# Show that a vector space has an eigenvalue if and only if there exists $\dim V-1$ invariant subspace [closed]

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 $$T^{**}$$.