T-annihilators and whether it kills the set Question : $V=F[t]v$ is T-cyclic, $g(t)$ is a polynomial. Show that if $g(T)v=0$, then $g(T)=0$.
Any ideas how to solve this?
plus, does this statement implies T-annihilators of v are exactly same as Ann(T)? I don't really get the differences between two concepts.

Firstly, note $V$ being $T$-cyclic means $V = \text{span}\{v, T(v), T^2(v), \cdots \}$, and that $\{v, T(v), T^2(v), \cdots \}$ form a basis for $V$. The moment you are given $g(T)v=0$, you can conclude that $V$must be finite dimensional, since you may write $g(T)v= \sum_i g_i T^i v = 0$, so you have finitely many basis vectors. (Here, $g_i$'s are coefficients of the polynomial $g$).
Now, if $g(T)v = 0$, then $g(T)(Tv) = 0$, $g(T)(T^2v) = 0$, and repeat the argument for each of the finitely many basis vectors. The (linear) operator $g(T)$ maps each basis element $(T^iv)$ to $0$, so it is the zero operator on $V$.
$Ann(T)= \{f(x) \mid f(T) = 0\}$ (set of all polynomials which, when $T$ is "plugged in", gives the zero operator).
$T$-annihilators of $v$ = $\{ f(x) \mid f(T)v = 0 \}$ (set of all polynomials which, when $T$ is "plugged in" and acts on $v$, gives the zero vector.
Normally, you can only conclude that $Ann(T) \subseteq $ $T$-annihilators of $v$. However, in this case, when $V$ is cyclic and you are specifically looking at the $T$-annihilators of $v$, the cyclic vector, an annihilator of $v$ is also a $T$ annihilator, so you're right in saying the two sets are equal.
