surjective, but not injective linear transformation $T$ is a transformation from the set of polynomials on $t$ to the set of polynomials on $t$.  So, the input to $T$ should be a polynomial, and the output should be some other polynomial.  Two common linear transformations are differentiation and integration from $t=0$.  Namely, we can describe differentiation operator $T(p) = \frac{dp}{dt}$ by saying that if $p(t) = a_0 + a_1 t + \cdots + a_n t^n$, then 
$$
T[p(t)] = a_1 + 2a_2 t + \cdots + na_n t^{n-1}
$$
Similarly, we can describe the operator $T(p) = \int_0^t p(x)\,dx$ by saying that if $p(t) = a_0 + a_1 t + \cdots + a_n t^n$, then 
$$
T[p(t)] = a_0t + \frac {a_1}2 t^2 + \cdots + \frac {a_n}{n+1} t^{n+1}
$$
How can i  prove that the first operator is surjective, but not injective, while the second is injective, but not surjective. Some help please.
 A: The surjective part is easy.  Given a polynomial $a_{0} + a_{1}t + a_{2}t^{2} + \cdots + a_{n}t^{n}$, which polynomial is being sent to this polynomial under the differentiation transformation $T$?  Well, any polynomial of the form $C + a_{0}t + a_{1}\frac{t^{2}}{2} + \cdots + a_{n-1}\frac{t^{n}}{n} + a_{n}\frac{t^{n + 1}}{n + 1}$, where $C$ is any constant.  Why?  Differentiate $C + a_{0}t + a_{1}\frac{t^{2}}{2} + \cdots + a_{n-1}\frac{t^{n}}{n} + a_{n}\frac{t^{n + 1}}{n + 1}$ and you will get exactly $a_{0} + a_{1}t + a_{2}t^{2} + \cdots + a_{n}t^{n}$.
Now, I said $C$ can be any constant, and that polynomial will still be sent to $a_{0} + a_{1}t + a_{2}t^{2} + \cdots + a_{n}t^{n}$.  In particular, if $C_{1}$ and $C_{2}$ are two distinct constants, both $C_{1} + a_{0}t + a_{1}\frac{t^{2}}{2} + \cdots + a_{n-1}\frac{t^{n}}{n} + a_{n}\frac{t^{n + 1}}{n + 1}$ and $C_{2} + a_{0}t + a_{1}\frac{t^{2}}{2} + \cdots + a_{n-1}\frac{t^{n}}{n} + a_{n}\frac{t^{n + 1}}{n + 1}$ are being sent to the same polynomial, even though they are distinct polynomials (they differ by the constants $C_{1}$ and $C_{2}$).  If you don't believe they are sent to the same polynomial under $T$, differentiate both of them and check to see that you get the same polynomial as an output of $T$.  So this is an example of $T(a) = T(b)$ but $a \neq b$, which means $T$ is not injective.
Try using the ideas I showed above and apply them to the integral linear operator.  Do you see why it is injective but not surjective?
A: Hints for the first.
To say that differentiation is surjective means that every polynomial is the derivative of some polynomial.  Can you see why this is true?
To say that differentiation is injective means that a polynomial cannot be the derivative of two different polynomials.  Can you see why this is false?
A: Find two polynomials that have equal derivatives (hint: constants are polynomials). That will show the first transformation to be non-injective. 
Stare at the definition of the second transformation. Do you think every polynomial is of that form? You should answer negatively (and again constants will help you), and this shows the transformation is not surjective. 
Finally, denoting the first transformation by $T_1$ and the second by $T_2$, note that $T_1
\circ T_2$ is the identity function. In general, if a composition $f\circ g$ of functions is injective, then $g$ is injective, while if the composition is surjective, then $g$ is surjective. Now apply this to your transformations using the fact that the identity function is both injective and surjective. 
