Is the Following argument correct?
Problem. Suppose that $V$ is finite dimenisonal, and $T\in\mathcal{L}(V)$ is such that every subspace of $V$ with dimension $\dim V - 1$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.
Proof. Owing to the trivial nature of the proof for the case $\dim V = 1$. We construct our argument to address only the case where $\dim V = n\ge 2$.
Let $v_0$ be an arbitrary non-zero vector in $V$, and extend the list $v_0$ to yield the basis $v_0,v_1,v_2,\dots,v_{\dim V-1}$ for $V$. Now define the subspaces $\mathcal{A}_0,\mathcal{A}_1,\mathcal{A}_2,\dots,\mathcal{A}_{\dim V-1}$ such that $A_k$ is the span of all basis vectors, excluding the basis vector $v_k$, that is to say $$ \begin{cases} \mathcal{A}_o = \operatorname{span}(v_1,v_2,\dots,v_{\dim V-1})\\ \mathcal{A}_1 = \operatorname{span}(v_0,v_2,\dots,v_{\dim V-1})\\ \mathcal{A}_2 = \operatorname{span}(v_0,v_1,\dots,v_{\dim V-1})\\ \vdots\\ \mathcal{A}_{\dim V-1} = \operatorname{span}(v_0,v_1,\dots,v_{\dim V-2})\\ \end{cases} $$ Prior to proceeding further with our argument, we first note that owing to the linear independence of $v_0,v_1,v_2,\dots,v_{\dim V-1}$, we have $\dim \mathcal{A}_k = \dim V-1,\forall k$, consequently all the subspaces $\mathcal{A}_0,\mathcal{A}_1,\mathcal{A}_2,\dots,\mathcal{A}_{\dim V-1}$ are invariant under $T$, an appeal to $\textbf{(5)}$ then implies that the subspace $\mathcal{K}$ described below, is a invariant under $T$. $$\mathcal{K} = \bigcap_{k=1}^{\dim V-1}\mathcal{A}_{k}$$ We now prove $\mathcal{K} = \operatorname{span}(v_0)$. Assume $v\in\mathcal{K}$, since $v_0,v_1,v_2,\dots,v_{\dim V-1}$ is a basis for $V$, we have $v = \lambda_0v_0+\lambda_1v_1+\cdots+\lambda_{\dim V-1}v_{\dim V-1}$ for some $\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{\dim V-1}\in\mathbf{F}$, now let $j\in\{1,2,\dots,\dim V-1\}$ since $v\in\mathcal{K}$, then in particular $v\in\mathcal{A}_j$ and so $v = \alpha_0v_1+\alpha_1v_1+\cdots+\alpha_{\dim V-1}v_{\dim V-1}$ for some scalars $\alpha_0,\alpha_1,\dots,\alpha_{\dim V-1}\in\mathbf{F}$, where ofcourse $\alpha_{j} = 0$, but then we have $$(\lambda_0-\alpha_0)v_0+(\lambda_1-\alpha_1)v_1+\cdots+\lambda_jv_j+\cdots+(\lambda_{\dim V-1}-\alpha_{\dim V-1})v_{\dim V-1} = 0$$ Now appealing to the fact that $v_0,v_1,v_2,\dots,v_{\dim V-1}$ is linearily independent it follows that it follows that $\lambda_j = 0$, since our choice of $j$ was arbitrary it follows that $\lambda_1 = \lambda_2 = \cdots = \lambda_{\dim V-1} = 0$, thus $v\in\operatorname{span}(v_0)$ and by extension $\mathcal{K}\subseteq\operatorname{span}(v_0)$. Since $\operatorname{span}(v_0)\subseteq\mathcal{K}$ is trivial it follows that $\mathcal{K} = \operatorname{span}(v_0)$.
The $T$-invariance of $\mathcal{K}$ in conjunction with the above result then implies that $v_0$ is an eigenvector of $T$, since our choice of the nonzero vector $v_0$ was arbitrary it follows every non-zero vector in $V$ is an eigenvector of $T$, appealing to $\textbf{(27)}$, then yields the required result.
$\blacksquare$
NOTE:
$T\in\mathcal{L}(V)$, denotes that $T$ is a linear map on the vector space $V$.
$(27)$ - If $T\in\mathcal{L}(V)$ and is such that every nonzero vector in $V$ is an eigenvector of $T$, then $T$ is a scalar mulitple of the identity operator.
$(5)$ - If $T\in\mathcal{L}(V)$. The intersection of any collection of $T$-invariant subspace is another $T$-invariant subspace.