Semisimple operator


In mathematics, a linear operator T on a vector space is semisimple if every T-invariant subspace has a complementary T-invariant subspace;[1] in other words, the vector space is a semisimple representation of the operator T. Equivalently, a linear operator is semisimple if the minimal polynomial of it is a product of distinct irreducible polynomials.[2]

A linear operator on a finite dimensional vector space over an algebraically closed field is semisimple if and only if it is diagonalizable.[1][3]

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.

See alsoEdit


  1. ^ a b Lam (2001), p. 39
  2. ^ Jacobson 1979, A paragraph before Ch. II, § 5, Theorem 11.
  3. ^ 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.


  • 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.
  • Lam, Tsit-Yuen (2001). A first course in noncommutative rings. Graduate texts in mathematics. Vol. 131 (2 ed.). Springer. ISBN 0-387-95183-0.