Problem Set 1

1 What is the rank of the matrix A?

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
# The rank of the matrix is determined by the number of number of independent rows. These are the non-zero rows in row echelon form

reA<-rref(A)
reA
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1
#This matrix has rank 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?

The maximum rank of a matrix is equal to the value of the lowest dimension of mxn

3. What is the rank of matrix B?

B<- matrix(c(1,2,1,3,6,3,2,4,2), nrow=3, byrow= TRUE)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
reB<- rref(B)
reB
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0
#The rank of this matrix is 1

Problem Set 2

Av − λv = (A − λI)v = 0

A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow=3, byrow=TRUE)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
# characteristic polynomial of a matrix A is given by the determinant of A- lambda * Identity matrix 
# For this example we will be using lambda to be x

A <- matrix(c('1-λ', 2, 3, 0, '4-λ', 5, 0, 0, '6-λ'), nrow=3, byrow=TRUE)
A
##      [,1]  [,2]  [,3] 
## [1,] "1-λ" "2"   "3"  
## [2,] "0"   "4-λ" "5"  
## [3,] "0"   "0"   "6-λ"

determinant of A : (1-λ)((4-λ)(6-λ) - 50) - 20*0 = (1-λ)((4-λ)(6-λ)) characteristic equation: -λ^3 + 10λ^2 - 24λ + 24 lambda values (1-λ)((4-λ)(6-λ)) = 0 ### eigeinvalues of matrix λ = 1, 4, 6

eigen vector

To find the eigen vector we know: Av − λv = (A − λI)v = 0

we now have find the eigen vectors

1. First eigenvector

λ = 1
A1 <- matrix(c(1-λ, 2, 3, 0, 4-λ, 5, 0, 0, 6-λ), nrow=3, byrow=TRUE)
A1
##      [,1] [,2] [,3]
## [1,]    0    2    3
## [2,]    0    3    5
## [3,]    0    0    5
# Thus we need a vector v that creates proves the statement (A - λI)v = 0

0v.1 +2v.2 +3v.3 = 0 0v.1 +3v.2 +5v.3 = 0 0v.1 +0v.2 +5v.3 = 0

V1 = 0, V2 = V3

Eigenvector for λ = 1 : (0, 0, 1)

2. Second Eigenvector

λ = 4
A1 <- matrix(c(1-λ, 2, 3, 0, 4-λ, 5, 0, 0, 6-λ), nrow=3, byrow=TRUE)
A1
##      [,1] [,2] [,3]
## [1,]   -3    2    3
## [2,]    0    0    5
## [3,]    0    0    2
# Thus we need a vector v that creates proves the statement (A - λI)v = 0

-3v.1 +2v.2 +3v.3 = 0 0v.1 +0v.2 +5v.3 = 0 0v.1 +0v.2 +2v.3 = 0

V1 = 0, V2 = 1, V3 = 0

Eigenvector for λ = 4 : (0, 1, 0)

3. Third Eigen vector

λ = 6
A1 <- matrix(c(1-λ, 2, 3, 0, 4-λ, 5, 0, 0, 6-λ), nrow=3, byrow=TRUE)
A1
##      [,1] [,2] [,3]
## [1,]   -5    2    3
## [2,]    0   -2    5
## [3,]    0    0    0
# Thus we need a vector v that creates proves the statement (A - λI)v = 0

-5v.1 +2v.2 +3v.3 = 0 0v.1 −2v.2 +5v.3 = 0 0v.1 +0v.2 +0v.3 = 0

V1 = 0, V2 = 0, V3 = 1

Eigenvector for λ = 6 (0, 0, 1)