Is $K_a$ subspace of vector space $V$? Let $\text{Hom}(V,V) = H$ be the set of linear transformations from $V$ to $V$. Let $a \in V, K_a \subset H$ such that for all $T \in K_a, T(a)=0.$ Is $K_a$ a subspace? Does there exist an $A$ such that $K_a=H$? Let $a_1, a_2, …, a_n$ be a basis $B$ for $V$. What is $\cap_{a_i\in B}K_{a_i}$?
Some thoughts that I have:
I think that $K_a$ is a set of transformations where it transforms all $a$'s into 0's. So $K_a$ is a subspace - I would be able to prove that it is additive and homogeneous. (I am not sure how exactly, however.) I do not think there exists an $a$ where $K_a = H$ because $H$ would never be a subset of $K_a$ for any $a$. Any transformation in $H$ that maps an element to a non-zero element would not exist in $K_a$. Thus, is the intersection also 0? (Also do not know how to prove this.)
Some hints that my professor gave me were what is $K_0$ and whether there are any linear transformations where $T(0)\neq0$.
Thanks!
 A: To prove something is a subspace of H, you must prove the properties of a vector space still hold and that it is a subset of H. By definition, $K_a \subset H$, so if you can show this, a lot of the properties transfer down. The main things left to prove are closure (does addition and scalar multiplication keep you in the subspace) and that zero exists. Keep reading if you want the answer, but everything above this should be enough if you just want hints:
Let $T = 0$ be the linear transformation that maps everything from $V$ to the $0$ in $V$. By definition, $T(a) = 0$, so it is in $K_a$.
Let $T_1, T_2 \in K_a, \alpha_1, \alpha_2 \in F$, where $F$ is whatever scalar field you are working on. Consider $\alpha_1 T_1 + \alpha_2 T_2$:
$(\alpha_1 T_1 + \alpha_2 T_2)(a) = \alpha_1 T_1(a) + \alpha_2 T_2(a) = \alpha_1 (0) + \alpha_2 (0) = 0 \therefore \alpha_1 T_1 + \alpha_2 T_2 \in K_a$
The above works because of linearity. Now, for the second part, if you pick an $a$ such that for every $T \in H$, $T(a) = 0$, then $K_a = H$. The only $a$ that would satisfy this is $0$.
