$T$ 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.
Best Answer
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?