Data 605 - Assignment 3

N Nedd

2018-02-19

Problem Set 1

Solution

  1. Each row/column is independent therefore, the rank of the matrix A is 4. This can be calculated in R as follows
A <- cbind(c(1,-1,0,5),c(2,0,1,4), c(3,1,-2,-2),c(4,3,1,-3))
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
Ans <- qr(A)
RankA <- Ans$rank
RankA
## [1] 4

  1. For a mxn matrix if m > n, the maximum rank is n. Assuming the matrix is non-zero, the minimum rank is 1.

  2. The rank of the matrix B is 1. The first column is independent. The second is twice the first column and therefore not independent. The third column is half the second column and therefore also not independent.

B <- cbind(c(1,3,2), c(2,6,4), c(1,3,2))
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
Ans <- qr(B)
RankB <- Ans$rank
RankB
## [1] 1

Problem Set 2

Solution

The characteristic polynomial is determined as follows:

\(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix}\)

\(det\begin{Bmatrix} \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{bmatrix} = 0 \end{Bmatrix}\)

\(det \begin{bmatrix} \lambda - 1 & 2 & 3 \\ 0 & \lambda - 4 & 5 \\ 0 & 0 & \lambda - 6 \end{bmatrix} = 0\)

\((\lambda-1)[(\lambda - 4)(\lambda - 6) - 0]-2(0)+3(0) = 0\)

\((\lambda - 1)(\lambda^2 - 10\lambda +24) = 0\)

\(\lambda^3 - 11\lambda^2+34\lambda-24 = 0\) => This is the characteristic polynomial

To find eigen values

If we back track to this line \((\lambda - 1)(\lambda^2 - 10\lambda +24) = 0\)

we can see that \(\lambda - 1 = 0\), therefore \(\lambda = 1\)

The other eigen values can be found by factorising the other part of the equation \((\lambda - 1)(\lambda^2 - 10\lambda +24) = 0\) and therefore \((\lambda^2 - 10\lambda +24) = 0\)

By factorisation \((\lambda - 6)(\lambda - 4) = 0\)

Therefore if \(\lambda - 6 = 0, \lambda = 6\) and if \(\lambda - 4 = 0, \lambda = 4\)

The eigen values are 1, 4, and 6

This can also be calculated using R. See below for calculation of eigen values and vectors.

Ab <- cbind(c(1,0,0),c(2,4,0),c(3,5,6))
Ab
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
Ans <- eigen(Ab)
Ans$values
## [1] 6 4 1
Ans$vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0