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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 $Tv = \lambda_0v,\forall v\in V$.

$\blacksquare$

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