1. Problem set 1

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

\[A = \begin{pmatrix} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \]

library(Matrix)
A = matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),nrow=4,ncol=4,byrow=FALSE)

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
#rank of the matrix
rankMatrix(A)[1]
## [1] 4
#determinant of the matrix
det(A)
## [1] -9

The rank is 4 because the determinant is -9 which is not equal to zero

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

The number of rows of a matrix is a limit on the rank of the matrix, which means the rank of the matrix cannot exceed the total number of rows in a matrix. However, the maximum rank for a non-square matrix is the value of the smaller dimension.

For the mxn matrix, the Minimum rank is 1 while the Maximum rank is n

\[What\ is\ the\ rank\ of\ matrix\ B?\ B = \begin{pmatrix} 1 & 2 & 1\\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{pmatrix} \]

library(Matrix)
B = matrix(c(1,3,2,2,6,4,1,3,2),nrow=3,ncol=3,byrow=FALSE)

B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rankMatrix(B)[1]
## [1] 1

1. 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.

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

Av = λv

A = Square matrix

λ = eigen value

v = eigen vector

However, λ is the eigen value of A if det(λ*In - A) = 0.

In = Identity matrix

\[det(λ*\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} - \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix}) = 0 \]

Simplify the above:

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

Simplify the determinant:

(λ-1)[(λ-4)(λ-6)−0]=0

= (λ-1)(λ-4)(λ-6)=0

\[ The\ characteristic\ polynomial:\ λ^3-11{λ}^2+34λ-24\]

λ=1, 4, and 6.

Confirmation of Eigen values using the R eigen() function:

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

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

To find the Eigen Vectors:

Eigen vector for λ=1;

(A−λI)x=0;

where x is the eigen vector

For λ=1;

\[Let\ x\ = \begin{pmatrix} u\\ v\\ w \end{pmatrix} \]

\[ (\begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix} - λ*\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}) * x = 0 \]

\[ \left( \begin{array}{cc} 1-1 & 2-0 & 3-0 \\ 0-0 & 1-4 & 5-0 \\ 0-0 & 0-0 & 1-6 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

\[ \left( \begin{array}{cc} 0 & 2 & 3 \\ 0 & -3 & 5 \\ 0 & 0 & -5 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

Simplifying further, gives the below eigen vector

u = 1; v = 0; w = 0

\[ \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = \left( \begin{array}{cc} 1 \\ 0 \\ 0 \end{array} \right) \]

For λ=4;

\[Let\ x_2\ = \begin{pmatrix} u\\ v\\ w \end{pmatrix} \]

\[ (\begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix} - λ*\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}) * x_2 = 0 \]

\[ \left( \begin{array}{cc} 4-1 & 2-0 & 3-0 \\ 0-0 & 4-4 & 5-0 \\ 0-0 & 0-0 & 4-6 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

\[ \left( \begin{array}{cc} 3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & -2 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

Simplifying further, gives the below eigen vector

u = 2; v = 3; w = 0

\[ \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = \left( \begin{array}{cc} 2 \\ 3 \\ 0 \end{array} \right) \]

For λ=6;

\[Let\ x_3\ = \begin{pmatrix} u\\ v\\ w \end{pmatrix} \]

\[ (\begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{pmatrix} - λ*\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}) * x_3 = 0 \]

\[ \left( \begin{array}{cc} 6-1 & 2-0 & 3-0 \\ 0-0 & 6-4 & 5-0 \\ 0-0 & 0-0 & 6-6 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

\[ \left( \begin{array}{cc} 5 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & 0 \end{array} \right) * \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = 0 \]

Simplifying further, gives the below eigen vector

u = 16; v = 25; w = 10

\[ \left( \begin{array}{cc} u \\ v \\ w \end{array} \right) = \left( \begin{array}{cc} 16 \\ 25 \\ 10 \end{array} \right) \]

\[ Eigen\ vectors\ are:\ x\ = \begin{pmatrix} x_1\\ x_2\\ x_3 \\ \end{pmatrix} = \begin{pmatrix} 1 & 2 & 16\\ 0 & 3 & 25 \\ 0 & 0 & 10 \\ \end{pmatrix} \]