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Hi, I am having troubles even starting this problem and any help would be appreciated.
1) The question is: If B is an nxn matrix, X = {A in M(mxn) matrix space | AB = 0} and Y = {AB | A in M(mxn) matrix space}, show that X and Y are subpaces of M(mxn) and that dim(X) + dim(Y) = mn.
What I have) From observation, I can see that X resembles a null space or kernel of a T, and Y resembles an image space or image of a T. And partially, I can see that dim(X) + dim(Y) = mn = dim(M(mxn) matrix space).
2) As a related question, can someone explain to me how to show if a transformation is one to one and/or onto? Suppose T: V-->W: I know that one to one means that the ker(T) = {0}, does that mean when a T is one to one, there are no vectors that meet the requirement T(v) = 0? And likewise, for onto, not every vector w in W can be mapped via T(v)?
As always, thank you for any help. My main focus is the initial problem, and the explanation for one to one/onto is secondary; if there are any sources you may point me to for the explanation, please do - I just hope for a different way to approach the thinking since I am quite confused.
As for the reason for repost: The thread got buried after I posted late last night and am desperate to get some help with this.
1) X=ker(T) and Y=Im(T), where T(A)=AB. I think this will help you do the rest.
2) It means the only solution to T(v)=0 is v=0. If you want to test if T is one to one you solve the equation T(v)=0 and see if there are any v different than zero. if there are, then it is not one to one. If there aren't, T is one to one.
Onto means that every vector can w can be expressed as T(v), for some v in V. You can also show that T is onto by showing that dim Im(T) = dim W.
Thank you very much hedi and ModusPonens, you have relieved much of my stress.
If you would not mind checking this as well, to see if I used one-to-one/onto correctly:
Suppose U -> V -> W, have linear transformations, from U to V is T, from V to W is S.
Show that if ST is one-to-one, and T is onto, then S is one-to-one:
(ST)(u) = 0 where u = 0, T(u) = v
S(T(u)) = S(v) = 0
S(v) = 0
I'm just not sure if I have proved or said anything.