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$S$ and $T$ are subspaces of $\mathbb{R}^{n}$ and is defined as $S+T = \{v+w \mid v \in S \; and \; w \in T\}$. I need to show that $S+T$ is a subspace of $\mathbb{R}^{n}$.

Instinctively, $S+T$ is definitely inside $\mathbb{R}^{n}$ since $S \in \mathbb{R}^{n}$ and $T \in \mathbb{R}^{n}$. So the sum of any vectors in $S$ and $T$, although may not be a vector in both $S$ and $T$, ie: not inside $S \cap T$, it is still inside $\mathbb{R}^{n}$. But this is just my intuition and I want to prove it formally.

I thought I could use the subspace criteria $cv+dw$ to prove it.

Since $v \in S$ and $w \in T$, $cv \in S$ and $dw \in T$ where $c,d \in \mathbb{R}$.Then $cv+dw \in S+T \in \mathbb{R}^{n}$.

But somehow, I find what I've done isn't a very precise and convincing proof. How should I prove this more precisely?

Update of my attempt in the proof (Is this right?)

Let $\vec{v},\vec{w} \in S+T$, then $\vec{v}=\vec{s_1} + \vec{t_1}$ and $\vec{w}=\vec{s_2} + \vec{t_2}$ where $\vec{s_i} \in S, \; \vec{t_i} \in T$.

This implies that $\vec{v} = \vec{s_1}+\vec{t_1} \in S+T$. Let $c, d,r \in \mathbb{R}$, then $r\vec{v}=c\vec{s_1}+d\vec{t_1}$. Since $\vec{s_1}+\vec{t_1} \in S+T$, $c\vec{s_1}+d\vec{t_1} \in S+T \Rightarrow r\vec{v} \in S+T $.

Similarly, $\vec{w} = \vec{s_2}+\vec{t_2} \in S+T$. Let $j,k,s \in \mathbb{R}$, then $s\vec{w}=j\vec{s_2}+k\vec{t_2}$. Since $\vec{s_2}+\vec{t_2} \in S+T$, $j\vec{s_2}+k\vec{t_2} \in S+T \Rightarrow s\vec{w} \in S+T$.

Since $r\vec{v}, s\vec{w} \in S+T$, $r\vec{v}+s\vec{w} \in S+T$, hence $S+T$ is a subspace.

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    For example, to show any linear combination of elements of $S+T$ is in $S+T$, let $v_1$ be one element of $S+T$, $v_2$ another. Then $v_1=s_1+t_1$ for some $s_1\in S$, $t_1\in T$. Also, $v_2=s_2+t_2$, similar conditions. Then $pv_1+qv_2=p(s_1+t_1)+q(s_2+t_2)=(ps_1+qs_2)+(pt_1+qt_2)$, done. This is essentially, in different words, what William Chan wrote. Except he went back to the definitions, it took a couple more steps, his way is better.2011-09-24

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Clearly, $0 \in S + T$ since $0 \in S$ and $0 \in T$ and $0 + 0 = 0$.

Suppose $a \in S + T$ and $b \in S + T$. Then a = s' + t' and b = s'' + t'' where s', s'' \in S and t', t'' \in T. a + b = s' + t' + s'' + t'' = (s' + s'') + (t' + t'') by commutativity of addition. s' + s'' \in S and t' + t'' \in T since $S$ and $T$ are subspaces. Thus $a + b \in S + T$.

Finally, suppose $a \in S + T$. So a = s' + t' as above. Let $c$ be some scalar. Then ca = c(s' + t') = cs' + ct' and cs' \in S and ct' \in T since $S$ and $T$ are subspaces. Thus $ca \in S + T$.

This proves that $S + T$ is a subspace.

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    oh..so looks like I cannot simply use the combined $cv+dw$. Instead, I have to split them up to prove $v+w$ first and then prove the closure under multiplication with $c(v+w)$?2011-09-24
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It's obvious that $S+T\subset \mathbb R^n$, that's not the problem. What you need to show is that it is a subspace (not just a subset). This means you need to show that if $v,w\in S+T$, then $rv+sw\in S+T$, where $r,s$ are scalars.

So let $v,w\in S+T$. Then $v=s_1+t_1$, $w=s_2+t_2$ where $s_i\in S$ and $t_i\in T$. Now how do you finish?

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    @xEnOn: Your new proof isn't very convincing. William Chan gave nice details.2011-09-24