1. Problem set 1

  1. What is the rank of the matrix A?
#loading required libraries
library(pracma)
A <-matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),nrow = 4,ncol = 4)
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
rref(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1
Rank(A)
## [1] 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? The max rank cannot be > than ‘n’. The min rank in this case (non-zero) is 1.

Let Ra = rank of matrix X, where matrix X is an m x n matrix with m>n Ra≤{min(row_rank,column_rank)Ra≤{min(m,n)} We know the maximum rank will be n and since the matrix is non_zero,then Ra≤nRa≥1 Hence minimum rank of matrix A would be 1

B <- matrix(c(1,3,2,2,6,4,1,3,2),nrow = 3, ncol= 3)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
Rank(B)
## [1] 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. \[ A=\left[\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{array}\right]\\ Ax=\lambda x\\ det(A-I\lambda)=0\\ \] Solve for the roots of the polynomials to obtain the eigen values \[ (1−\lambda)((4−\lambda)(6−\lambda))+2(0−0)+3(0−0)=0\\ (1−\lambda)(4−\lambda)(6−\lambda)−0+0=0\\ (1−\lambda)(24−4\lambda−6\lambda+λ^2)=0\\ (1−\lambda)(24−10\lambda+\lambda^2)=0\\ 24−10\lambda+\lambda ^2−24λ+10\lambda ^2−\lambda ^3=0\\ −\lambda ^3+11\lambda ^2−34\lambda +24=0\\ \] Divide every term on the left and right side of the equation by a -1 to make the leading term positive \[ \lambda ^3-11\lambda ^2+34\lambda-24=0\\ \] Factoring by grouping \[ (\lambda^3-11\lambda^2)+(34\lambda-24)=0\\ \lambda ^ 2(\lambda-11)+2(17\lambda-12)=0\\ \] This needs further factorization to solve. To simplify getting the roots of this equation, I will use R programming
Verifying the characteristic polynomial:

A <- matrix(c(1,2,3,0,4,5,0,0,6), 3, 3, byrow=TRUE)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
library(pracma)
charpoly(A)
## [1]   1 -11  34 -24

Eigen values and vectors

eigen(A)
## eigen() decomposition
## $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