Difference between revisions of "Semi-simple endomorphism"
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An endomorphism $\alpha$ of $V$ with the following property: For any $\alpha$-invariant subspace $W$ of $V$ there exists an $\alpha$-invariant subspace $W'$ such that $V$ is the direct sum of $W$ and $W'$. In other words, $V$ should be a [[semi-simple module]] over the ring $K[X]$ with $X$ acting as $\alpha$. For example, any orthogonal, symmetric or skew-symmetric linear transformation of a finite-dimensional Euclidean space, and also any diagonalizable (i.e. representable by a diagonal matrix with respect to some basis) linear transformation of a finite-dimensional vector space, is a semi-simple endomorphism. The semi-simplicity of an endomorphism is preserved by passage to an invariant subspace $W \subset V$, and to the quotient space $V/W$. | An endomorphism $\alpha$ of $V$ with the following property: For any $\alpha$-invariant subspace $W$ of $V$ there exists an $\alpha$-invariant subspace $W'$ such that $V$ is the direct sum of $W$ and $W'$. In other words, $V$ should be a [[semi-simple module]] over the ring $K[X]$ with $X$ acting as $\alpha$. For example, any orthogonal, symmetric or skew-symmetric linear transformation of a finite-dimensional Euclidean space, and also any diagonalizable (i.e. representable by a diagonal matrix with respect to some basis) linear transformation of a finite-dimensional vector space, is a semi-simple endomorphism. The semi-simplicity of an endomorphism is preserved by passage to an invariant subspace $W \subset V$, and to the quotient space $V/W$. | ||
| − | Let $\dim V < \infty$. An endomorphism $\alpha$ is semi-simple if and only if its | + | Let $\dim V < \infty$. An endomorphism $\alpha$ is semi-simple if and only if its minimal polynomial (cf. [[Minimal polynomial of a matrix]]) has no multiple factors. Let $L$ be an extension of the field $K$ and let $\alpha_L$ be the extension of the endomorphism $\alpha \otimes 1$ to the space $V_L = V \otimes L$. If $\alpha_L$ is semi-simple, then $\alpha$ is also semi-simple, and if $L$ is separable over $K$, then the converse is true. An endomorphism $\alpha$ is called absolutely semi-simple if $\alpha_L$ is semi-simple for any extension $L/K$; for this it is necessary and sufficient that the minimal polynomial has no multiple roots in the algebraic closure $\bar K$ of $K$, that is, that the endomorphism $\alpha_{\bar K}$ is diagonalizable. |
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| − | Over an algebraically closed field any endomorphism $\alpha$ of a finite-dimensional vector space can be decomposed into a sum $\alpha = | + | Over an algebraically closed field any endomorphism $\alpha$ of a finite-dimensional vector space can be decomposed into a sum $\alpha = \sigma + \nu$ of a semi-simple endomorphism $\sigma$ and a nilpotent one $\nu$ such that $\sigma \nu = \nu \sigma$; cf. [[Jordan decomposition]], 2). |
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| + | The matrix of a semi-simple endomorphism of a finite-dimensional vector space with respect to any basis is a [[semi-simple matrix]]. | ||
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Latest revision as of 18:14, 12 November 2017
semi-simple linear transformation, of a vector space $V$ over a field $K$
An endomorphism $\alpha$ of $V$ with the following property: For any $\alpha$-invariant subspace $W$ of $V$ there exists an $\alpha$-invariant subspace $W'$ such that $V$ is the direct sum of $W$ and $W'$. In other words, $V$ should be a semi-simple module over the ring $K[X]$ with $X$ acting as $\alpha$. For example, any orthogonal, symmetric or skew-symmetric linear transformation of a finite-dimensional Euclidean space, and also any diagonalizable (i.e. representable by a diagonal matrix with respect to some basis) linear transformation of a finite-dimensional vector space, is a semi-simple endomorphism. The semi-simplicity of an endomorphism is preserved by passage to an invariant subspace $W \subset V$, and to the quotient space $V/W$.
Let $\dim V < \infty$. An endomorphism $\alpha$ is semi-simple if and only if its minimal polynomial (cf. Minimal polynomial of a matrix) has no multiple factors. Let $L$ be an extension of the field $K$ and let $\alpha_L$ be the extension of the endomorphism $\alpha \otimes 1$ to the space $V_L = V \otimes L$. If $\alpha_L$ is semi-simple, then $\alpha$ is also semi-simple, and if $L$ is separable over $K$, then the converse is true. An endomorphism $\alpha$ is called absolutely semi-simple if $\alpha_L$ is semi-simple for any extension $L/K$; for this it is necessary and sufficient that the minimal polynomial has no multiple roots in the algebraic closure $\bar K$ of $K$, that is, that the endomorphism $\alpha_{\bar K}$ is diagonalizable.
References
| [1] | N. Bourbaki, "Algèbre" , Eléments de mathématiques , Hermann (1970) pp. Chapts. I-III |
Comments
Over an algebraically closed field any endomorphism $\alpha$ of a finite-dimensional vector space can be decomposed into a sum $\alpha = \sigma + \nu$ of a semi-simple endomorphism $\sigma$ and a nilpotent one $\nu$ such that $\sigma \nu = \nu \sigma$; cf. Jordan decomposition, 2).
The matrix of a semi-simple endomorphism of a finite-dimensional vector space with respect to any basis is a semi-simple matrix.
Semi-simple endomorphism. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Semi-simple_endomorphism&oldid=36910