Data 605 Week 2 Discussion

Excercise C10 Page 249

C10y Find a basis for S, where S =

ss <- matrix(c(1,3,2,1,1,2,1,1,1,1,0,1,1,2,2,1,3,4,1,3),nrow=4, ncol=5)
ss

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    1    3
## [2,]    3    2    1    2    4
## [3,]    2    1    0    2    1
## [4,]    1    1    1    1    3

Set to reduce row echelon form:

s<-ss
s[2,]<-(3*s[1,]-s[2,])
s

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    1    3
## [2,]    0    1    2    1    5
## [3,]    2    1    0    2    1
## [4,]    1    1    1    1    3

s[3,]<-(2*s[1,]-s[3,])
s

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    1    3
## [2,]    0    1    2    1    5
## [3,]    0    1    2    0    5
## [4,]    1    1    1    1    3

s[3,]<-(s[2,]-s[3,])
s

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    1    3
## [2,]    0    1    2    1    5
## [3,]    0    0    0    1    0
## [4,]    1    1    1    1    3

s[4,]<-(s[1,]-s[4,])
s

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    1    3
## [2,]    0    1    2    1    5
## [3,]    0    0    0    1    0
## [4,]    0    0    0    0    0

Basis for S are our pivot columns which are vectors 1 and 2:

ss[,1:2]

##      [,1] [,2]
## [1,]    1    1
## [2,]    3    2
## [3,]    2    1
## [4,]    1    1