What are $S$ and $T$ in $C(T) = \{S \in L(V,V)\mid ST=TS\}$? Let $C(T) \subseteq L(V, V)$ be the subset of all linear operators which commute with $T$, i.e.:
  $$C(T) = \{S \in L(V, V ) \mid ST = TS\}$$
  Prove that: $C(T)$ is a subspace of $L(V, V)$, $C(T)$ contains the identity operator $I_V$, $C(T)$ is closed under composition of operators, and $T$ is an isomorphism if the linear map is a nonzero linear map. 


Am I supposed to assume that $T$ is also in $L(V, V)$ so that $S: V \to V$ and $S = T$ or that $S$ is an inverse of $T$? I feel like if I know what $S$ and $T$ are, I can solve the question. Please help.
You need to show that $C(T)$ is closed under addition and scalar multiplication and also contains the identity. Clearly, $C(T)$ contains the identity since the identity commutes with all operators.  To show that $C(T)$ is closed under addition, let $U, V \in C(T)$.  Then, $(U + V)T = UT + VT = TU + TV = T(U + V)$.
You need to show that $C(T)$ is closed under addition and scalar multiplication and also contains the identity. Clearly, $C(T)$ contains the identity since the identity commutes with all operators.  To show that $C(T)$ is closed under addition, let $U, V \in C(T)$.  Then, $(U + V)T = UT + VT = TU + TV = T(U + V)$.