# A linear transformation on a finite dimentional nontrivial vector space has at least one eigenvalue.

Put content here**Theorem:** Every linear transformation \(T: V 	o V\) on a finite-dimensional nontrivial vector space (\(\dim V \geq 1\)) over an algebraically closed field (such as \(\mathbb{C}\)) has at least one eigenvalue.
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**Proof sketch:** The characteristic polynomial \(p_T(\lambda) = \det(A - \lambda I)\) is a polynomial of degree \(n = \dim V \geq 1\). By the Fundamental Theorem of Algebra, every non-constant polynomial over \(\mathbb{C}\) has at least one root. That root is an eigenvalue.
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**Important caveat:** Over \(\mathbb{R}\), this may fail. For example, the rotation \(T(x,y) = (-y, x)\) in \(\mathbb{R}^2\) has characteristic polynomial \(\lambda^2 + 1\), which has no real roots. The same transformation over \(\mathbb{C}\) has eigenvalues \(i\) and \(-i\).
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**Consequence:** This theorem guarantees that eigenvalue analysis is always possible when working over \(\mathbb{C}\).

# Parents

* Eigenvalues and eigenvectors