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Vector Spaces and Dimensions

  • Thread starter RVP91
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1. Homework Statement



2. Homework Equations
From my notes I'm aware of the following equation: dim(A + B) = dimA + dimB − dim(A ∩ B).


3. The Attempt at a Solution
I'm assuming part of the solution involves the equation above and rearranging it but I'm not sure how I would determine dim(A + B). I also know A + B := <A ∪ B> = = {a + b: a ∈ A, b ∈ B} but I can't see where to go.

Thanks in advance!
 
Last edited:

HallsofIvy

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Since dim(B)> dim(A), it is not possible for B to be a subset of A but consider the situation of A being a subset of B. That will give the largest possible size of [itex]A\subset B[/itex]. To find the smallest possible size think about how separate A and B can be. For example if U had only dimension 9, all of A would have to be in B! But if U had dimension 17= 8+ 9, A and be could be "completely" separate, having only the 0 vector in common.
 
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I'm not too sure where to go from what you said.

Following from having only the 0 vector in common this would mean A and B would only contain one element and so have dimension 1?

Also is the equation I stated relevant because then would I also need to work out A + B and thus dim(A + B).

I'm sure what you said probably is a big hint but I can't seem to figure out where to go, could you please provide more help?

Thanks
 

HallsofIvy

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No, that was not my point! A and B cannot have "only the zero vector in common" since then the dimension of their direct sum would be 8+ 9= 17 which is impossible in only 10 dimensions. That's the key idea! As you say, [itex]dim(A+ B)= dim(A)+ dim(B)- dim(A\cap B)[/itex]. Here dim(A)+ dim(B)= 8+ 9= 17 while dim(A+ B) cannot be larger than 10. So, at most, [itex]10= 8+ 9- dim(A\cap B)[/itex].

(The dimension of the "space" containing only the zero vector is 0, not 1, by the way.)
 
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Oh right I see. So dim(A∩B) must be at least 7 and at most 17?
 

HallsofIvy

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No, that's not at all right! [itex]A\cap B[/itex] is a subset of A so it can't possibly have dimension greater than the dimension of A.
 
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Apologies for causing any frustration but after having a rethink I have now arrived at the following(which could be wrong again),

A+B is a subset of V, therefore,
dim(A+B) <= dim(V) = 10, therefore,
dim(A+B) <= 10.

Then, dim(A+B) = dim(A) + dim(B) - dim(A∩B) = 17-dim(A∩B) <= 10
and so, dim(A∩B) >= 7.

But A∩B is a subset of A and a subset of B, thus
dim(A∩B) <= dim(A) = 8
and dim(A∩B) <= dim(B) =9.

Therefore, dim(A∩B) <= 8.

Giving, dim(A∩B) = 7 or 8?
 
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Could someone check what I wrote above please?

Thanks.
 
21,992
3,272
What you wrote is correct.

But please wait before bumping your thread after 24 hours have passed, thank you.
 
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Okay thanks and apologies.

Could anyone help me out with the part concerning "Give an example illustrating each
possible value." What are they actually asking for?
 
21,992
3,272
They want you to give specific examples of V, A, B such that dim(A)=8, dim(B)=9 and [itex]dim(A\cap B)[/itex] is 8 or 9.

I suggest taking [itex]V=\mathbb{R}^{10}[/itex], then dim(V)=10 as required. Now, how would you pick A and B?
 
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I'm not totally sure but could you take A = {(a,b,c,d,e,f,g,h,0,0) | a,b,c,d,e,f,g,h are Real} and B = {(a,b,c,d,e,f,g,h,i,0) |a,b,c,d,e,f,g,h,i are Real}?

Somehow I think that's wrong though :s
 
21,992
3,272
I'm not totally sure but could you take A = {(a,b,c,d,e,f,g,h,0,0) | a,b,c,d,e,f,g,h are Real} and B = {(a,b,c,d,e,f,g,h,i,0) |a,b,c,d,e,f,g,h,i are Real}?

Somehow I think that's wrong though :s
OK, that's good. What is [itex]dim(A\cap B)[/itex] in this case?
 
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Would it equal to dim(A) and so 8?
 
21,992
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Would it equal to dim(A) and so 8?
Correct!! So that's one example.

Now, can you find an example where [itex]dim(A\cap B)=7[/itex]??
Hint: A and B have to span V.
 
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Does it not need to equal 7? As my answer before was dim(A∩B) = 7 or 8, not 8 or 9?
 
21,992
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For the answer of 7 would the following be okay?

A={(a,b,0,c,d,e,f,g,0,h) |a,b,c,d,e,f,g,h are Real}
B={(a,b,0,c,d,e,f,g,h,i) |a,b,c,d,e,f,g,h,i are Real}
Then dim A = 8, dim B = 9,
but A∩B = ({a,b,0,c,d,e,f,g,0,0)|a,b,c,d,e,f,g are Real} and this has dimension 7?

Also going back to my earlier working was it correct to say A∩B was a subset of A of should I have said it is a subspace of A?

Thanks for all the help so far it much appreciated.
 
21,992
3,272
For the answer of 7 would the following be okay?

A={(a,b,0,c,d,e,f,g,0,h) |a,b,c,d,e,f,g,h are Real}
B={(a,b,0,c,d,e,f,g,h,i) |a,b,c,d,e,f,g,h,i are Real}
Then dim A = 8, dim B = 9,
but A∩B = ({a,b,0,c,d,e,f,g,0,0)|a,b,c,d,e,f,g are Real} and this has dimension 7?
This is incorrect. The correct intersection is

[tex]A\cap B=\{(a,b,0,c,d,e,f,g,0,h)|a,b,c,d,e,f,g,h\in \mathbb{R}\}[/tex]

(notice the h in the last coordinate which is nonzero unlike your example).

Like I said, you need to make sure that A and B together span the space.

Also going back to my earlier working was it correct to say A∩B was a subset of A of should I have said it is a subspace of A?
It's both a subset and a subspace, so saying it's a subset isn't wrong. But it indeed seems more appropriate to call it a subspace here.
 
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I'm not getting very far, could you give another hint perhaps?

Thanks.
 
21,992
3,272
With your previous example:

A={(a,b,0,c,d,e,f,g,0,h) |a,b,c,d,e,f,g,h are Real}
B={(a,b,0,c,d,e,f,g,h,i) |a,b,c,d,e,f,g,h,i are Real}
You put a 0 in the third place in both A and B. What if you don't put a 0 in the same place?
 
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Would this work?

A={(a,b,c,d,0,e,f,g,h,0)}
B={(a,0,b,c,d,e,f,g,h,i)}

Then A and B span R^10 since a linear combiantion Q(a,b,c,d,0,e,f,g,h,0) + P(a,0,b,c,d,e,f,g,h,i) = (Qa+Qb,Qb,Qc+Pb,....) = (x1,x2,x3,x4,x5,x6,x7,x8,x9,x10).

And also A∩B i think is equal to {(a,0,b,c,0,d,e,f,g,0).

Is that correct?
 
21,992
3,272
Would this work?

A={(a,b,c,d,0,e,f,g,h,0)}
B={(a,0,b,c,d,e,f,g,h,i)}

Then A and B span R^10 since a linear combiantion Q(a,b,c,d,0,e,f,g,h,0) + P(a,0,b,c,d,e,f,g,h,i) = (Qa+Qb,Qb,Qc+Pb,....) = (x1,x2,x3,x4,x5,x6,x7,x8,x9,x10).

And also A∩B i think is equal to {(a,0,b,c,0,d,e,f,g,0).

Is that correct?
That's good!!
 
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So is that all correct now?

Also thank you so much for all the help!
 

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