Exploring Vector Spaces and Dimensions: A Homework Challenge

[RVP91]

Homework Statement

Homework Equations

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

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!

 

[HallsofIvy]

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 $A\subset B$. 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.

 

[RVP91]

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]

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, $dim(A+ B)= dim(A)+ dim(B)- dim(A\cap B)$. Here dim(A)+ dim(B)= 8+ 9= 17 while dim(A+ B) cannot be larger than 10. So, at most, $10= 8+ 9- dim(A\cap B)$.

(The dimension of the "space" containing only the zero vector is 0, not 1, by the way.)

 

[RVP91]

Oh right I see. So dim(A∩B) must be at least 7 and at most 17?

 

[HallsofIvy]

No, that's not at all right! $A\cap B$ is a subset of A so it can't possibly have dimension greater than the dimension of A.

 

[RVP91]

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?

 

[RVP91]

Could someone check what I wrote above please?

Thanks.

 

[micromass]

What you wrote is correct.

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

 

[RVP91]

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?

 

[micromass]

They want you to give specific examples of V, A, B such that dim(A)=8, dim(B)=9 and $dim(A\cap B)$ is 8 or 9.

I suggest taking $V=\mathbb{R}^{10}$, then dim(V)=10 as required. Now, how would you pick A and B?

 

[RVP91]

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

 

[micromass]

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 $dim(A\cap B)$ in this case?

 

[RVP91]

Would it equal to dim(A) and so 8?

 

[micromass]

Would it equal to dim(A) and so 8?

Correct! So that's one example.

Now, can you find an example where $dim(A\cap B)=7$??
Hint: A and B have to span V.

 

[RVP91]

Does it not need to equal 7? As my answer before was dim(A∩B) = 7 or 8, not 8 or 9?

 

[micromass]

Does it not need to equal 7? As my answer before was dim(A∩B) = 7 or 8, not 8 or 9?

Yes, sorry.

 

[RVP91]

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.

 

[micromass]

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

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

(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.

 

[RVP91]

I'm not getting very far, could you give another hint perhaps?

Thanks.

 

[micromass]

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?

 

[RVP91]

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?

 

[micromass]

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!

 

[RVP91]

So is that all correct now?

Also thank you so much for all the help!