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[再寄小读者之数学篇](2014-05-27 矩阵的迹与 Jacobian)

2024-07-06 06:14:28   223人阅读

(from MathFlow) 设 A=(aij)<script id="MathJax-Element-1" type="math/tex">A=(a_{ij})</script>, 且定义

?Af(A)=(?f?aij).
<script id="MathJax-Element-2" type="math/tex; mode=display">\bex \n_A f(A)=\sex{\cfrac{\p f}{\p a_{ij}}}. \eex</script> 试证: (1) ?Atr(AB)=Bt<script id="MathJax-Element-3" type="math/tex">\n_A\tr (AB)=B^t</script>; (2) ?Atr(ABAtC)=CAB+CtABt<script id="MathJax-Element-4" type="math/tex">\n_A \tr(ABA^tC)=CAB+C^tAB^t</script>.

证明: (1)

?Atr(AB)=????aij∑m,namnbnm??=??∑m,nδmiδnjbnm??=(bji)=Bt.
<script id="MathJax-Element-5" type="math/tex; mode=display">\beex \bea \n_A\tr (AB) &=\sex{\cfrac{\p }{\p a_{ij}}\sum_{m,n}a_{mn}b_{nm}}\\ &=\sex{\sum_{m,n} \delta_{mi}\delta_{nj}b_{nm}}\\ &=\sex{b_{ji}}\\ &=B^t. \eea \eeex</script> (2)
?Atr(ABAtC)=????aij∑m,n,p,qamnbnpaqpcqm??=??∑m,n,p,qδmiδnjbnpaqpcqm+∑m,n,p,qamnbnpδqiδpjcqm??=??∑p,qbjpaqpcqi+∑m,namnbnjcim??=??∑p,qcqiaqpbjp+∑m,ncimamnbnj??=CtABt+CAB.
<script id="MathJax-Element-6" type="math/tex; mode=display">\beex \bea \n_A\tr (ABA^tC) &=\sex{\cfrac{\p }{\p a_{ij}} \sum_{m,n,p,q} a_{mn}b_{np}a_{qp}c_{qm} }\\ &=\sex{ \sum_{m,n,p,q} \delta_{mi}\delta_{nj}b_{np}a_{qp}c_{qm} +\sum_{m,n,p,q} a_{mn}b_{np}\delta_{qi}\delta_{pj}c_{qm} }\\ &=\sex{ \sum_{p,q} b_{jp}a_{qp}c_{qi} +\sum_{m,n} a_{mn}b_{nj}c_{im} }\\ &=\sex{ \sum_{p,q} c_{qi}a_{qp}b_{jp} +\sum_{m,n} c_{im}a_{mn}b_{nj} }\\ &=C^tAB^t+CAB. \eea \eeex</script>


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