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Is the given Proof Correct?

Theorem. Given a vector space $V$ over a scalar field $\mathbf{F}$ and a list of vectors $v_1,v_2,...,v_{m-1},v_m$ The span of a list of vectors is the smallest subspace to contain all the vectors in the above list.

Proof. For the purpose of this proof we use $\mathcal{S}(U)$ to represent the proposition that $U$ is a subspace of $V$. Consider now the following sets $$W=\left\{U\subseteq V\middle|\mathcal{S}(U)\ and\ v_1,v_2,...,v_{m-1},v_m\in U\right\}$$ $$span(v_1,v_2,...,v_{m-1},v_m)=\left\{\sum_{i=1}^{m}a_iv_i\middle|\forall i\in\{1,2,...,m-1,m\}(a_i\in\mathbf{F})\right\}$$ Consider now the following Lemmas.

Lemma $(1)$. $span(v_1,v_2,...,v_{m-1},v_m)\in W$.

Proof. Let $\sum_{i=1}^{m}a_iv_i$ and $\sum_{i=1}^{m}b_iv_i$ be arbitrary members of $span(v_1,v_2,...,v_{m-1},v_m)$, evidently $\sum_{i=1}^{m}(a_i+b_i)v_i\in\ span(v_1,v_2,...,v_{m-1},v_m)$ implying that $span(v_1,v_2,...,v_{m-1},v_m)$ is closed under addition.

Consider now an arbitrary $\lambda\in\mathbf{F}$ and $\sum_{i=1}^{m}c_iv_i\in span(v_1,v_2,...,v_{m-1},v_m)$ evidently $\lambda\sum_{i=1}^{m}c_iv_i=\sum_{i=1}^{m}\lambda c_iv_i\in span(v_1,v_2,...,v_{m-1},v_m)$ implying that $span(v_1,v_2,...,v_{m-1},v_m)$ is closed under scalar multiplication.

Consider now the linear combination $\sum_{i=1}^{m}d_iv_i$ where $\forall i\in\{1,2,...,m-1,m\}(d_i=0)$ consequently $\sum_{i=1}^{m}d_iv_i=\sum_{i=1}^{m}0v_i=0$ since $0v_i=0$.

In a similar fashion we can see that $\forall i\in\{1,2,...,m-1,m\}(v_i\in\ span(v_1,v_2,...,v_{m-1},v_{m}))$.

Lemma $(2)$. $\forall U\in W(span(v_1,v_2,...,v_{m-1},v_m)\subseteq U)$.

Proof. Consider now an arbitrary $U\in W$ and assume that $\sum_{i=1}^{m}a_iv_i\in span(v_1,v_2,...,v_{m-1},v_{m})$ since $v_1,v_2,...,v_{m-1},v_m\in U$ closure under scalar multiplication implies that $a_1v_1,a_2v_2,...,a_{m-1}v_{m-1},a_{m}v_m\in U$ and since $U$ is closed under addition it follows that $\sum_{i=1}^{m}a_iv_i\in U$, consequently $span(v_1,v_2,...,v_{m-1},v_{m})\subseteq U$.

Taking the above two lemmas together we have the required result.

$\blacksquare$

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  • $\begingroup$ This is perfect. $\endgroup$ – ervx Jul 23 '17 at 16:54
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Your proof is indeed correct. I will just make a small amount of suggestions:

  1. There is no need to introduce the concept of $\mathcal{S}(U)$. Simply writing $U\leqslant V$ means that $U$ is a vector subspace of $V$.
  2. There was no need, within the proof of lemma 1, of proving that $0\in\operatorname{span}(v_1,\ldots,v_m)$. When you have a subset $U$ of a vector space which is closed under addition and multiplication by a scalar, the only thing that is still needed in order to prove that $U$ is a subspace is that $U\neq\emptyset$. But you proved right after that $v_1,\ldots,v_n$ belong to your space. Therefore, it cannot be empty.
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  • $\begingroup$ Is the reason behind the second point the fact that $0v=0$ is implied by closure under scalar multiplication and is therefore an equivalent condition to that of the existence of the $0$ vector. $\endgroup$ – Atif Farooq Jul 23 '17 at 17:07
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    $\begingroup$ @AtifFarooq I know. But if $U\neq\emptyset$, then the existence of $0$ is automatic. indeed, if $v\in U$, then $0=v+(-1)v\in U$. $\endgroup$ – José Carlos Santos Jul 23 '17 at 17:10

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