Given $V = Fn(x)$ a vector space. With derivative operator $T: V \rightarrow V$, $T(p(x)) = p'(x)$
I proved that $V$ is a vector space by showing that all the axioms are true, and after that, I found the only eigenvalue of $T$ which is $\lambda = 0$, now I'm asked to find the Vector space of $\lambda$ which is $V_\lambda$, any idea or a hint for how to find it?
You are looking at the kernel of the derivative operator $T$.
If $Fn(x)$ is the set of polynomials with coefficients in a field $k$ of degree $n$ (with $a_n\ne0$ ad $a_i=0$ for each $i>n$) and the constant $0$, there is a problem with the definition of $T$ (because if you derive a degree $n$ polynomial, you get a degree $\leqslant n-1$ polynomial.
So let me assume that $Fn(x)$ is the set of polynomial with degree $\leqslant n$. Then a polynomial is of the form $$ p(x)=\sum_{i=0}^n a_i x^i. $$ And $$ p'(x)=\sum_{i=0}^{n-1} (i+1)a_{i+1} x^{i}. $$ So the kernel is the set of polynomials s.t. $$ (i)a_{i}=0\quad \text{for each }i\in\{ 1,\cdots, n\}. $$ If the characteristic of the field $k$ is $0$ you get that the kernel is done by the constant polynomial, so $$ V_0=k. $$ If the characteristic of the field $k$ is $p$ you get that the kernel is done by polynomial of the form $$ \sum_{i=0}^s a_{ip} x^{ip}\quad \text{s.t. }sp\leqslant n\text{ and } a_{ip}\in k. $$ So $$ V_0=\bigoplus_{s=0}^{\lfloor \frac{n}{p}\rfloor} \{a_{sp} x^{sp}\mid a_{sp}\in k\}=\bigoplus_{s=0}^{\lfloor \frac{n}{p}\rfloor} k<x^{sp}>. $$ (where $k<x^{sp}>$ means the $k$-vector space generated by $x^{sp}$).
I hope that my definition is equal to your definition.