Is the following argument correct?
Problem. Suppose $T\in\mathcal{L}(V)$ and is such that every nonzero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar mulitple of the identity operator.
Proof. Let $v_1,v_2$ be any two non-zero vectors in $V$, then from hypothesis there exists $\lambda_1,\lambda_2\in\mathbf{F}$ such that $Tv_1 = \lambda_1v_1$ and $Tv_2 = \lambda_2v_2$. Now either $v_2\in\operatorname{span}(v_1)$ or $v_2\not\in\operatorname{span}(v_1)$, in the event of the former, $v_2 = \alpha v_1$ for some $\alpha\in\mathbf{F}$, consequently $Tv_2 = T(\alpha v_1) = \alpha Tv_1 = \alpha(\lambda_1v_1) = \lambda_1(\alpha v_1) = \lambda_1v_2$, but $Tv_2 = \alpha_2v_2$ implying $(\lambda_1-\lambda_2)v = 0$ and by extension $\lambda_1 = \lambda_2$.
We now address the latter case, since $v_2\not\in\operatorname{span}(v_1)$, it follows that $v_1$ and $v_2$ are linearily independent, consequently $v_1+v_2\neq 0$, and so by hypothesis $T(v_1+v_2) = \lambda_3(v_1+v_2)$, problem $\textbf{(25)}$ then implies that $\lambda_1 = \lambda_2$.
In summary then there exists a unique scalar $\lambda_0\in\mathbf{F}$, such that $Tv = \lambda_0v,\forall v\in V-\{0\}$, this in conjunction with theorem $\textbf{3.11}$, then implies that $Tv = \lambda_0v,\forall v\in V$.
$\blacksquare$
Note :
theorem $3.11$ is the result that every linear map on a vector space $V$ maps $0_V$ to $0_V$.
the statement of problem $(25)$ is presented here Implications of $T(u+v)$ when $u$ and $v$ are eigenvectors
I would also appreciate it if the anyone checking the above argument would also take a look at $T = \lambda I_V$ for some $\lambda\in\mathbf{F}$ whenever every subspace of $V$ with dimension $\dim V-1$ is $T$-invariant. , since that problem makes use of the exercise solved in this post.