(1) What is the rank of the matrix A?
#The original matrix
A<- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3
#Multiply the 1st row by -1 and subtract it from the 2nd row
A<- matrix(c(1,2,3,4,0,2,4,7,0,1,-2,1,5,4,-2,-3), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3
#Multiply the 1st row by 5 and subtract it from the 4th row
A<- matrix(c(1,2,3,4,0,2,4,7,0,1,-2,1,0,-6,-17,-23), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 0 -6 -17 -23
#Divide the 2nd row by 2 and subtract it from the 3rd row
A<- matrix(c(1,2,3,4,0,1,2,7/2,0,0,-4,-5/2,0,-6,-17,-23), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.0
## [2,] 0 1 2 3.5
## [3,] 0 0 -4 -2.5
## [4,] 0 -6 -17 -23.0
#Multiply the 2nd row by -6 and subtract it from the 4th row
A<- matrix(c(1,2,3,4,0,1,2,7/2,0,0,-4,-5/2,0,0,-5,-2), nrow= 4 ,ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.0
## [2,] 0 1 2 3.5
## [3,] 0 0 -4 -2.5
## [4,] 0 0 -5 -2.0
#Restore the 2nd row to the original by multiplying it by 2
A<- matrix(c(1,2,3,4,0,2,4,7,0,0,-4,-5/2,0,0,-5,-2), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.0
## [2,] 0 2 4 7.0
## [3,] 0 0 -4 -2.5
## [4,] 0 0 -5 -2.0
#Divide the 3rd row by -4 and multiply the 3rd row by -5
A<- matrix(c(1,2,3,4,0,2,4,7,0,0,-5,-25/8,0,0,-5,-2), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -5 -3.125
## [4,] 0 0 -5 -2.000
#Subtract the 3rd row from the 4th row and divide 3rd row by -5
A<- matrix(c(1,2,3,4,0,2,4,7,0,0,1,5/8,0,0,0,9/8), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 1 0.625
## [4,] 0 0 0 1.125
#Restore the 3rd row to the original view
A<- matrix(c(1,2,3,4,0,2,4,7,0,0,-4,-5/2,0,0,0,9/8), nrow= 4 ,ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -4 -2.500
## [4,] 0 0 0 1.125
#Calculate the number of linearly independent rows
A<- matrix(c(1,2,3,4,0,2,4,7,0,0,-4,-5/2,0,0,0,9/8), nrow= 4, ncol=4, byrow= TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -4 -2.500
## [4,] 0 0 0 1.125
# Since the reduced echelon matrix has four pivots and linearly independent, hence the Rank is 4
(2) Given an mxn matrix where m > n, what can be the maximum rank? The minimum
rank, assuming that the matrix is non-zero?
#let A be an mxn matrix over field F. Then the row rank of A is equal
#to the maximum number of linear independent row vectors of A. hence
#the row-rank of mxn matrix is at most n.
#Similarly, the column-rank of A is the maximum number of linear
#independent columns of A. Hence the column-rank of A is at most m.
#Since for any matrix A mxn over field F, the row-rank = column-rank. Therefore
#to be the common value of row-rank A and column-rank A, clearly rank A has to be
#equal or less of Min (m,n) for any mxn matrix
# from the above, for any m x n matrix, if m is greater than n, then the maximum rank of the matrix is n.
# and if m is less than n, then the maximum rank of the matrix is m.
# The rank of a matrix would be zero only if the matrix had no elements.
# However, If a matrix had even one element, its minimum rank would be one.
(3) What is the rank of matrix B?
# The original matrix
B<- matrix(c(11,2,1,3,6,3,2,4,2), nrow=3, ncol=3, byrow=TRUE)
B
## [,1] [,2] [,3]
## [1,] 11 2 1
## [2,] 3 6 3
## [3,] 2 4 2
# Multiply the 1st row by 3 and subtract it from the 2nd
B<- matrix(c(1,2,1,0,0,0,2,4,2), nrow=3, ncol=3, byrow=TRUE)
B
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 2 4 2
#Multiply the 1st row by 2 and subtract it from the 3rd row
B<- matrix(c(1,2,1,0,0,0,0,0,0), nrow=3, ncol=3, byrow=TRUE)
B
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0
#Calculate the number of linearly independent rows
B<- matrix(c(1,2,1,0,0,0,0,0,0), nrow=3, ncol=3, byrow=TRUE)
B
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0
# Hence the rank is 1.
Problem set 2
Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your
work. You’ll need to write out the characteristic polynomial and show your solution.
