Find $\dim(\ker T_1 \cap \ker T_2)$ if $T_1,T_2:V\to F$, $T_1 \neq 0$, $T_2\neq 0$, $\dim V=n$ and $\ker T_1\neq \ker T_2$ 
Find $\dim(\ker T_1 \cap \ker T_2)$ if:
$T_1,T_2:V\to \mathbb F$
$T_1 \neq 0$, $T_2\neq 0$
$\dim V=n$ and $\ker T_1\neq \ker T_2$. $\mathbb F$ can be any field.

Let $\ker T_1=S_1$ and $\ker T_2=S_2$.
We know that $\operatorname{Image}T_1 \subseteq \mathbb F$ but also it is given that $T_1 \neq 0$ so $\dim(\operatorname{Image}T)=1$.
Then $\dim S_1=\dim V-\dim(\operatorname{Image}T)=n-1$.
We can get the same result for $T_2$.
$\dim(S_1\cap S_2) \neq n-1$ because if it were then $S_1=S_2$ which contradicts the given.
Therefore $\dim(S_1\cap S_2) \le n-2$.
We also know that $\dim(S_1 + S_2)\le n. \tag{$\ast$}$
Finally from the dimension theorem and $(\ast)$ we get:
$$
\dim(S_1+S_2)=\dim S_1+\dim S_2-\dim(S_1 \cap S_2) \Rightarrow\\
\dim(S_1+S_2)=2n-2-\dim(S_1 \cap S_2) \le n \tag{$\ast\ast$}
$$
We know that $\dim(S_1 \cap S_2) \le n-2$ but if we take for example $\dim(S_1 \cap S_2)=n-3$ then the inequality $(\ast\ast)$ won't hold so $\text{dim}(S_1 \cap S_2)=n-2$.
Is my reasoning correct?
 A: Lemma: Intersection of two different hyperplanes of a $n$-dimensional vector space $V$ is a linear subspace $W \subset V$ of $\dim W=n-2$.
Proof: For Grassmann formula
$\dim S \cap T=\dim S+\dim T-\dim(S+T)=2n-2-n=n-2$ ($\dim(S+T)=n$ because $S$ and $T$ are different hyperplanes of $V$).

Your reasoning is correct.
I would say that $S_1$ and $S_2$ are two linear subspace of $V$ of dimension n-1 (from the dimension theorem and noticing that $\dim(\operatorname{Im}(T_i))=1)$.
We have that $S_1$ and $S_2$ are DIFFERENT hyperplanes of V.
Intersection of two different hyperplanes of a vector space is a linear subspace of $V$ with dimension $n-2.$
A: Let 
$$\Phi: V\to F^2, x\mapsto (T_1(x),T_2(x))$$
then $\Phi$ is clearly linear and $\ker \Phi=\ker T_1\cap\ker T_2$. Since $T_1\ne0$ and $T_2\ne0$ then $\operatorname{rank}\Phi\ne0$. If $\operatorname{rank}\Phi=1$ then there is $\lambda\in F\setminus\{0\}$ such that $T_1=\lambda T_2$ (why?). But in this case $\ker T_1=\ker T_2$ which is a contradiction. Finally, $\operatorname{rank}\Phi=2$ and the rank-nullity theorem gives
$$\dim(\ker T_1\cap\ker T_2)=n-2$$
