1. Problem set 1

(1) What is the rank of the matrix A?

A1 <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, byrow = T)

A1
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
#After the row reduced echelon matrix
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

Ans: From the above matrix, its known that its dimension is 4x4 square matrix and lineary independent, therefore it rank is 4

  1. Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Ans: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.

  1. What is the rank of matrix B? B =
B <- matrix(c(1,2,1,3,6,3,2,4,2), nrow = 3, byrow = T)

B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
dim(B)
## [1] 3 3
R1 <- B[1, ]
R2 <- B[2, ]
R3 <- B[3, ]

a <- R1-(1/3)%*%R2
b <- R3-(2/3)%*%R2


M <- matrix(c(a,b,R2), nrow = 3, byrow = T)

M
##      [,1] [,2] [,3]
## [1,]    0    0    0
## [2,]    0    0    0
## [3,]    3    6    3

Answer: Since the rank is number of non zero row, rank is 1.

2. 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.

B <- matrix(c(1,2,3,0,4,5,0,0,6), nrow = 3, byrow = T)

B
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

\[ pB(\lambda) = det(B-(\lambda I3)) \] \[ = \begin{pmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda \\ \end{pmatrix} \] \[ = (1-\lambda)[(4-\lambda)(6-\lambda)-(0)(5)] - 2[0] + 3[0] \] \[ = (1-\lambda)[(4-\lambda)(6-\lambda)] \]

\[ \lambda = 1, 4 and 6 \] \[ Substituting \lambda = 1, Eigen vectors are = \Bigg[\begin{pmatrix} &1\\ &0\\ &0\\ \end{pmatrix} \Bigg] \] \[ Substituting \lambda = 4, Eigen vectors are = \Bigg[\begin{pmatrix} &1.6\\ &2.5\\ &1\\ \end{pmatrix} \Bigg] \]

\[ Substituting \lambda = 6, Eigen vectors are = \Bigg[\begin{pmatrix} &0.6\\ &1\\ &0\\ \end{pmatrix} \Bigg] \]

eigen(B, only.values = FALSE, EISPACK = TRUE)
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0