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GR0568 Q. 18

Posted: Fri Oct 31, 2008 11:43 am
by amateur
Question: Let V be the real vector space of all real 2x3 matrices and let W be the real vector space of all real 4x1 column vectors. If T is a linear transformation from V onto W, what is the dimension of the subspace { v e V: T(v) = 0}?

I am confused wrt to T. If it is a linear transformation surely it must have some matrix representation. Now how can a matrix applied to the set of all 2x3 matrices produce the set of all 4x1 matrices (or column vectors)?

Since matrix product is only defined for m x n (for T) and n x p (for V)matrices and results in and m x p (for W) matrix if I take m = 4, p = 1 then n x p cannot be 2 x 3.

I hope I am not on the right track and some one could provide a hint.

P.S. I found something interesting:

Let A = [ 1 1 0 , 1 2 1 ], and let B = [ 2 1 , -1 -1 , 0 2]

Then AB = I2, although both A and B are not invertible since they are not square. Hence we can conclude that the statement

"If A and B are two matrices such that AB = the identity matrix, then the inverse of A exists" is false.

Posted: Fri Oct 31, 2008 7:44 pm
by ralphhumacho
dim V = 6, dim(Img(T)) = 4. Here they ask for the dimension of ker(T). Use rank nullity theorem for operators.

To find At, just find what happens to the basis vectors of W and stick them in a matrix.