Basis in the vector space of all polynomials Let $V$ vector space of all polynomials  $p(t) = a_0 + a_1t + \cdots + a_nt^n$,$\forall  n \in\mathbb{N}$ and $a_0,\ldots,a_n \in\mathbb{R}$. How can I prove that $ \gamma = \{1,t,t^2,\ldots\}$ is a basis of $V$, and use it to find a linear transformation $T:V \rightarrow V$ such that $T$ is surjective but not injective. Is $T(x) = x^2$ an example of this transformation? Some help please.

You seem to have a fundamental misunderstanding about this question.
$T$ 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 
$$
T[p(t)] = a_0t + \frac {a_1}2 t^2 + \cdots + \frac {a_n}{n+1} t^{n+1}
$$
Try to prove that the first operator is surjective, but not injective, while the second is injective, but not surjective.

As for your first question: you should look up the definition of a basis, and verify that $\gamma$ satisfies that definition.
Is it true that every member $v \in V$ can be written as a finite sum
$$
v = \sum_{i=1}^n a_i v_i
$$
where $v_i$ are elements of $\gamma$?  Is it true that if
$$
\sum_{i=1}^n a_i v_i = 0
$$
for some $a_i$, then all $a_i$ have to be equal to zero?  If so, then $\gamma$ is a basis.  If you show that this is the case, then you have proven $\gamma$ to be a basis of $V$.
