First show that $Tv$ is scalar multiple of $v$ for all $v\in V$. You can do this by assuming, for the sake of contradiction, that there exists $v$ such that the vectors $v$ and $Tv=w$ are linearly independent. Form a basis $\mathcal B$ which includes both $v$ and $w$. Now, define a new linear operator $L:V\to V$ which maps every basis element to $0$ except $w$, which we map to $v$. You will get a contradiction.
Now, you must show that there is just one scalar $\lambda$ such that $Tv=\lambda v$ for all $v$. Say $Tv_1=\lambda_1v_1$ and $Tv_2=\lambda_2v_2$ where neither $v_1$ nor $v_2$ is a scalar multiple of the other--i.e. they are linearly indepdendent--notice that the case where they are linearly dependent is easily handled. Write $T(v_1+v_2)=\lambda_3(v_1+v_2)$. Then the equation $$(\lambda_3-\lambda_1)v_1+(\lambda_3-\lambda_2)v_2=0$$ implies $\lambda_1=\lambda_3=\lambda_2$ as desired.