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Suppose $V$ is finite-dimensional and $T \in L(V)$. Prove that $T$ has an eigenvalue if and only if there exists a subspace of $V$ of dimension $V-1$ that is invariant under $T$.

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Since $T$ and its transpose $T^{*}$ have the same characteristic polynomial, $T$ has an eigenvalue if and only if $T^{*}$ has one. If $T^{*}$ has an eigenvalue $w_1$, then extend to a basis $(w_1, w_2, \ldots, w_n)$ of $V^{*}$ and note that the subspace generated by $\{w_2^{*}, \ldots, w_n^{*}\}$ in $(V^{*})^{*}$ in $(V^{*})^{*}$ is invariant. Then we finish by identifying $V^{**}$ with $V$. On the other hand, if $T^{*}$ has an invariant subspace with basis $(u_2, \ldots, u_n)$, extend with $u_1$ to a basis of $V^{*}$ and note that $u_1^{*}$ is an eigenvector of $T^{**}$.

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Massimiliano Foschi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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