Forma escalonada de escala reducida

Question

¿Existe una función en R que produzca el reduced row echelon form de una matriz ?. This referencia dice que no hay. ¿Estás de acuerdo?

Answer 1

El paquete pracma también contiene una implementación. Ver pracma :: rref.

Answer 2

Parece que no hay uno incorporado, pero encontré esta función rref en la página this.

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)))) 
} 

Answer 3

No tengo suficientes representantes para comentar, pero la función dada anteriormente en la respuesta aceptada es incorrecta: no maneja matrices donde la solución RREF tiene ceros en su diagonal principal. Pruebe, por ejemplo,

m < -matrix (c (1,0,1,0,0,2), byrow = TRUE, nrow = 2) rref (m)

y tenga en cuenta que la salida no está en RREF .

creo que tengo que trabajar, pero es posible que desee comprobar las salidas por sí mismo:

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 
    # Modified by Geoffrey Brent 2014-12-17 to fix a bug 
    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) 
    x.position<-1 
    y.position<-1 
    # change loop: 
    while((x.position<=m) & (y.position<=n)){ 
    col <- A[,x.position] 
    col[1:n < y.position] <- 0 
    # find maximum pivot in current column at or below current row 
    which <- which.max(abs(col)) 
    pivot <- col[which] 
    if (abs(pivot) <= tol) x.position<-x.position+1  # check for 0 pivot 
    else{ 
     if (which > y.position) { A[c(y.position,which),]<-A[c(which,y.position),] } # exchange rows 
     A[y.position,]<-A[y.position,]/pivot # pivot 
     row <-A[y.position,] 
     A <- A - outer(A[,x.position],row) # sweep 
     A[y.position,]<-row # restore current row 
     if (verbose) 
     if (fractions) print(fractions(A)) 
     else print(round(A,round(abs(log(tol,10))))) 
     x.position<-x.position+1 
     y.position<-y.position+1 
    } 
    } 
    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)))) 
} 

Answer 4

También hay un paquete de reciente desarrollado para enseñanza Álgebra Lineal (matlib), que tanto calcula la forma escalonada de una matriz, y muestra los pasos utilizados en el camino.

Ejemplo de la reference docs:

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