Linear Transformations and Compositions


Sep 2012
Hi guys, I'd like some help with this question. It's been incredibly frustrating trying to solve this for hours, but getting nowhere. (Headbang)

Suppose U -> V -> W, with transformations T (U to V) and S (V to W).

1) Suppose ST is one-to-one and T is onto, show that S is one-to-one.

My answer 1)
Suppose (ST)(u) is one-to-one, and suppose T is onto:
(ST)(u) = S(T(0)) = 0 = S(T(u)) = S(v)
Therefore v = 0.

2) Suppose ST is onto and S is one-to-one, show that T is onto.

This one has me beat and I'm not sure how to approach it at all.

Can someone please (hopefully) verify that I did 1) correctly, and help with 2)?

Thank you!


MHF Hall of Honor
Mar 2011
your proof of one is hard to follow.

we want to prove that S is 1-1, which means proving S(v) = 0 implies v = 0.

suppose S(v) = 0. since v is in V, and T is onto v, v = T(u) for some u in U.

therefore: S(v) = S(T(u)) = ST(u).

since S(v) = 0, ST(u) = 0.

since ST is 1-1, u = 0.

thus v = T(u) = T(0) = 0, which is what we desired to prove.

2) here, we must find for any given v in V, some u in U with T(u) = v.

now, we know that given any w in W, we have u in U, with ST(u) = w.

so consider w = S(v), for our given v above.

we have w = S(v) = ST(u), for some u in U (since ST is onto).

since S is 1-1, and S(v) = ST(u) = S(T(u)), v = T(u), and we are done.
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