library(Matrix)
library(pracma)
## Warning: package 'pracma' was built under R version 3.6.3
##
## Attaching package: 'pracma'
## The following objects are masked from 'package:Matrix':
##
##     expm, lu, tril, triu
A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow=4, byrow=TRUE)
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

# Problem Set 1

#### (1)

What is the rank of the matrix A?

rankMatrix(A)
##  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?

The maximum rank is the minimum of the column and row rank. In this case the maxium column and row rank is equal to M and N. As N < M, we know the the maximum rank is N.

The minimum rank of a non zero matrix is 1.

#### (3)

What is the rank of matrix B?

B <- matrix(c(1,2,1,3,6,3,2,4,2), nrow=3, byrow=TRUE)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rankMatrix(B)
##  1

# Problem Set 2

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

Compute the eigenvalues and eigenvectors of the matrix A.

eigen(A)\$values
##  6 4 1

Therefore, the charecteristic polynomial is:

(1−λ)(4−λ)(6−λ)

#### Eigen Vectors:

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

With V1=1, we know the Eigen Vector is 1, 0, 0

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

Here we see V1=V1; V2 = 3*V1/2; V3 = 0

With V2=1, we know the Eigen Vector is 1, 3/2, 0

##### λ=6
rref(A - 6 * diag(3))
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

We see V1=1.6*V3, V2=2.3V3, V3=V3

With V3=1, the Eigen Vector is 1.6, 2.5, 1