# Is this proof of a subspace property correct?

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?

Tags :

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

Fred
July 12, 2019 08:12 AM

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 $$L(S)\subseteq T$$

Federico Fallucca
July 12, 2019 08:12 AM

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.

Wuestenfux
July 12, 2019 08:16 AM

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