Problem Set 1

(1) What is the rank of the matrix A? \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]

reforder = function(x) {
  pivot = 1
  xrows = nrow(x)
  xcolumns = ncol(x)
  
  for(rows in 1:xrows) {
    if ( xcolumns <= pivot)
      break;
    
    newrow = rows
    
    while( x[newrow, pivot] == 0 ) {
      newrow = newrow + 1
      if ( xrows == newrow ) {
        newrow = rows
        pivot = pivot + 1
        if ( xcolumns == pivot ) return(x)
      }
    }
    
    # swapping rows
    placeholder = x[newrow, ]
    x[newrow, ] = x[rows, ]
    x[rows, ] = placeholder
    
    # reducing equation
    x[rows, ] = x[rows, ] / x[rows, pivot]
    for(newrow in 1:xrows) {
      if ( newrow != rows )
        x[newrow, ] = x[newrow, ] - x[rows, ] * x[newrow, pivot]
    }
    pivot = pivot + 1
  }
  
  # ordering the matrix
  x = x[order(x[ ,1],decreasing = T), ]
  
  # counting the order
  count = 0
  for (rows in 1:xrows){
    if ( sum(x[rows, 1:xcolumns] ) != 0) {
      count = count + 1
    }
  }
  printed = sprintf("The order is: %s", count)
  output = list(x, printed)
  return(output)
}

A = matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5, 4, -2, -3), byrow=TRUE, nrow = 4)

reforder(A)
## [[1]]
##      [,1] [,2] [,3]   [,4]
## [1,]    1    0    0 -2.375
## [2,]    0    1    0  2.250
## [3,]    0    0    1  0.625
## [4,]    0    0    0  1.125
## 
## [[2]]
## [1] "The order is: 4"

The rank of the matrix is determined by the number of non-zero rows in a matrix. In this example, matrix A does not have any non-zero rows, therefore the rank is four.

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

Since the rank of a matrix is dependent on the number of non-zero rows, the highest the rank can be in an mxn matrix (where m is the number of rows and n is the number of columns) is m and the lowest the rank can be (if it is a non-zero matrix) is one.

(3) What is the rank of matrix B? \[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right] \]

B = matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), byrow=TRUE, nrow = 3)

reforder(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Matrix B’s rank is one because out of the three rows reduced to row echelon form, only one is non-zero.

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. \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] \]

A = matrix(c(1,2,3,0,4,5,0,0,6), byrow=TRUE, nrow = 3)

\[det(A-\lambda I) = det(\left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right]- \left[\begin{array} {rrr} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array}\right])= det(\left[\begin{array} {rrr} (1-\lambda) & -2 & -3 \\ 0 & (4-\lambda) & -5 \\ 0 & 0 & (6-\lambda) \end{array}\right]) \] To solve, we keep in mind that \(det(A-\lambda) = 0\) so that

\[det(\left[\begin{array} {rrr} (1-\lambda) & -2 & -3 \\ 0 & (4-\lambda) & -5 \\ 0 & 0 & (6-\lambda) \end{array}\right])= ((1-\lambda))[(4-\lambda)(6-\lambda)-(-5)(0)]-(-2)[(0)(6-\lambda)-(-5)(0)]+(-3)[0(0)-(4-lambda)(0)]=(1-\lambda)(4-\lambda)(6-\lambda) \] so

\(p(\lambda)= -\lambda^3+11\lambda^2-34\lambda+24=0\) or \((1-\lambda)(4-\lambda)(6-\lambda)=0\), which makes the eigenvalues \(\lambda= 1, 4, 6\).

A = matrix(c(1,2,3,0,4,5,0,0,6), byrow=TRUE, nrow = 3)

l1 = diag(1, dim(A), dim(A))
l4 = diag(4, dim(A), dim(A))
l6 = diag(6, dim(A), dim(A))

lambda1 = A-l1
lambda4 = A-l4
lambda6 = A-l6

lambda1
##      [,1] [,2] [,3]
## [1,]    0    2    3
## [2,]    0    3    5
## [3,]    0    0    5
lambda4
##      [,1] [,2] [,3]
## [1,]   -3    2    3
## [2,]    0    0    5
## [3,]    0    0    2
lambda6
##      [,1] [,2] [,3]
## [1,]   -5    2    3
## [2,]    0   -2    5
## [3,]    0    0    0
library("pracma")
## Warning: package 'pracma' was built under R version 3.3.3
rref(lambda1)
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0
rref(lambda4)
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0
rref(lambda6)
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

For \(\lambda=1\):

\[ E(\lambda=1)=\left[\begin{array} {rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array}\right]= \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \] \(v3=0\), \(v2=0\), \(v1=t\) \[ v=\left[\begin{array} {rrr} t \\ 0 \\ 0 \end{array}\right]= \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right]t \]

For \(\lambda=4\): \[ E(\lambda=4)=\left[\begin{array} {rrr} 0 & -0.67 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array}\right]= \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \] \(v3=0\), \(v2=t\), \(v1=0.67t\) \[ v=\left[\begin{array} {rrr} 0.67t \\ t \\ 0 \end{array}\right]= \left[\begin{array} {rrr} 0.67 \\ 1 \\ 0 \end{array}\right]t \]

For \(\lambda=6\):

\[ E(\lambda=6)=\left[\begin{array} {rrr} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array}\right]= \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array}\right] \] \(v3=t\), \(v2=2.5t\), \(v1=1.6t\) \[ v=\left[\begin{array} {rrr} 1.6t \\ 2.5t \\ t \end{array}\right]= \left[\begin{array} {rrr} 1.6 \\ 2.5 \\ 1 \end{array}\right]t \]