Re: Solving complicated matrix equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg132047] Re: Solving complicated matrix equations*From*: Ray Koopman <koopman at sfu.ca>*Date*: Mon, 25 Nov 2013 23:58:13 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net

The difference between * and . does matter. If we change the * to a . then we have M.(v-m1) - u.M = 0, which is a form of the Sylvester equation AX + XB = C for which efficient methods are known. ----- Dmitry Smirnov <dsmirnov90 at gmail.com> wrote: > > I do really mean M*m1, not M.m1 . However, I believe that this is not > crucial for my example . Let, for simplicity, m1, u, v be real full rank > matrices. > > The only way, that I know, to solve the problem is to define every element > of M as a variable and then to solve a huge system of linear equations. > > 2013/11/23 Ray Koopman <koopman at sfu.ca> > >> Do you really mean M*m1, or should that be M.m1 ? >> Also, what can you tell us about m1, u, v ? >> Are they Real? Full rank? Etc. >> >> ----- dsmirnov90 at gmail.com wrote: >>> Hi! >>> >>> I have to solve analytically an equation like this: >>> >>> M*m1 = M.v - u.M, >>> >>> where m1, v, u are the given matrices and M should be found. Is there >>> any standard function in Mathematica to do this? >>> >>> Thanks, >>> Dmitry.