ASSIGNMENT 3 IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - 2014

#install.packages('pracma')

library(pracma)
library(matlib)

1 Problem set 1

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

Note: Rank is equal to the number of non-zero rows in the reduced row echelon form (RREF). \[A=\begin{Bmatrix} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{Bmatrix}\]

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

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 = 4

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

Note: Rank is equal to the number of non-zero rows of a RREF matrix. For a rectangular matrix of the form m x n the max number of non-zero rows is equal the the smaller of the two(columns and rows) and in this case is equal to n.

For a non zero matrix, the minimum rank is atleast 1.

(3) What is the rank of matrix B?

\[B = \begin{Bmatrix} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2\\ \end{Bmatrix}\]

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

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

Rank = 1. There is only one non zero row.

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. \[A =\begin{Bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{Bmatrix}\]

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

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

Step by step solution: Note: For a triangular matrix the eigenvalues are the values of its diagonal.

eigen(A)$values
## [1] 6 4 1

\[Thus: \lambda_1=1 , \lambda_2=4 and \lambda_3=6\]

The characteristic polynomial takes the form: \[pA(\lambda)=(1-\lambda)(4-\lambda(6-\lambda)\]

and thus:

\[pA(\lambda)=24-34\lambda+11\lambda^2-\lambda^3\]

lets take a case of

\[\lambda=1\]

then

\[A-1I_3\]

row-reduces to:

rref(A - 1*diag(3))
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

in this case set the matrix vector product to equal zero vector. \[\begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0\\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix}\]

where if:

\[\lambda = 4\] Then: \[A - 4I_3\]

row reduces to: See R computation below

rref(A - 4 * diag(3))
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

Now solve for:

\[x_1, x_2 and x_3\]

\[\begin{pmatrix} 1& \frac{-2}{3} &0\\ 0&0&1\\ 0&0&0\\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix}\]

Solution: \[x_1 = 2/3, x_2 = 0 and x_3 = 0\]

the eigenspace is thus

\[E_{\lambda=4}=\begin{Bmatrix} 1\\ 1.5\\ 0\\ \end{Bmatrix}\] Repeat the step above:

For:

\[\lambda = 6\]

Then:

\[A - 6I_3\]

row reduces to: See R computation below

#rref(A - 6 * diag(3))

b = c(0,0,0)

echelon((A - 6 * diag(3)), b, reduced=TRUE, verbose=TRUE, fractions=TRUE)
## 
## Initial matrix:
##      [,1] [,2] [,3] [,4]
## [1,] -5    2    3    0  
## [2,]  0   -2    5    0  
## [3,]  0    0    0    0  
## 
## row: 1 
## 
##  multiply row 1 by -1/5 
##      [,1] [,2] [,3] [,4]
## [1,]    1 -2/5 -3/5    0
## [2,]    0   -2    5    0
## [3,]    0    0    0    0
## 
## row: 2 
## 
##  multiply row 2 by -1/2 
##      [,1] [,2] [,3] [,4]
## [1,]    1 -2/5 -3/5    0
## [2,]    0    1 -5/2    0
## [3,]    0    0    0    0
## 
##  multiply row 2 by 2/5 and add to row 1 
##      [,1] [,2] [,3] [,4]
## [1,]    1    0 -8/5    0
## [2,]    0    1 -5/2    0
## [3,]    0    0    0    0
## 
## row: 3

Now solve for:

\[x_1, x_2 and x_3\]

\[\begin{Bmatrix} 1 & 0 & -1.6\\ 0 & 1 & -2.5\\ 0 & 0 & 0\\ \end{Bmatrix} \begin{Bmatrix} x_1\\ x_2\\ x_3\\ \end{Bmatrix} = \begin{Bmatrix} 0\\ 0\\ 0\\ \end{Bmatrix}\]

Solution: \[x_1 = 1.6, x_2 = 2.5 and x_3 = 1\]

for lambda = 6 the eigenvector is

\[E_{\lambda=6}= \begin{Bmatrix} 1.6\\ 2.5\\ 1\\ \end{Bmatrix}\] *** The same applies to the case of lambda = 1. Please show your work using an R-markdown document. Please name your assignment submission with your first initial and last name.***

*** See problem set 2 by hand calculation.***