In mathematics, a linear operator T : V → V on a vector space V is semisimple if every T-invariant subspace has a complementary T-invariant subspace. If T is a semisimple linear operator on V, then V is a semisimple representation of T. Equivalently, a linear operator is semisimple if its minimal polynomial is a product of distinct irreducible polynomials.
A linear operator on a finite-dimensional vector space over an algebraically closed field is semisimple if and only if it is diagonalizable.
Over a perfect field, the Jordan–Chevalley decomposition expresses an endomorphism as a sum of a semisimple endomorphism s and a nilpotent endomorphism n such that both s and n are polynomials in x.
as a sum of a semisimple endomorphism s and a Notes
- ^ Lam (2001), p. 39
- Jacobson 1979, A paragraph before Ch. II, § 5, Theorem 11.
- This is trivial by the definition in terms of a minimal polynomial but can be seen more directly as follows. Such an operator always has an eigenvector; if it is, in addition, semi-simple, then it has a complementary invariant hyperplane, which itself has an eigenvector, and thus by induction is diagonalizable. Conversely, diagonalizable operators are easily seen to be semi-simple, as invariant subspaces are direct sums of eigenspaces, and any basis for this space can be extended to an eigenbasis.
References
- Hoffman, Kenneth; Kunze, Ray (1971). "Semi-Simple operators". Linear algebra (2nd ed.). Englewood Cliffs, N.J.: Prentice-Hall, Inc. MR 0276251.
- Jacobson, Nathan (1979). Lie algebras. New York. ISBN 0-486-63832-4. OCLC 6499793.
{{cite book}}: CS1 maint: location missing publisher (link) - Lam, Tsit-Yuen (2001). A first course in noncommutative rings. Graduate texts in mathematics. Vol. 131 (2 ed.). Springer. ISBN 0-387-95183-0.
 | This linear algebra-related article is a stub. You can help Misplaced Pages by expanding it. |