GR9768 Q 28

Forum for the GRE subject test in mathematics.
mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

GR9768 Q 28

Postby mathQ » Tue Apr 06, 2010 3:18 pm

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

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: GR9768 Q 28

Postby EugeneKudashev » Tue Apr 06, 2010 3:24 pm

take two 6-dimensional vectors and look at their intersection. it is plain that it contains no less than 2 components.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

Re: GR9768 Q 28

Postby mathQ » Tue Apr 06, 2010 3:40 pm

EugeneKudashev wrote:take two 6-dimensional vectors and look at their intersection. it is plain that it contains no less than 2 components.


How did you figured out so quickly ? Any tricks ?

which vectors you took ?
thanks

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: GR9768 Q 28

Postby EugeneKudashev » Tue Apr 06, 2010 3:46 pm

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.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

Re: GR9768 Q 28

Postby mathQ » Tue Apr 06, 2010 3:59 pm

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.


indeed..u have just explained in the right and quick manner. Awesome again!!!
Thanks

Btw, where are you from Eugene ?

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: GR9768 Q 28

Postby EugeneKudashev » Tue Apr 06, 2010 4:12 pm

you're welcome!
I'm from Moscow, Russia, will graduate from Moscow State University this June. you?

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: GR9768 Q 28

Postby Hom » Sat Oct 01, 2011 3:29 am

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.

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: GR9768 Q 28

Postby Hom » Sat Oct 01, 2011 3:55 am

Hom 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.


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 intersection

cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

Re: GR9768 Q 28

Postby cauchy2012 » Wed Oct 19, 2011 10:49 am

We have a formula in the textbook of linear algebra: dim(V1)+dim(V2)=dim(V1+V2)+dim(V1intersectV2),so 6+6-10=2 :D

goombayao
Posts: 53
Joined: Sun Oct 16, 2011 9:17 am

Re: GR9768 Q 28

Postby goombayao » Wed Oct 19, 2011 3:31 pm

You can also notice that the kernel of the intersection is at most 8 dimensional. Thus the image must be at least 2 dimensional.




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