# Proving $S+T$ is a subspace of $\mathbb{R}^{n}$

$S$ 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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You want a formal proof that sum of two vectors, one from $S$ and the other from $T$, lies in $\mathbb R^n$? Note that the vectors lie in $\mathbb R^n$ as well (since $S,T \subseteq \mathbb R^n$) and $\mathbb R^n$, being a vector space, is closed under addition of vectors. (And saying $S \in \mathbb R^n$ and $T \in \mathbb R^n$ is plain wrong. You want to use $S \subseteq \mathbb R^n$ instead.) – Srivatsan Sep 24 '11 at 15:20
Use the criteria is a good idea, but you have to check that for $v$ and $w\in S+T$, not only for $v\in S$ and $w\in T$. If $v\in S+T$, then we can write $v=v_1+v_2$ with $v_1\in S$ and $v_2\in T$, and now continue. – Davide Giraudo Sep 24 '11 at 15:21
You do not always use mathematical language correctly. For example, early on, you write $S\in \mathbb{R}^n$. But $S$ is not an element of $\mathbb{R}^n$, it is a subset of $\mathbb{R}^n$. Remember, vector spaces are sets. – André Nicolas Sep 24 '11 at 15:25
@André In fact, not just early on. The OP is consistent in using $S \in \mathbb R^n$, but of course, consistency does not mean correct. – Srivatsan Sep 24 '11 at 15:31
@xEnOn: Most proofs that you will be asked to do, at least early on in a subject, basically write themselves. If you understand and use the language correctly, what needs to be done usually comes straight from the definitions. The actual verification details are usually straightforward. – André Nicolas Sep 24 '11 at 15:46

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 $c(v+w)$? – xenon Sep 24 '11 at 16:14

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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I tried to continue from here. I updated my question with my attempt. Did I complete the proof correctly? – xenon Sep 24 '11 at 15:56
@xEnOn: Your new proof isn't very convincing. William Chan gave nice details. – Grumpy Parsnip Sep 24 '11 at 17:36