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This result is only true if V is a vector space over an algebraically closed field, such as the complex numbers. For example, the map $T:\mathbb{R}^2 \to \mathbb{R}^2$ with matrix $\begin{bmatrix}0&1 \\-1&0\end{bmatrix}$ represents the operation of rotation through a right angle, and it is fairly obvious that there are no nonzero vectors in $\mathbb{R}^2$ whose direction is left unchanged by this map. However, if you allow complex scalars then $(1,i)$ is an eigenvector, with eigenvalue $i$, because $T(1,i) = (i,-1) = i(1,i)$.From wikipedia I read that every linear map from T:V->V, where V is finite dimensional and dim(V) > 1 has an eigenvector. What is the proof ?

So assume that V is a complex vector space. The definition of an eigenvector $x$, with eigenvalue $\lambda$, is that $x\ne0$ and $Tx = \lambda x$. Then the equation $(T-\lambda I)x = 0$ has the nonzero solution $x$, the condition for which is that $\det(T-\lambda I) = 0$. But that is a polynomial equation of degree $n$, and therefore has a root (because $\mathbb{C}$ is algebraically closed).

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