Proving composition of one-to-one linear transformations is one-to-one

[angela123]

Question: Let T:V-->W and S:W-->U be linear transformation.Show that
1) If T and S are one-to-one,then ST is one-to one
2) If ST is one-to-one,then T is one-to-one
3)Give example of two linear transformations T and S, such
that ST is one-to-one ,but S is not.

For 1),Since T is one-to-one,there is some vector in V,such that T(v)=O
Since S is one-to-one,there is some vector in W,such that S(w)=O
If ST is one-to-one,then there is some vector t in T,such that
ST(t)=S(T(t))=O,so how could I proof T(t)=w?

How about question 2) &3)

 

[Hurkyl]

Since T is one-to-one,there is some vector in V,such that T(v)=O

How many are there?

 

[angela123]

It could be more than One vector in V, such that T(v)=O,I know Ker T=O


Angela

 

[angela123]

Sorry, I made a mistake,It has only One vector in V, such that T(v)=O




Angela

 

[ircdan]

Here is one way to do:
2) If ST is one-to-one,then T is one-to-one

Proof: Suppose T is not one-to-one. Then there exists x, y in V, x != y, such that T(x) = T(y).Then ST(x) = S(T(x)) = S(T(y)) = ST(y) but x != y, and thus ST is not one-to-one.

You should try to give a direct proof for practice. Also notice the fact that S and T are linear transformations was never used and so this result holds for functions in general.