Give an example of a linear transformation T: V --> V which is one-to-one but not onto.?

Define V and T clearly, and justify your choice.





I know the integral is an example of this but I'm looking for a clear, simple explanation. My attempt so far is to say that T= integral from o to x f(t) dt and that V could be polynomials of degree 2 for example. Then it is one to one because each poly in this space will have a unique integral and then I could show how it acts on the standard ordered basis {1, x, x^2} Obviously each vector in basis will have unique integral but also any vectors generated from this set. For example, each constant will have unique integral. As why its not onto, I'm less clear. I was told by classmate that because of the plus C that comes from integrating that each element in range may not have pre-image, but I don't really see it. Also, the teacher made sure to say its the DEFINITE integral so I'm not sure about the plus C.


Also, same question for example onto but not one to one. ex. derivitive. I see why not one to one but why is it onto?

Give an example of a linear transformation T: V --%26gt; V which is one-to-one but not onto.?
if V is finite dimensional, then the rank nullity theorem tells us that any injective linear transformation on V is surjective. so we need V to be infinite dimensional. consider the vector space of polynomials over the real numbers. it is infinite dimensional, with basis {1,x,x^2,...}. consider the map T:V--%26gt;V given by T(P(x)) = xP(x). you can check that this map is injective. it's also linear: if a is a real number and P,Q two polynomials in V, then T(aP) = x(aP) = a(xP) = aT(P) and T(P+Q) = x(P+Q) = xP+xQ = T(P)+T(Q). but T is not surjective because none of the constant polynomials lie in the image.