# The original matrix
A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, ncol = 3, byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
# To get the eigenvalues, we need to solve the determinant of of teh below matrxi:
col1<- c('1- lambda',0,3)
col2<- c(2,'4-lambda',0)
col3<- c(3,5,'6-lambda')
#Find the determinat of the below:
cbind(col1,col2,col3)
## col1 col2 col3
## [1,] "1- lambda" "2" "3"
## [2,] "0" "4-lambda" "5"
## [3,] "3" "0" "6-lambda"
#Which gives us the characteristic polynomial as
Det<- c(('-lambda^3)+11*lambda^2-34*lambda+24 = 0'))
Det
## [1] "-lambda^3)+11*lambda^2-34*lambda+24 = 0"
# which factorizes as
Det <- c('-(lambda-1)(lambda-4)(lambda-6)')
Det
## [1] "-(lambda-1)(lambda-4)(lambda-6)"
# Which gives us Three eigenvalues as lambda1 =1, lambda2 = 4, or lambda3 =6.
# When you use lambda in the definition of eigenvalue, you get Av ??? lambdav = (A ??? lambda I)v = 0
when lambda1 = 1:
# The original matrix
A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, ncol = 3, byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
##### when lambda1 = 1, we get:
A<- matrix(c(0,2,3,0,3,5,0,0,5), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 2 3
## [2,] 0 3 5
## [3,] 0 0 5
# devide row1 by row2
A<- matrix(c(0,1,3/2,0,3,5,0,0,5), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 3 5.0
## [3,] 0 0 5.0
# substract 3 times row1 from row2
A<- matrix (c(0,1,3/2,0,0,1/2,0,0,5), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 0 0.5
## [3,] 0 0 5.0
# devide row 2 by 1/2
A<- matrix(c(0,1,3/2,0,0,1,0,0,5), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 0 1.0
## [3,] 0 0 5.0
# substract 5 times row 2 from row 3
A<- matrix(c(0,1,3/2,0,0,1,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 1 1.5
## [2,] 0 0 1.0
## [3,] 0 0 0.0
# substrct 3/2 times row 2 from row 1
# final equation
A<- matrix(c(0,1,0,0,0,1,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0
# Hence from the final equation above, for lambda1 =1, we have x2 = 0, x3=0, and x1 = x1 ( x1 can be any variable).
# The eigenvector is ( x1, 0, 0)
###
when lambda2 = 4:
# The original matrix
A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, ncol = 3, byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
#### when lambda2 = 4, we get:
A<- matrix (c(-3,2,3,0,0,5,0,0,2), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] -3 2 3
## [2,] 0 0 5
## [3,] 0 0 2
# devide row1 by -3
A<- matrix(c(1,-2/3,-1,0,0,5,0,0,2), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.6666667 -1
## [2,] 0 0.0000000 5
## [3,] 0 0.0000000 2
# devide row2 by 5
A<- matrix(c(1,-2/3,-1,0,0,1,0,0,2), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.6666667 -1
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 2
# substract row2 times 2 from row3
A<- matrix(c(1,-2/3,-1,0,0,1,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.6666667 -1
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0
# substract -1 times row2 from row1
A<- matrix(c(1,-2/3,0,0,0,1,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0
# Hence from the final equation above, for lambda2 = 4, we have x3=0, x1=2/3*x2, and x2 = x2.
# The eigenvector is (2/3*x2, x2, 0)
####
for lambda3 = 6
# The original matrix
A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, ncol = 3, byrow=TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
#### for lambda3 = 6, we get:
A<- matrix(c(-5,2,3,0,-2,5,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] -5 2 3
## [2,] 0 -2 5
## [3,] 0 0 0
# Devide row1 by -5
A<- matrix(c(1,-2/5,-3/5,0,-2,5,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.4 -0.6
## [2,] 0 -2.0 5.0
## [3,] 0 0.0 0.0
# Devide row2 by -2
A<- matrix(c(1,-2/5,-3/5,0,1,-5/2,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 -0.4 -0.6
## [2,] 0 1.0 -2.5
## [3,] 0 0.0 0.0
# substract -2/5 times row2 from row1
A<- matrix(c(1,0,-8/5,0,1,-5/2,0,0,0), nrow=3, ncol=3, byrow= TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0
# Hence from the final equation above, for lambda3 =6, x2=5/2*x3, x1=8/5*x3.
# The eigenvector is ( 8/5*x3, x2=5/2*x3, x3 )