statement theorem
Existence of an eigenvalue over an algebraically closed field
Statement
Let be an algebraically closed field, let be a finite-dimensional vector space over with , and let be a linear map. Then has at least one eigenvalue in .
Remarks
The characteristic polynomial of has degree , so it has a root in . The statement applies to , by the fundamental theorem of algebra. It fails over : the rotation of by degrees has no real eigenvalue. It also fails in infinite dimensions: the map on has no eigenvalue.
Explanations
L1 human
Over the complex numbers, every square matrix has at least one eigenvector. The reason is that every nonconstant complex polynomial has a root. Over the real numbers, a rotation can move every line, so no real eigenvector exists.
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What this needs
To read this node, read these 21 nodes in this order. The foundations come first, and the list ends with this node.
- no evidence level
- Identity matrix object
- Matrix definition
- Vector space definition
- Square matrix definition
- Linear combination definition
- Linear map definition
- Linear subspace definition
- Linear independence definition
- Endomorphism definition
- Linear span definition
- Multilinear map definition
- Eigenvalue definition
- Spanning set definition
- Alternating multilinear map definition
- Basis definition
- Determinant definition
- Coordinate vector definition
- Characteristic polynomial definition
- Matrix of a linear map definition
- Characteristic polynomial of a linear map definition
- Existence of an eigenvalue over an algebraically closed field statement
Sources
- Web https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors, Eigenvalues and characteristic polynomial
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