Need help on matrix prove?

Question

Theroem: Let T:R^n->R^n be linear. Then T one to one-> onto Try to prove

Give with explanations
i) an example of a one to one function f:R->R everywhere differentiable but onto
ii) Example g:R->R(differentiable) onto but no one to one

Answer 1

Since T is one-to-one, the kernal of T is just the 0-dimensional subspace consisting of just the 0-vector. The dimension of the kernal plus the dimension of the image must be equal to the dimension of the domain, so the dimension of the image must be n (because the dim of the domain is n and the dimension of the kernal is 0) Also, the image must be a SUBSPACE of the target since T is linear, and since the target space is n-dimensional, the only n-dimensional subspace is the entire space, so T must be onto.

As for the two examples you asked for, assuming the the first one is correctly asked, how about just the identity function, f(x)=x. It is certainly everywhere differentiable and it is certainly onto. For example 2, try g(x)=x^3-x. Since it is a polynomial it is differentiable, since it is an odd degree polynomial, it's range is all of R, so it is onto, and since
g(0)=g(1)=0, it is certainly not one-to-one.

Answer 2

By dimension thm , n=Ker(T)+Rank(T) ==> Rank(T)=n
==> T is onto

(i) f(x)=x

(ii) g(x) =x sin(x)