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I'm trying to understand how to complete this proof about subspaces. I understand the basics about the definition of a subspace (i.e. the zero matrix must exist, and addition and multiplication must be closed within the subspace). But I'm confused as to how to show that the addition of two elements from completely different sets somehow are preserved under the same subspace.

I'm pretty sure the zero vector exists because the zero vector is within C and D, but I'm unsure about the other two conditions. The complete problem is listed below.

Problem: Let W be a vector space and C,D be two subspaces of W. Prove or disprove that { a + b | a $\in$ C, b $\in$ D} is also a subspace of W.

Any help would be appreciated.

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2 Answers

up vote 2 down vote accepted

Hint: For $a_1,a_2\in C,b_1,b_2\in D,\alpha\in F$, where $F$ is the field, $$(a_1+b_1)+(a_2+b_2)=(a_1+a_2)+(b_1+b_2)$$ and $$\alpha (a_1+b_1)=(\alpha a_1)+(\alpha b_1)$$

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So can we just say that since ($a_{1}$ + $a_{2}$) is preserved within the subspace, by adding ($b_{1}$ + $b_{2}$), which is also preserved, and rearranging terms you prove that the addition of elements is also within the subspace? If so I guess that makes sense. –  Math_Illiterate Dec 7 '12 at 12:16
    
@Ockham: Yes, you are correct. Only one reminder: rearranging is permitted due to commutativity and associativity; try to see how. –  pritam Dec 7 '12 at 12:20
    
Is it because C and D are subspaces which means that the operations available in vector spaces is also available (eg commutative, associative properties). Can we make that claim even though the elements are from different sets? –  Math_Illiterate Dec 7 '12 at 12:23
    
@Ockham: Yes, why not ? use commutativity and associativity of the parent vector space, note that the elements belong to the parent vector space as well. –  pritam Dec 7 '12 at 12:28
    
Ok that makes sense, thanks –  Math_Illiterate Dec 7 '12 at 12:29
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You have to show (among other things) that $Z=\{{\,a+b:a\in C,b\in D\,\}}$ is closed under addition. So, let $x$ and $y$ be in $Z$; you have to show $x+y$ is in $Z$. So, what is $x$? Well, it's in $Z$, so $x=a+b$ for some $a$ in $C$ and some $b$ in $D$. What's $y$? Well, it's also in $Z$, so $y=r+s$ for some $r$ in $C$ and some $s$ in $D$. Now what's $x+y$? Can you take it from here?

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