1. Problem set 1


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

\(\mathbf{A}\) = \(\left[ \begin{array}{ccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3 \end{array} \right]\)

Rank of matrix A by Row Echelon Form and by Determinant Method

REDF <- function (M)
{
r = dim(M)[1]
for (p in 1:(r-1))
{
 for (i in p:(r-1))                            #row
 {
  mult = ((M[i+1,p])/(M[p,p])) 
  print (mult)
  for (j in p:r)                               #column
  {  
    M[i+1,j] = M[i+1,j] - M[p,j]*mult 
   }
 }
}
return(M)
}


A = matrix(c(1, -1, 0, 5, 2, 0, 1, 4, 3, 1, -2, -2, 4, 3, 1, -3), 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
det(A)
## [1] -9

Since determinant is non-zero, rank of Matrix A is 4.
Also by Reduced Row Echelon Form Method to find rank of matrix

s <- REDF(A)
## [1] -1
## [1] 0
## [1] 5
## [1] 0.5
## [1] -3
## [1] 1.25
print(s)
##      [,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
REDF2 <- function (M)
{
r = nrow(M)
c = ncol(M)
for (p in 1:(c-1))
{
 for (i in p:(r-1))                            #row
 {
  mult = ((M[i+1,p])/(M[p,p])) 
  print (mult)
  for (j in p:c)                               #column
  {  
    M[i+1,j] = M[i+1,j] - M[p,j]*mult 
   }
 }
}
return(M)
}

Q = matrix(c(s[,2],s[,3],s[,4]), ncol=3)
Q
##      [,1] [,2]   [,3]
## [1,]    2    3  4.000
## [2,]    2    4  7.000
## [3,]    0   -4 -2.500
## [4,]    0    0  1.125
s <- REDF2(Q)
## [1] 1
## [1] 0
## [1] 0
## [1] -4
## [1] 0
print(s)
##      [,1] [,2]  [,3]
## [1,]    2    3 4.000
## [2,]    0    1 3.000
## [3,]    0    0 9.500
## [4,]    0    0 1.125
Q = matrix(c(s[,2],s[,3]), ncol=2)
Q
##      [,1]  [,2]
## [1,]    3 4.000
## [2,]    1 3.000
## [3,]    0 9.500
## [4,]    0 1.125
s <- REDF2(Q)
## [1] 0.3333333
## [1] 0
## [1] 0
print(s)
##      [,1]     [,2]
## [1,]    3 4.000000
## [2,]    0 1.666667
## [3,]    0 9.500000
## [4,]    0 1.125000

Since all rows/columns are non-zero in reduced echelon form, rank of square matrix A is same as its dimensions which is 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?


Since m(rows) > n(columns), maximum rank of the matrix is n (number of columns)

Minimum rank of a non-zero matrisx is 1.



(3) What is the rank of matrix B?

\(\mathbf{B}\) = \(\left[ \begin{array}{ccc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{array} \right]\)

__Rank of matrix B by Row Echelon Form and by Determinant Method_

B = matrix(c(1,2,1,3,6,3,2,4,2),nrow=3,byrow=TRUE)
det(B)
## [1] 0

Since determinant is 0, rank of matrix B is less than rows/columns of 3.

Using the Row Echelon Form method

s <- REDF2(B)
## [1] 3
## [1] 2
## [1] NaN
print(s)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0  NaN  NaN
C = matrix(c(B[1,],B[3,]),nrow=2,byrow=TRUE)
C
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    2    4    2

Since there is only 1 independent row, rank of the matris 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.

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

Let \(\lambda\) be an eigenvalue of \(\mathbf{A}\) where \(\mathbf{A}\vec{v}\) = \(\lambda\vec{v}\) for non-zero \(\vec{v}\),

(\(\lambda\)\(I_n\)-\(\mathbf{A}\))\(\vec{v}\)=\(\vec{0}\)

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

det \(\left| \begin{array}{ccc} \lambda - 1 & -2 & -3\\ 0 & \lambda - 4 & -5\\ 0 & 0 & \lambda - 6 \end{array} \right|\) = (\(\lambda\)-1) (\(\lambda\)-4) (\(\lambda\)-6)

(\(\lambda\)-1) = 1, 4 and 6 (eigenvalues)

Eigenvectors -
For \(\lambda_1\) -

= \(\left[ \begin{array}{ccc} 0 & -2 & -3\\ 0 & -3 & -5\\ 0 & 0 & -5 \end{array} \right]\) \(\vec{v}\)

Reduced Row Echelon Form to find the vector for eigenvalue \(\lambda_1\)

C = matrix(c(0, -2, -3, 0, -3, -5, 0, 0, -5),nrow=3,byrow=TRUE)
C
##      [,1] [,2] [,3]
## [1,]    0   -2   -3
## [2,]    0   -3   -5
## [3,]    0    0   -5
s <- REDF2(C)
## [1] NaN
## [1] NaN
## [1] NaN
print(s)
##      [,1] [,2] [,3]
## [1,]    0   -2   -3
## [2,]  NaN  NaN  NaN
## [3,]  NaN  NaN  NaN

0\(V_1\) -2\(V_2\) - 3\(V_3\) = 0

            -2$V_2$ = 3$V_3$  
            
             $V_2$ = -1.5$V_3$