope so. – P. Siehr Aug 17 '17 at 08:08
  • How do you $W$ is larger? – Jyotishraj Thoudam Aug 17 '17 at 08:19
  • How do you know? $W$ is an arbitrarily chosen subspace that contains $\text{span}(v_1,…,v_k)$ - that is what the Theorem states, right? So it is at least as large as $\text{span}(v_1,…,v_k)$. At least as large meaning that $W$ could possibly contain more elements. Here is a small example. Take $V=ℝ^3$, $k=1$, $v_1=[1,0,0]^{\top}$. One example for $W$ is $$W=\text{span}([1,0,0]^{\top},[0,1,0]^{\top})$$. That space contains $\text{span}(v_1)$ and is larger than $\text{span}(v_1)$, right? – P. Siehr Aug 17 '17 at 08:25