GR9768 Q 28
GR9768 Q 28
If V1 and V2 are 6 dimensional subspaces of a 10 dimensional vector space V, what is the smallest possible dimension that V1 intersection V2 can have ?
A) 0
B) 1
C) 2
D) 4
E) 6
A) 0
B) 1
C) 2
D) 4
E) 6
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Re: GR9768 Q 28
take two 6-dimensional vectors and look at their intersection. it is plain that it contains no less than 2 components.
Re: GR9768 Q 28
How did you figured out so quickly ? Any tricks ?EugeneKudashev wrote:take two 6-dimensional vectors and look at their intersection. it is plain that it contains no less than 2 components.
which vectors you took ?
thanks
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- Posts: 27
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Re: GR9768 Q 28
perhaps I'll sketch it like that:
| | | | | | x x x x
x x x x | | | | | |
1 2 3 4 5 6 7 8 9 10
first row is your first subspace ( | stands for element, x for emptiness) and the second row is your second subspace. both have dim=6 and as you can see their intersection should not "exceed" dim=10. thus, intersection has dim no less than 2. I hope that is understandable.
| | | | | | x x x x
x x x x | | | | | |
1 2 3 4 5 6 7 8 9 10
first row is your first subspace ( | stands for element, x for emptiness) and the second row is your second subspace. both have dim=6 and as you can see their intersection should not "exceed" dim=10. thus, intersection has dim no less than 2. I hope that is understandable.
Re: GR9768 Q 28
indeed..u have just explained in the right and quick manner. Awesome again!!!EugeneKudashev wrote:perhaps I'll sketch it like that:
| | | | | | x x x x
x x x x | | | | | |
1 2 3 4 5 6 7 8 9 10
first row is your first subspace ( | stands for element, x for emptiness) and the second row is your second subspace. both have dim=6 and as you can see their intersection should not "exceed" dim=10. thus, intersection has dim no less than 2. I hope that is understandable.
Thanks
Btw, where are you from Eugene ?
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- Posts: 27
- Joined: Tue Apr 06, 2010 8:22 am
Re: GR9768 Q 28
you're welcome!
I'm from Moscow, Russia, will graduate from Moscow State University this June. you?
I'm from Moscow, Russia, will graduate from Moscow State University this June. you?
Re: GR9768 Q 28
Hi, i was thinking the same way at first but then I came up with another thought. Not sure if it's valid.
Let a 2D plane p1 to be x=0, and the other plane p2 to be x=2. See p1 and p2 are 2D space of 3D space. But there is no intersection between p1 and p2.
Please comment.
Let a 2D plane p1 to be x=0, and the other plane p2 to be x=2. See p1 and p2 are 2D space of 3D space. But there is no intersection between p1 and p2.
Please comment.
Re: GR9768 Q 28
Sorry, I forgot the vector space has to contain the origin. x=2 is not a valid 2D space then. p1 and p2 containing the origin will have intersectionHom wrote:Hi, i was thinking the same way at first but then I came up with another thought. Not sure if it's valid.
Let a 2D plane p1 to be x=0, and the other plane p2 to be x=2. See p1 and p2 are 2D space of 3D space. But there is no intersection between p1 and p2.
Please comment.
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- Joined: Sat Aug 13, 2011 9:04 am
Re: GR9768 Q 28
We have a formula in the textbook of linear algebra: dim(V1)+dim(V2)=dim(V1+V2)+dim(V1intersectV2),so 6+6-10=2 ![Very Happy :D](./images/smilies/icon_biggrin.gif)
![Very Happy :D](./images/smilies/icon_biggrin.gif)
Re: GR9768 Q 28
You can also notice that the kernel of the intersection is at most 8 dimensional. Thus the image must be at least 2 dimensional.