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| − | ''semi-simple linear transformation, of a vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843601.png" /> over a field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843602.png" />'' | + | ''semi-simple linear transformation, of a vector space $V$ over a field $K$'' |
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| − | An endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843603.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843604.png" /> with the following property: For any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843605.png" />-invariant subspace <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843606.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843607.png" /> there exists an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843608.png" />-invariant subspace <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s0843609.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436010.png" /> is the direct sum of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436011.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436012.png" />. In other words, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436013.png" /> should be a [[Semi-simple module|semi-simple module]] over the ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436014.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436015.png" /> acting as <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436016.png" />. 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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436017.png" />, and to the quotient space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436018.png" />. | + | 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 $V/W$. |
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| − | Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436019.png" />. An endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436020.png" /> is semi-simple if and only if its minimum polynomial (cf. [[Matrix|Matrix]]) has no multiple factors. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436021.png" /> be an extension of the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436022.png" /> and let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436023.png" /> be the extension of the endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436024.png" /> to the space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436025.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436026.png" /> is semi-simple, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436027.png" /> is also semi-simple, and if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436028.png" /> is separable over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436029.png" />, then the converse is true. An endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436030.png" /> is called absolutely semi-simple if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436031.png" /> is semi-simple for any extension <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436032.png" />; for this it is necessary and sufficient that the minimum polynomial has no multiple roots in the algebraic closure <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436033.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436034.png" />, that is, that the endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436035.png" /> is diagonalizable. | + | 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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| | ====References==== | | ====References==== |
| − | <table><TR><TD valign="top">[1]</TD> <TD valign="top"> N. Bourbaki, "Algèbre" , ''Eléments de mathématiques'' , Hermann (1970) pp. Chapts. I-III</TD></TR></table> | + | <table> |
| | + | <TR><TD valign="top">[1]</TD> <TD valign="top"> N. Bourbaki, "Algèbre" , ''Eléments de mathématiques'' , Hermann (1970) pp. Chapts. I-III</TD></TR> |
| | + | </table> |
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| | + | ====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). |
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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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| − | ====Comments====
| + | {{TEX|done}} |
| − | Over an algebraically closed field any endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436036.png" /> of a finite-dimensional vector space can be decomposed into a sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436037.png" /> of a semi-simple endomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436038.png" /> and a nilpotent one <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436039.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s084/s084360/s08436040.png" />; cf. [[Jordan decomposition|Jordan decomposition]], 2).
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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 |
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.
How to Cite This Entry:
Semi-simple endomorphism. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Semi-simple_endomorphism&oldid=16870
This article was adapted from an original article by A.L. Onishchik (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article