John Suh

September 15, 2019

Problem Set 1

1. What is the rank of the matrix A? $$A=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 0& 1& 3\\ 0& 1& 2& 1\\ 5& 4& 2& 3\end{array}\right)$$

requires pracma and Matrix packages

library(matrixcalc)
library(pracma)
x<-matrix(c(1, 2, 3, 4, 1, 0, 1, 3, 0, 1, 2, 1, 5, 4, 2, 3),4,4,byrow=TRUE)
x

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

r<-matrix.rank(x)
r

## [1] 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 for a mxn matrix where m>n would be n.

The minimum rank for a mxn matrix where m>n would be 1

3. What is the rank of matrix B? $$B=\left(\begin{array}{ccc}1& 2& 1\\ 3& 6& 3\\ 2& 4& 2\end{array}\right)$$

x<-matrix(c(1,3,2,2,6,4,1,3,2),3,3,byrow=TRUE)
x

##      [,1] [,2] [,3]
## [1,]    1    3    2
## [2,]    2    6    4
## [3,]    1    3    2

mrank<-matrix.rank(x)
mrank

## [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.

$$B=\left(\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 0& 0& 6\end{array}\right)$$

x<-matrix(c(1,2,3,0,4,5,0,0,6),3,3,byrow=TRUE)
x

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

eg<-eigen(x)
eg

## eigen() decomposition
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0

From the Khan Academy Videos on EigenValues and Eigen Vectors.

We would need to solve for det(A - λI) = 0

The A Matrix would be the 3 by 3 given to us for this problem set and then the identity matrix would have lambda set across diagonally with zeros. The A matrix would be subtracting from the identity Matrix.

First row as an example would be 1-λ,2,3. That would be done across for all 3 rows.

Once thats done the Khan Video uses the rule of Saurus which will find the determinant.

(1−λ)((4−λ)(6−λ))+2(0−0)+3(0−0) (1−λ)(4−λ)(6−λ)−0+0 (1−λ)(24−4λ−6λ+λ2) (1−λ)(24−10λ+λ2) 24−10λ+λ2−24λ+10λ2−λ3

characteristic polynomial:

−λ3+11λ2−34λ+24

c<-charpoly(x)
c

## [1]   1 -11  34 -24

Factor:

(-x+1)(x-6)(x-4)

Eigen Values: 1,4,6

Then to find the eigenvectors, we plug the eigen values for each of the lamda equations. This I have not computed to compare with the eigenvector output from the eigen function.

eg

## eigen() decomposition
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0