Re: Solutions that are not solutions
- To: mathgroup at smc.vnet.net
- Subject: [mg32093] Re: [mg32049] Solutions that are not solutions
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Sat, 22 Dec 2001 04:23:25 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
You can see more clearly what is going on by using Reduce instead of Solve: In[1]:= X = {{0, y, z}, {y, x, t}, {u, v, w}}; A = {{1, 1, a}, {0, 1, 0}, {0, 0, 1}}; Reduce[{X . A == Transpose[A] . X, Det[X] == 1}, {x, y, z, t, u, v, w}, VerifySolutions -> True] Out[3]= a == 0 && u == 0 && w == -(1/y^2) && z == 0 || u == a*y && x == (-1 + a*t*y^2 + a*v*y^2 - w*y^2)/ (a^2*y^2) && z == a*y && a != 0 && y != 0 You see that the first solution works only if a=0. So the "spurious" solution that you get actually is a solution that works only in the special case a==0. In general Solve tries to avoid such non-generic solutions. However, in some situations it is possible that certain variables get eliminated during the process of finding a solution and then one may obtain non-generic solutions with respect to these eliminated variables (parameters). Note that in your equation the number of solve variables is larger than the number independent variables. You can avoid getting this non-generic solution by reducing the number of solve variables in your equations to just three and treating all the others as parameters, e.g. : In[35]:= Solve[{X . A == Transpose[A] . X, Det[X] == 1}, {x, y, z}, VerifySolutions -> True] Out[35]= {{x -> (-a^2 + a*t*u^2 + a*u^2*v - u^2*w)/(a^2*u^2), z -> u, y -> u/a}} In[36]:= Simplify[{X . A == Transpose[A] . X, Det[X] == 1} /. %] Out[36]= {{True, True}} Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Friday, December 21, 2001, at 05:57 PM, PSi wrote: > I'm trying to solve a system as follows: > X={{0,y,z},{y,x,t},{u,v,w}} > A={{1,1,a},{0,1,0},{0,0,1}} > Solve[{X.A==Transpose[A].X, Det[X]==1}, > {x,y,z,t,u,v,w}] > > Mathematica 4.1 gives two solutions: > X1={{0,y,0},{y,x,t},{0,v,-1/y^2}} > and > X2={{0,y,a*y},{y,(a*t*y^2-1+a*v*y^2-w*y^2)/(a*y)^2,t}, > {a*y,v,w}} > > However, X1 is not a solution: > X1.A-Transpose[A].X1={{0,0,0},{0,0,a*y},{0,-a*y,0}} > > Could anybody explain this behaviour? > Thanks > PSi > > > > > >