by John Arg
Last Updated July 12, 2019 08:20 AM

The question states:

Prove that if $S\subseteq T \subseteq V$ and if $T$ is a subspace of $V$, then $L(S)\subseteq T$.

My proof is as follows:

All elements in $S$ are in $T$. Now $T$ is a subspace, $\therefore$ all elements in $S$ spanning $L(S)$ are independent elements in $T$.

$\therefore T$ is spanned by $S$ if $\dim T = n$ and there are $n$ elements in $S$; or $T$ is spanned by a set $Q$ containing all independent elements in $S$.

In both cases, the independent elements in $S$ (which span $L(S)$) form part of a basis for $T$.

$\therefore L(S) \subseteq T$.

Is it correct?

Since $ S \subseteq T$ , we have $L(S) \subseteq L(T)$. Since $T$ is a supspace of $V$, we get $T=L(T).$ Hence $L(S) \subseteq T.$

Let $v\in L(S)$ then there exists $v_1,\dots , v_n\in S$ and $a_1,\dots , a_n\in \mathbb{R}$ such that

$v=\sum_{i=1}^n a_iv_i$

But $v_i\in S\subseteq T$ and $T$ is a subspace so each linear combination of element of $T$ is in $T$ then

$v\in T$ so $L(S)\subseteq T$

First, the span operator is monotonous w.r.t. set inclusion. That is, if $S,T$ are subsets of a $K$-vector space $V$, then $$S\subseteq T\Rightarrow L(S)\subseteq L(T).$$

Second, the span operator is idempotent. That is, if $S$ is a subset of a $K$-vector space $V$, then $$L(L(S)) =L(S).$$

Both properties suffice to prove your claim.

Updated September 22, 2018 13:20 PM

Updated September 26, 2018 12:20 PM

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