Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$S$ 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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer 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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.