My Math Forum notation help..

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 October 31st, 2009, 04:52 PM #1 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 notation help.. My book says dim(W1+W2) and span(W1) + span (W2).. do they mean dim(W1 U W2) and span(W1) U span (W2) ? thanks a lot
 November 1st, 2009, 06:10 PM #2 Member   Joined: Oct 2009 Posts: 64 Thanks: 0 Re: notation help.. I think they really mean what they say. Take the union of two number lines, and you do NOT get a two-dimensional vector space. You need to have a notion of how to add elements from the different sets together, and every such sum must be an element in the new vector space.
 November 2nd, 2009, 06:37 PM #3 Newbie   Joined: Aug 2009 Posts: 16 Thanks: 0 Re: notation help.. yup, the above comment goes right. also, it is not uncommon to use the notation $A+B= \{a+b \qquad | \qquad a \in A, b \in B\}$ when A and B are subsets of a set with addition. those are used often and can be drastically different from the union A U B, so a notation to that comes in handy. but usually the authors defines that before using. hope that helps. cheers!
 November 3rd, 2009, 03:23 PM #4 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 Re: notation help.. thanks a lot guys, I assume that they are referring to the addition of vectors then.. but how does it make sense then? dim(W1 U W2) = dim(W1) + dim(W2) - dim(W1 intersect W2) makes sense but I'm not sure what to make out of dim(W1 + W2) = dim(W1) + dim(W2) - dim(W1 intersect W2)
 November 3rd, 2009, 04:23 PM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Generally speaking, $\mathcal{A}+\mathcal{B}=\big\{\ (A,B)\ |\ A\in\mathcal{A},\ B\in\mathcal{B}\ \big\}.$
November 3rd, 2009, 07:00 PM   #6
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Quote:
 Originally Posted by mattpi Generally speaking, $\mathcal{A}+\mathcal{B}=\big\{\ (A,B)\ |\ A\in\mathcal{A},\ B\in\mathcal{B}\ \big\}.$

thanks, then would the dimension of (A+B) "count" (a,b) where a and b are common to both A and B? I don't see why I wouldn't

 November 4th, 2009, 03:40 AM #7 Newbie   Joined: Aug 2009 Posts: 16 Thanks: 0 Re: notation help.. well my dear, when you say dimW1 we must assume W1 is a vector subspace, right? so, taking X is a big vector space, W1 and W2 are subspaces of X. dim(W1 U W2) will not make much sense, for it will not necessarily be a subspace. take for example X=R^3, W1 = { (x,0,0) , x real } and W2 = { (0,y,0), y real}. Then W1 U W2 will be two lines, not a subspace, and W1 + W2 will be the xy-plane subspace. what you must do is the following. take W1 and W2 and name V its intersection, V1 = W1-V, V2=W2-V. Then V, V1,V2 will be disjoint subspaces. Take their basis B,B1,B2, say. Then W1 + W2 will be the span of B U B1 U B2, and therefore have the dimension dimW1 + dimW2 - dim(W1 inter W2). You have to work out the proofs, but I think it will not be troublesome. Should you get in trouble, think of subspaces of R^n! cheers.
 November 4th, 2009, 12:03 PM #8 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 Re: notation help.. I think I might on the right track but I have some doubts.. I might have been thinking of W1 + W2 while thinking that it is W1 U W2, since I have successfully proven the theorems.. anyway, would Dim(W1 + W2) be the same as dim(w1) or dim(w2) except only adding the number of basis elements not present in the bases of dim(W1) or dim(W2)? for example, W1 has 4 elements, W2 has 5 elements so dim(W1+W2) has 5 elements??
 November 5th, 2009, 01:32 AM #9 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: notation help.. I hate to say it, but you seem really confused here... Ok. W1 U W2 is all vectors that are in W1 or in W2... This set may not be closed under addition. W1+W2 is the closure of this under addition (That might be fun to prove.) I'm guessing that you just assumed it was closed. Now, about dimension.... What is the dimension of a vector space? What I mean is: What does it count? Now, if you have two vector spaces, W1 and W2, which may overlap, and you want to find the dimension of their sum. max{ dim(W1), dim(W2)} is pretty obviously not the answer-- Take R^6, and take the subspaces W1={(a,b,c,0,0,0) | a,b,c in R} and W2={(0,0,0,x,y,z)|x,y,z in R}-- their sum is R^6, which has dimension 6, while each has dimension 3. What's happening here is the two vector spaces aren't overlapping at all, so we need all the basis elements. On the other hand, the dimension of the sum must be *at least* the dimension of W1 (and W2), since it must contain both. As an upper bound, we have: dim(W1)+dim(W2) as a good starting guess, but what happens if the two spaces overlap? You have common vectors-- so there should end up being common basis elements. How many should there be? (It is worth pointing out that the intersection of two vector spaces is a vector space, unlike the union.)
November 5th, 2009, 01:35 PM   #10
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Re: notation help..

Quote:
 Originally Posted by cknapp I hate to say it, but you seem really confused here... Ok. W1 U W2 is all vectors that are in W1 or in W2... This set may not be closed under addition. W1+W2 is the closure of this under addition (That might be fun to prove.) I'm guessing that you just assumed it was closed. Now, about dimension.... What is the dimension of a vector space? What I mean is: What does it count? Now, if you have two vector spaces, W1 and W2, which may overlap, and you want to find the dimension of their sum. max{ dim(W1), dim(W2)} is pretty obviously not the answer-- Take R^6, and take the subspaces W1={(a,b,c,0,0,0) | a,b,c in R} and W2={(0,0,0,x,y,z)|x,y,z in R}-- their sum is R^6, which has dimension 6, while each has dimension 3. What's happening here is the two vector spaces aren't overlapping at all, so we need all the basis elements. On the other hand, the dimension of the sum must be *at least* the dimension of W1 (and W2), since it must contain both. As an upper bound, we have: dim(W1)+dim(W2) as a good starting guess, but what happens if the two spaces overlap? You have common vectors-- so there should end up being common basis elements. How many should there be? (It is worth pointing out that the intersection of two vector spaces is a vector space, unlike the union.)
I think I see what's going on here, my intuitive initial idea was that the dim(W1+W2) should consider the dim(W1) + dim(W2) if there are no overlapping vectors, and the dim(W1) + dim(W2) if there are overlapping vectors but only considering 1 vector w1+w2 over 2 vectors w1, w2 where w1 = w2 .. so dim(W1) + dim(W2) - dim( W1 intersect W2). also, W1UW2 might not be a v.s. unless w1 is contained in w2 and w2 is contained in w1, so it makes no sense to say dim(W1 U W2)

so now that I think about it, it makes more sense to say dim(W1+W2) instead

thanks!

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