A fun one for all you kiddies...(In Linear Algebra)?

Question

Suppose T: Rn ->Rm (please ignore the crudity of the expression of this) is a linear transformation. Provide counterexamples to each of the following statements, with an explanation of why each is correct. Note: you may not use the zero transformation, and T must be a linear transformation.

a) If m>n or m=n, T is surjective.
b) If m
have fun with this!

Answer 1

Let T: R^2 -> R^2 be defined by T([x y]) = [x 0]. Then T is obviously a linear transformation because it is an orthogonal projection. Or, if you like, because its standard matrix is
[1 0]
[0 0]

However, [0 1] does not get mapped to by T, so T is not surjective.

Also, both [0 0] and [1 0] map to the zero vector, so T is not injective.

Answer 2

this isn't an quite finished of life answer, even with the undeniable fact that the gist of the proofs may bypass alongside those lines. 7. The rank is the length of a dull ringer for A. If the length of the photo is decrease than m, then it may't span R^m. It obviously won't be able to be any more beneficial than m because the photo is only vectors of length m. So it must be m. 8. If S isn't linearly autonomous, then the rank of the matrix operator A in accordance to S is decrease than 3, and subsequently a dull ringer for A, it really is the set of all linear combos of the vectors in S, is likewise decrease than 3, so it may't span R3.

Answer 3

All you have to do is show a matrix that has a null space.

For m=n=2

1 0
0 0

Maps (x,y) into (x,0) thus it is not surjective.