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I got the following problem

Let $V$ be a vector space over field $\mathbb{F}$ and let $T:V \to V$ be a linear transformation such that every subspace of $V$ is invariant under $T$, Show that there exist a scalar $\alpha \in \mathbb{F}$ such that $T = \alpha I$ (meaning $T$ is a scalar transformation)

I tried to show it but I got that the matrix for $T$ in some basis $B$ is a diagonal matrix where each entry on the main diagonal is some eigenvalue $\lambda _i$, How do I show that all the eigenvalues are equal to each other?

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    $\begingroup$ Hint: You can assume the dimension of $V$ is at least $2$. If $v$ and $w$ and independent vectors, consider the subspaces generated by $v$, $w$ and $v+w$. $\endgroup$ – Tobias Kildetoft Jan 8 '14 at 13:01
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For each vector $v \in V$, there exists $\alpha_v$ such that $Tv=\alpha_v v$ (because $Tv \in span({v})$). Suppose there are vectors $u,v \in V$ such that $\alpha_u \neq \alpha_v$. Then $$T(u+v)=Tu+Tv \Rightarrow \alpha_{u+v} (u+v)=\alpha_u u +\alpha_v v \Rightarrow (\alpha_{u+v}-\alpha_u)u+(\alpha_{u+v}-\alpha_v)v=0$$ which means $(u,v)$ is linearly dependent (since both coefficients cannot be zero simultaneously, by assumption). Therefore, either $\alpha_u=\alpha_v$, and $T$ is a scalar transformation, or $V$ has a dimension of at most 1, which also implies $Tx=\alpha x$, with constant $\alpha$.

Hence, $T=\alpha I, \alpha \in \mathbb{F}$

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