GRE 0568 - Problem 18

Forum for the GRE subject test in mathematics.
EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

GRE 0568 - Problem 18

Postby EmLasker » Sun Oct 12, 2008 7:27 pm

I think I have this one figured out, but I'm not exactly sure... maybe somebody can clarify things a bit for me.

The problem is: Let V be the vector space consisting of all real 2x3 matrices and W be the vector space consisting of all 4x1 matrices. If T is a linear transformation from V onto W, what is the dimension of the subspace {v in V, such that T(v) = 0}.

This is of course the kernel of T. The rank nullity theorem gives that Dim(Ker(T))+Dim(Im(T)) = Dim(V). Since Dim(Im(T)) = Dim(W) = 4x1 = 4 and Dim(V) = 2x3 = 6, we have that Dim(Ker(T)) = Dim(V)-Dim(W) = 6 - 4 = 2... right?

Peter

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Sun Oct 12, 2008 9:36 pm

Absolutely ! :D

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Sun Oct 12, 2008 10:37 pm

Well, that's a relief... Thank you for the replay. :lol:

thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

Re: GRE 0568 - Problem 18

Postby thmsrhn » Fri Mar 26, 2010 8:06 am

I m sorry i didn t follow
Could you elaborate?

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: GRE 0568 - Problem 18

Postby origin415 » Fri Mar 26, 2010 8:58 am

Here is another, more expository, way, if this helps:
Choose a basis of W w1, w2, w3, w4. T goes from a space of dimension 6 to a space of dimension 4, so it is surjective. Then we can find v1, v2, v3, v4 such that T(vi) = wi. One can show that those vi's are linearly independent, so they can be extended to a basis by appending u1, u2. If T(ui) isn't zero, it can be represented by some combination of wi's, so ui can be represented by some linear combination of vi's. By contradiction, T(ui) = 0, so these are are the basis of the kernel of T, which is then of dimension 2.




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