Proof verification: linear span is the smallest subspace containing vectors

linear algebrasolution-verification

I've already read several answers to this very same question. Although I understand the proof, I came up with one slightly different (and shorter I think) for the smallest subspace part, and I'd like to be sure it has no gaps. Since I want to focus on the smallest subspace part, I left out the proof that the span is a subspace (which is not very difficult).

Proposition. The linear span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.

Proof. Suppose $v_1,\dots,v_m$ is a list of vectors in $V$. Let $S=\text{span}(v_1,\ldots,v_m)$.

[proof that span is subspace left out]

Let $T$ be an arbitrary subspace of $V$. Suppose $v_1,\ldots,v_m \in T$. To show $S \subseteq T$, take some vector $w \in S$. Then, there exists scalars $\alpha_1,\ldots,\alpha_m$ such that
$$w=\alpha_1v_1+\ldots+\alpha_mv_m.$$
Setting a $\alpha_j=1$ and $\alpha_i=0$ for all $i \neq j$, we have that $w=v_j$. Since $T$ contains $v_j$, then $w \in T$. Therefore, $S \subseteq T$, as desired. QED

I read that some people prove that $S$ contains $v_1,\ldots,v_m$ and that it is smallest separately. For the former, the argument is the one I used above, whereas for the latter they use the fact that $T$ is closed under addition and scalar multiplication, and, therefore, $T$ contains all linear combinations of the list of vectors. Because of this, I felt I might have a flaw in my proof.

Does my proof have any gaps? Does the way I wrote prove $S$ contains the vectors AND is the smallest subspace containing these?

Best Answer

Your proof is incorrect. In order to show that $S\subseteq T$ you need to prove that for any $w\in S$ you have $w\in T$. So you cannot set $v_j=1$ and $v_i=0$. It does not make sense, and you cannot have any constraint on the $\alpha_i$'s. (I guess this is what you meant)

Here you basically said : Assume that $v_i$ is in T, then $v_i$ is in T. That's it.

Edit : the correct argument goes as:

  • Assume $w\in S$.
  • By definition of $S$ as linear span, there exist $\alpha_i$'s such that $w=\alpha_1v_1+\ldots, \alpha_mv_m$.
  • By Assumption, $T$ contains all $v_i$'s.
  • As $T$ is a vector space, it is closed under vector addition and scalar multiplication, i.e. $\alpha_1v_1+\ldots, \alpha_mv_m \in T$
  • Therefore $w\in T$ and $S\subseteq T$

As $T$ is the smallest subspace of $V$ containing all $v_i$'s, and $S$ is a subspace of $V$ containing all $v_i$'s with $S\subseteq T$, then $T=S$.

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