Load the packages

suppressWarnings(suppressMessages(library(Matrix))) # eigen and rankMatrix functions will be used 
suppressWarnings(suppressMessages(library(pracma))) #charpoly function will be used

1. Problem set 1

  1. What is the rank of the matrix A?
\[\begin{equation} A= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 0 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 5 & 4 & 2 & 3 \end{pmatrix} \label{eq:1} \end{equation}\] 
(A <- matrix(c(1, 2, 3, 4, 1, 0, 1, 3, 0, 1, 2, 1, 5, 4, 2, 3),4,byrow = T))
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    1    0    1    3
## [3,]    0    1    2    1
## [4,]    5    4    2    3
rank_a <- rankMatrix(A)
print(paste("rank of matrix A is ",rank_a))
## [1] "rank of matrix A 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?

For an mxn matrix where m > n, let B be a row-equivalent matrix in reduced row-echelon form. According to Theorem CRN(Computing Rank and Nullity), Let r denote the number of pivot columns (or the number of nonzero rows),then r(A) = r. For matrix B, the left most nonzero entry of each nonzero row is equal to 1 and it is the only nonzero entry in its column. The pivot columns could form a indentity matrix if we rearrange matrix B. If this indentity matrix is in size kxk, then it is obvious that the maximum of k is equal to m or n which ever is smaller. Here m > n, so we know r < n. For nonzero matrix, the minimum number of pivot columns would be 1. So r >= 1. The minimum rank could be 1.

  1. What is the rank of matrix B? \[\begin{equation} B= \begin{pmatrix} 1&2&1\\ 3&6&3\\ 2&4&2 \end{pmatrix} \label{eq:2} \end{equation}\]
(B <- matrix(c(1,2,1,3,6,3,2,4,2),3,byrow = T))
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rank_b <- rankMatrix(B)
print(paste("rank of matrix A is ",rank_b))
## [1] "rank of matrix A 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.

\[\begin{equation} A= \begin{pmatrix} 1&2&3\\ 0&4&5\\ 0&0&6 \end{pmatrix} \label{eq:3} \end{equation}\]

eigenvalues

(C <- matrix(c(1,2,3,0,4,5,0,0,6),3,byrow = T))
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
eigen_c <- eigen(C)
eigenvalue <- eigen_c$values
print(paste("eigenvalues includes:",list(round(eigenvalue))))
## [1] "eigenvalues includes: c(6, 4, 1)"

eigenvectors

I3 <- matrix(c(1,0,0,0,1,0,0,0,1),3, byrow=T)
(NS_6 <- nullspace(C-6*I3)) # eigenvectors for # eigenvalue = 6
##           [,1]
## [1,] 0.5108407
## [2,] 0.7981886
## [3,] 0.3192754
(NS_4 <- nullspace(C-4*I3)) # eigenvectors for # eigenvalue = 4
##              [,1]
## [1,] 5.547002e-01
## [2,] 8.320503e-01
## [3,] 5.551115e-17
(NS_1 <- round(nullspace(C-1*I3))) # eigenvectors for # eigenvalue = 1
##      [,1]
## [1,]   -1
## [2,]    0
## [3,]    0
# or eigenvectors could be calculated as follows:
(reduce_6 <- rref(C-6*I3))
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0
(NS_six <- matrix(c(1.6,2.5,1),3,byrow=T)) # eigenvectors for # eigenvalue = 6
##      [,1]
## [1,]  1.6
## [2,]  2.5
## [3,]  1.0
(reduce_4 <- rref(C-4*I3))
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0
(NS_four <- matrix(c(0.67,1,0),3,byrow=T)) # eigenvectors for # eigenvalue = 4
##      [,1]
## [1,] 0.67
## [2,] 1.00
## [3,] 0.00
(reduce_1 <- rref(C-1*I3))
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0
(NS_one <- matrix(c(1,0,0),3,byrow=T)) # eigenvectors for # eigenvalue = 1
##      [,1]
## [1,]    1
## [2,]    0
## [3,]    0

Characteristic Polynomials

(polyn <- charpoly(C,info=T))
## Error term: 0
## $cp
## [1]   1 -11  34 -24
## 
## $det
## [1] 24
## 
## $inv
##      [,1]  [,2]        [,3]
## [1,]    1 -0.50 -0.08333333
## [2,]    0  0.25 -0.20833333
## [3,]    0  0.00  0.16666667

So the Characteristic Polynomials is : \(x^3-11x^2+34x+24=(x-6)(x-4)(x-1)\)

The solution of the Characteristic Polynomial

print(paste("The solution of the Characteristic Polynomial are: ",list(roots(polyn$cp))))
## [1] "The solution of the Characteristic Polynomial are:  c(6, 4, 1)"