Re: Re: Re: Creating a symmetric matrix

*To*: mathgroup at smc.vnet.net*Subject*: [mg46897] Re: [mg46876] Re: [mg46857] Re: [mg46853] Creating a symmetric matrix*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 14 Mar 2004 03:24:18 -0500 (EST)*References*: <200403110850.DAA13986@smc.vnet.net> <200403120702.CAA25478@smc.vnet.net> <200403130439.XAA15074@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

An equivalent but perhaps even simpler version is: A + Transpose[A] - IdentityMatrix[First[Dimensions[A]]*A Andrzej On 13 Mar 2004, at 05:39, Andrzej Kozlowski wrote: > It seems that I was the only one who thinks that a "triangular matrix" > is a "strick triangular matrix", that is with zeros on the diagonal. > had I not made this assumption I would have suggested: > > > A+Transpose[A]*Array[KroneckerDelta[#1 != #2,True]&,Dimensions[A]] > > > Andrzej > > > On 12 Mar 2004, at 08:02, Andrzej Kozlowski wrote: > >> >> On 11 Mar 2004, at 09:50, Mark Coleman wrote: >> >>> Greetings, >>> >>> How can I efficiently build a symmetric matrix from an upper >>> triangular >>> one, i.e., extract the upper triangular elements and insert them into >>> the lower triangle in such a way as to make the resulting square >>> matrix >>> symmetric? >>> >>> Thanks, >>> >>> Mark >>> >>> >>> >> >> The most natural way must be >> >> A+Tranpose[A] >> >> e.g. >> >> A = Array[KroneckerDelta[#1 < #2, True] & , {3, 3}]; >> >> >> {{0, 1, 1}, {0, 0, 1}, {0, 0, 0}} >> >> >> A + Transpose[A] >> >> >> {{0, 1, 1}, {1, 0, 1}, {1, 1, 0}} >> >> >> >> Andrzej Kozlowski >> Chiba, Japan >> http://www.mimuw.edu.pl/~akoz/ >> >> > >

**References**:**Creating a symmetric matrix***From:*Mark Coleman <mark@markscoleman.com>

**Re: Creating a symmetric matrix***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: Creating a symmetric matrix***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>