If an operator $T \in L(V)$ satisfies $Tv = 0$ for every $v \in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,…0), (0,1,…0), … (0,0,…1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
Best Answer
The question is poorly worded, because it could be interpreted as "Prove that the zero operator is the zero operator" which is circular (as has already been pointed out).
But the intention of the questions could be "prove that the zero operator $T$ on a finite dimensional vector space $V$, defined by $Tv=0 \space \forall v \in V$, is a linear operator and can be represented with respect to any basis of $V$ by the $0$ matrix". In which case your proof is fine.