KNOWPIA
WELCOME TO KNOWPIA

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*.

- ^
^{a}^{b}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.

- 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.