Re: SingularValueDecomposition

*To*: mathgroup at smc.vnet.net*Subject*: [mg119723] Re: SingularValueDecomposition*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sat, 18 Jun 2011 19:57:20 -0400 (EDT)*References*: <201106181015.GAA15305@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

SingularValueDecomposition seems to HANG for that matrix, so my next try was solving directly: v = Array[x, 14] {x[1], x[2], x[3], x[4], x[5], x[6], x[7], x[8], x[9], x[10], x[11], x[12], x[13], x[14]} soln = Quiet@Solve[Flatten@{Thread[v - m.v == 0], Plus @@ v == 1}, v] soln[[1, All, -1]] // Total {{x[2] -> 132/7967 + (6119 x[1])/7967, x[3] -> x[1], x[4] -> 193/7967 + (5265 x[1])/7967, x[5] -> 396/7967 + (2423 x[1])/7967, x[6] -> 193/7967 + (5265 x[1])/7967, x[7] -> 508/7967 + (855 x[1])/7967, x[8] -> 802/7967 - (3261 x[1])/7967, x[9] -> 508/7967 + (855 x[1])/7967, x[10] -> 1047/7967 - (6691 x[1])/7967, x[11] -> 1390/7967 - (11493 x[1])/7967, x[12] -> 1047/7967 - (6691 x[1])/7967, x[13] -> -(325/7967) + (12517 x[1])/7967, x[14] -> 2076/7967 - (21097 x[1])/7967}} 1 - x[1] x[1] has to be zero, and that doesn't seem like what we're looking for, if all states communicate. (Do they?) You can try something with SingularValueDecomposition@N@m (which doesn't hang), but this may also be instructive: {values, vectors} = Eigensystem@m; vectors = Normalize[#, Total] & /@ vectors; p = Transpose@vectors; d = DiagonalMatrix@values; m.p == p.d True steady = vectors[[Flatten[Position[values, 1, 1], 1]]]; m.# == # & /@ steady Total /@ steady {True, True} {1, 1} With TWO unit eigenvalues, this markov chain is not of the sort you may want, with a single steady state reached from any initial conditions. p is also not invertible: Det@p 0 Hence, we can't use the handy trick that PowerMatrix[m, n] == p . d^n . Inverse@p Good luck. Bobby On Sat, 18 Jun 2011 05:15:18 -0500, John Snyder <jsnyder at wi.rr.com> wrote: > I read Jon McLoone's recent post on the WolframBlog concerning the > solution of the drunken sailor's walk problem by using a Markov chain > transition probability matrix. He mentions that it may also be possible > to solve the problem using the SingularValueDecomposition function, but > he does not illustrate this. I am trying to figure out how this could be > done. > > Here is a simple "toy" example. Assume that I have the following Markov > chain transition probability matrix m where each row sums to 1: > > > m={{1/4,1/4,0,1/4,0,0,0,0,0,0,0,0,1/4,0},{1/4,1/4,1/4,0,1/4,0,0,0,0,0,0,0,0,0},{0,1/4,1/4,0,0,1/4,0,0,0,0,0,0,1/4,0},{0,0,0,1/4,1/4,0,1/4,0,0,0,0,0,1/4,0},{0,0,0,1/4,1/4,1/4,0,1/4,0,0,0,0,0,0},{0,0,0,0,1/4,1/4,0,0,1/4,0,0,0,1/4,0},{0,0,0,0,0,0,1/4,1/4,0,1/4,0,0,1/4,0},{0,0,0,0,0,0,1/4,1/4,1/4,0,1/4,0,0,0},{0,0,0,0,0,0,0,1/4,1/4,0,0,1/4,1/4,0},{0,0,0,0,0,0,0,0,0,1/4,1/4,0,1/4,1/4},{0,0,0,0,0,0,0,0,0,1/4,1/4,1/4,0,1/4},{0,0,0,0,0,0,0,0,0,0,1/4,1/4,1/4,1/4},{0,0,0,0,0,0,0,0,0,0,0,0,1,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,1}}; > > Assuming that I start in the position 2 (column 2 out of 3, in the first > of 4 rows) I want to find the so-called "fixed point", the ultimate > state density function, as the number of steps goes to infinity. I know > that I can do this numerically using MatrixPower as follows (here is 100 > steps which appears to be more than enough in this case): > > In[19]:= MatrixPower[N[m],100][[2]]//Chop > Out[19]= {0,0,0,0,0,0,0,0,0,0,0,0,0.809663,0.190337} > > I believe that it is also possible to get this same result by using the > SingularValueDecomposition function, but I cannot figure out how to get > this to work. Can someone please show me how to use > SingularValueDecomposition to get the same answer to this question? I > know there are other ways to solve this, but I am really interested in > using SingularValueDecomposition in this case. Thanks. > > -- DrMajorBob at yahoo.com

**References**:**SingularValueDecomposition***From:*John Snyder <jsnyder@wi.rr.com>

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