r - 削減された列のエシュロン形式

Question

reduced row echelon form行列のを生成する R の関数はありますか? この参照は、そうではないと言っています。同意しますか？

Answer 1

pracma パッケージには実装も含まれています。pracma::rref を参照してください。

Answer 2

また、線形代数を教えるために開発された最近のパッケージ ( matlib ) もあります。これは、行列のエシェロン形式を計算し、途中で使用されるステップを示します。

リファレンス ドキュメントの例:

library('matlib')
A <- matrix(c(2, 1, -1,-3, -1, 2,-2, 1, 2), 3, 3, byrow=TRUE)
b <- c(8, -11, -3)
echelon(A, b, verbose=TRUE, fractions=TRUE)

Initial matrix:
     [,1] [,2] [,3] [,4]
[1,]   2    1   -1    8 
[2,]  -3   -1    2  -11 
[3,]  -2    1    2   -3 

row: 1 

 exchange rows 1 and 2 
     [,1] [,2] [,3] [,4]
[1,]  -3   -1    2  -11 
[2,]   2    1   -1    8 
[3,]  -2    1    2   -3 

 multiply row 1 by -1/3 
     [,1] [,2] [,3] [,4]
[1,]    1  1/3 -2/3 11/3
[2,]    2    1   -1    8
[3,]   -2    1    2   -3

 multiply row 1 by 2 and subtract from row 2 
     [,1] [,2] [,3] [,4]
[1,]    1  1/3 -2/3 11/3
[2,]    0  1/3  1/3  2/3
[3,]   -2    1    2   -3

 multiply row 1 by 2 and add to row 3 
     [,1] [,2] [,3] [,4]
[1,]    1  1/3 -2/3 11/3
[2,]    0  1/3  1/3  2/3
[3,]    0  5/3  2/3 13/3

row: 2 

 exchange rows 2 and 3 
     [,1] [,2] [,3] [,4]
[1,]    1  1/3 -2/3 11/3
[2,]    0  5/3  2/3 13/3
[3,]    0  1/3  1/3  2/3

 multiply row 2 by 3/5 
     [,1] [,2] [,3] [,4]
[1,]    1  1/3 -2/3 11/3
[2,]    0    1  2/5 13/5
[3,]    0  1/3  1/3  2/3

 multiply row 2 by 1/3 and subtract from row 1 
     [,1] [,2] [,3] [,4]
[1,]    1    0 -4/5 14/5
[2,]    0    1  2/5 13/5
[3,]    0  1/3  1/3  2/3

 multiply row 2 by 1/3 and subtract from row 3 
     [,1] [,2] [,3] [,4]
[1,]    1    0 -4/5 14/5
[2,]    0    1  2/5 13/5
[3,]    0    0  1/5 -1/5

row: 3 

 multiply row 3 by 5 
     [,1] [,2] [,3] [,4]
[1,]    1    0 -4/5 14/5
[2,]    0    1  2/5 13/5
[3,]    0    0    1   -1

 multiply row 3 by 4/5 and add to row 1 
     [,1] [,2] [,3] [,4]
[1,]    1    0    0    2
[2,]    0    1  2/5 13/5
[3,]    0    0    1   -1

 multiply row 3 by 2/5 and subtract from row 2 
     [,1] [,2] [,3] [,4]
[1,]  1    0    0    2  
[2,]  0    1    0    3  
[3,]  0    0    1   -1  

Answer 3

組み込まれているようには見えませんが、このページでこのrref関数を見つけました。

 rref <- function(A, tol=sqrt(.Machine$double.eps),verbose=FALSE,
                 fractions=FALSE){
  ## A: coefficient matrix
  ## tol: tolerance for checking for 0 pivot
  ## verbose: if TRUE, print intermediate steps
  ## fractions: try to express nonintegers as rational numbers
  ## Written by John Fox
  if (fractions) {
    mass <- require(MASS)
    if (!mass) stop("fractions=TRUE needs MASS package")
  }
  if ((!is.matrix(A)) || (!is.numeric(A)))
    stop("argument must be a numeric matrix")
  n <- nrow(A)
  m <- ncol(A)
  for (i in 1:min(c(m, n))){
    col <- A[,i]
    col[1:n < i] <- 0
    # find maximum pivot in current column at or below current row
    which <- which.max(abs(col))
    pivot <- A[which, i]
    if (abs(pivot) <= tol) next     # check for 0 pivot
    if (which > i) A[c(i, which),] <- A[c(which, i),]  # exchange rows
    A[i,] <- A[i,]/pivot            # pivot
    row <- A[i,]
    A <- A - outer(A[,i], row)      # sweep
    A[i,] <- row                    # restore current row
    if (verbose)
      if (fractions) print(fractions(A))
      else print(round(A,round(abs(log(tol,10)))))
  }
  for (i in 1:n)
    if (max(abs(A[i,1:m])) <= tol)
      A[c(i,n),] <- A[c(n,i),] # 0 rows to bottom
  if (fractions) fractions (A)
  else round(A, round(abs(log(tol,10))))
}