PROOFS! Help Please!!! Suppose T: U→V and S: V→W are linear transformations...?

a. It suffices to show that ker(S) (the kernel of S) is trivial, i.e ker(S) = {0}.

Let x be in ker(S) then S(x) = 0. Since x is in V and T is onto then we can find u in U such that T(u)=x.

Applying S on both sides gives: ST(u) = S(x). But S(x) = 0 and therefore ST(u)=0 which implies that u is an element of ker(ST). By assumption ST is one-to-one, hence ker(ST) = {0}.

Thus u=0 and since T is linear then T(u)=T(0) =0. But T(u) = x so x = 0, i.e ker(S) is trivial and we're done.

b. Let v be an element of V. We need to show there is some u in U such that T(u) = v.

Note that ST is a map from U to W. Since v is in V then S(v) is an element of W. Using the fact that ST is onto it follows there is some u in U such that ST(u) = S(v). By assumption S is one-to-one and therefore T(u) = v, as desired.

Hope this helps

it is sufficient to instruct, for this reason, that there exists an x=(x1, x2) and y=(y1, y2) such that T(x+y) does no longer equivalent T(x) + T(y). enable's attempt something for x and y. enable x = (0, 0) and y = (a million, 0). Then T(x) = (0, -3, 0) and T(x+y) = T(y) = (a million, -2, 2). yet T(x) + T(y) = (a million, -5, 2) so T isn't linear.