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Difference between revisions of "Trace of a square matrix"

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The sum of the entries on the main diagonal of this [[Matrix|matrix]]. The trace of a matrix <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935501.png" /> is denoted by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935502.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935503.png" /> or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935504.png" />:
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{{TEX|done}}
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{{MSC|15A15}}
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$
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\newcommand{\tr}{\mathop{\mathrm{tr}}}
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\newcommand{\Tr}{\mathop{\mathrm{Tr}}}
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\newcommand{\Sp}{\mathop{\mathrm{Sp}}}
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$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935505.png" /></td> </tr></table>
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The sum of the entries on the main diagonal of this [[Matrix|matrix]]. The trace of a matrix $A = [a_{ij}]$ is denoted by $\tr A$, $\Tr A$ or $\Sp A$:
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$$
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\tr A = \sum_{i=0}^n a_{ii}
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$$
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Let $A$ be a square matrix of order $n$ over a [[Field|field]] $k$. The trace of $A$ coincides with the sum of the roots of the [[Characteristic polynomial|characteristic polynomial]] of $A$. If $k$ is a field of characteristic 0, then the $n$ traces $\tr A, \ldots \tr A^n$ uniquely determine the characteristic polynomial of $A$. In particular, $A$ is nilpotent if and only if $\tr A^m = 0$ for all $m=1,\ldots,n$.
  
Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935506.png" /> be a square matrix of order <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935507.png" /> over a field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935508.png" />. The trace of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t0935509.png" /> coincides with the sum of the roots of the [[Characteristic polynomial|characteristic polynomial]] of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355010.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355011.png" /> is a field of characteristic 0, then the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355012.png" /> traces <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355013.png" /> uniquely determine the characteristic polynomial of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355014.png" />. In particular, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355015.png" /> is nilpotent if and only if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355016.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355017.png" />.
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If $A$ and $B$ are square matrices of the same order over $k$, and $\alpha,\beta \in k$, then
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$$
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\tr(\alpha A + \beta B) = \alpha \tr A + \beta \tr B, \quad
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\tr AB = \tr BA,
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$$
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while if $\det B \neq 0$,
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$$
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\tr(BAB^{-1}) = \tr A.
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$$
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The trace of the [[Tensor product|tensor (Kronecker) product]] of square matrices over a field is equal to the product of the traces of the factors.
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355018.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355019.png" /> are square matrices of the same order over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355020.png" />, and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355021.png" />, then
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====References====  
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355022.png" /></td> </tr></table>
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{|
 
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|-
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355023.png" /></td> </tr></table>
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|valign="top"|{{Ref|Co}}||valign="top"| P.M. Cohn, "Algebra", '''1''', Wiley (1982) pp. 336
 
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while if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355024.png" />,
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|valign="top"|{{Ref|Ga}}||valign="top"| F.R. [F.R. Gantmakher] Gantmacher, "The theory of matrices", '''1''', Chelsea, reprint (1959) (Translated from Russian)
 
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t093/t093550/t09355025.png" /></td> </tr></table>
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|}
 
 
The trace of the tensor (Kronecker) product of square matrices over a field is equal to the product of the traces of the factors.
 
 
 
 
 
 
 
====Comments====
 
 
 
 
 
====References====
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top"P.M. Cohn,   "Algebra" , '''1''' , Wiley (1982) pp. 336</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top"F.R. [F.R. Gantmakher] Gantmacher,   "The theory of matrices" , '''1''' , Chelsea, reprint (1959) (Translated from Russian)</TD></TR></table>
 

Revision as of 21:32, 3 May 2012

2010 Mathematics Subject Classification: Primary: 15A15 [MSN][ZBL] $ \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\Tr}{\mathop{\mathrm{Tr}}} \newcommand{\Sp}{\mathop{\mathrm{Sp}}} $

The sum of the entries on the main diagonal of this matrix. The trace of a matrix $A = [a_{ij}]$ is denoted by $\tr A$, $\Tr A$ or $\Sp A$: $$ \tr A = \sum_{i=0}^n a_{ii} $$ Let $A$ be a square matrix of order $n$ over a field $k$. The trace of $A$ coincides with the sum of the roots of the characteristic polynomial of $A$. If $k$ is a field of characteristic 0, then the $n$ traces $\tr A, \ldots \tr A^n$ uniquely determine the characteristic polynomial of $A$. In particular, $A$ is nilpotent if and only if $\tr A^m = 0$ for all $m=1,\ldots,n$.

If $A$ and $B$ are square matrices of the same order over $k$, and $\alpha,\beta \in k$, then $$ \tr(\alpha A + \beta B) = \alpha \tr A + \beta \tr B, \quad \tr AB = \tr BA, $$ while if $\det B \neq 0$, $$ \tr(BAB^{-1}) = \tr A. $$ The trace of the tensor (Kronecker) product of square matrices over a field is equal to the product of the traces of the factors.

References

[Co] P.M. Cohn, "Algebra", 1, Wiley (1982) pp. 336
[Ga] F.R. [F.R. Gantmakher] Gantmacher, "The theory of matrices", 1, Chelsea, reprint (1959) (Translated from Russian)
How to Cite This Entry:
Trace of a square matrix. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Trace_of_a_square_matrix&oldid=25902
This article was adapted from an original article by D.A. Suprunenko (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article