Solutions of equation systems

From: Moreno Mancosu <nomero at tiscali.it>
Subject: [R] Solutions of equation systems
To: r-help at r-project.org
Received: Friday, 14 August, 2009, 2:29 AM
Hello all!

Maybe it's a newbie question(in fact I _am_, a newbie), but
I hope you'll find the solution.
I have an equation system which has k equation and n
variables (k<n).
I would like to obtain a description of the solutions
(which can be the equation of lines or a plane, or
everything else) with the lesser degree of freedom,
obviously using R.
In other words, I would like to obtain a result which is
very similar to the results of Maxima or similar software.
I tried to search on internet and i found the systemfit
package, but I think it is not related with my problem.
Any idea?

Thanks in advance,

Moreno

Is your system of equations linear?
  

?solve

  

Dear Greg,

I tried to use function "solve", but I have some problems with it.
Let's try with a simple equation:

x + 3y -2z = 5
3x + 5y + 6z =7

 > a<-matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3)
 > a

     [,1] [,2] [,3]
[1,]    1   -2    5
[2,]    3    3    6

 > b<-matrix(c(5,7),nrow=2,ncol=1)
 > b

     [,1]
[1,]    5
[2,]    7

When I try to solve this system, I find this error:

 > solve(a,b)

Error in solve.default(a, b) : 'b' must be compatible with 'a'

Also, when i try to solve a system with n equation and n variables, 
solve works perfectly.

 
 
 Moreno Mancosu wrote:

 > 
 > Dear Greg,
 > 
 > I tried to use function "solve", but I have some problems with it.
 > Let's try with a simple equation:
 > 
 > x + 3y -2z = 5
 > 3x + 5y + 6z =7
 > 

 >  > a<-matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3)
 >  > a

 >      [,1] [,2] [,3]
 > [1,]    1   -2    5
 > [2,]    3    3    6

 >  > b<-matrix(c(5,7),nrow=2,ncol=1)
 >  > b

 >      [,1]
 > [1,]    5
 > [2,]    7
 > 
 > When I try to solve this system, I find this error:
 > 

 >  > solve(a,b)

 > Error in solve.default(a, b) : 'b' must be compatible with 'a'
 > 
 > Also, when i try to solve a system with n equation and n variables, 

 > solve works perfectly.
 > 

 
 If you do ?solve then you will see that 
 the matrix a must be non square (it says so in the description of a in
 section Arguments)/
 and in section Details right at the end it is written that "qr.solve 
can
 handle non-square systems".
 
 Your example can be solved in several ways.
 See this
 
 <example>
 A <- matrix(c(1,2,3,5,6,7),nrow=2,byrow=T)
 b <- c(2,8) 
 
 A <- matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3) 
 b <- matrix(c(5,7),nrow=2,ncol=1)
 
 # a particular solution
 xsolve <- qr.solve(A,b)
 xsolve
 
 # another particular solution
 
 xany <- solve(qr(A,LAPACK=T),b)   
 xany
 
 # QR decomposition of A-transposed
 
 ATQR <- qr(t(A),LAPACK=TRUE)
 ATQR
 
 qr.Q(ATQR)
 qr.R(ATQR)
 
 # minimum norm solution comes from QR decomposition of t(A)
 # and solving t(R) %*% t(Q) %*% x = b in two steps
 # ! Lapack does a pivoted QR
 
 z <- solve(t(qr.R(ATQR)),b[ATQR$pivot])
 xmin <- qr.Q(ATQR) %*% z
 xmin
 
 # null space of A  
 library(MASS)
 A.null <- Null(t(A))
 A.null
 
 # test
 A %*% (xmin+5*A.null)  
 </example>
 
 ! Use the manual
 
 Berend
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