1 Problem Set

1.1

What is the rank of matrix A?

Solution:

\(A=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array} \right]\rightarrow\) \(R2+R1, R4-5*R1\)

\(=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{array} \right]\rightarrow\) \(R2*0.5\)

\(=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3.5 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{array} \right]\rightarrow R3-R2\) \(R4+6*R2\)

\(=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3.5 \\ 0 & 0 & 4 & -2.5 \\ 0 & 0 & -5 & -2 \end{array} \right]\rightarrow R3*0.25\)

\(=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3.5 \\ 0 & 0 & 1 & -0.625 \\ 0 & 0 & -5 & -2 \end{array} \right]\rightarrow R4+5*R3\)

\(=\left[ \begin{array}{cccccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3.5 \\ 0 & 0 & 1 & -0.625 \\ 0 & 0 & 0 & -6.125 \end{array} \right]\)

Answer: Rank is 4.

Checking in R.

a <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol = 4, nrow = 4)

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
#install.packages("Matrix")

library(Matrix)

rankMatrix(a)
## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16

1.2

Maximum rank is the minimum of m and n. In our case, it will be n. The minimum rank will be 1(all columns are lineary dependent) .

1.3

What is the rank of matrix B?

Solution:

\(B=\left[ \begin{array}{cccccc} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array} \right]\)

Answer: Rank is 1. All 3 columns are linearly dependent

Checking in R.

b <- matrix(c(1,3,2,2,6,4,1,3,2), ncol = 3, nrow = 3)

b
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rankMatrix(b)
## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16

2 Problem Set

Compute the eigenvalues and eigenvectors of the matrix A.

Solution:

\(A=\left[ \begin{array}{cccccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array} \right]\)

\(p_a(x)=Det(A-xI_3)\)

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

\(=(1-x)*\left[ \begin{array}{cccccc} 4-x & 5 \\ 0 & 6-x \end{array} \right]\)

\(=(1-x)(6-x)(4-x)=(6-x-6x+x^2)(4-x)=(6-7x+x^2)(4-x)=24-28x+4x^2-6x+7x^2-x^3=24-34x+11x^2-x^3\)

Checking in R.

a <- matrix(c(1,0,0,2,4,0,3,5,6), nrow = 3, ncol = 3)

#install.packages("pracma")

library(pracma)
## Warning: package 'pracma' was built under R version 3.5.2
## 
## Attaching package: 'pracma'
## The following objects are masked from 'package:Matrix':
## 
##     expm, lu, tril, triu
b<-charpoly(a)

b
## [1]   1 -11  34 -24

\(x=1,x=6,x=4\)

\(x=1\)

\(=\left[ \begin{array}{cccccc} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{array} \right]\rightarrow\) \(R3\times0.2\), \(R2-5\times R3\), \(R1-3\times R3\)

\(=\left[ \begin{array}{cccccc} 0 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{array} \right]\rightarrow\) \(R2\times (1/3)\), \(R1-2\times R2\)

\(=\left[ \begin{array}{cccccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\)

\(\varepsilon_A(1)=\left\langle\Bigg\{\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\ \Bigg\}\right\rangle\)

\(x=6\)

\(=\left[ \begin{array}{cccccc} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{array} \right]\) \(R2\times-0.5\), \(R1-2\times R2\)

\(=\left[ \begin{array}{cccccc} -5 & 0 & 8 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array} \right]\rightarrow\) \(R1\times -0.2\)

\(=\left[ \begin{array}{cccccc} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array} \right]\)

\(\varepsilon_A(1)=\left\langle\Bigg\{\begin{bmatrix}16 \\ 25 \\ 10\end{bmatrix}\ \Bigg\}\right\rangle\)

\(x=4\)

\(=\left[ \begin{array}{cccccc} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{array} \right]\rightarrow\) \(R3\times0.5\), \(R2-5\times R3\), \(R1-3\times R3\)

\(=\left[ \begin{array}{cccccc} -3 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]\rightarrow\) \(R1\times -(1/3)\)

\(=\left[ \begin{array}{cccccc} 1 & -2/3 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]\)

\(\varepsilon_A(1)=\left\langle\Bigg\{\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}\ \Bigg\}\right\rangle\)

ev <- eigen(a)

ev$values
## [1] 6 4 1
ev$vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0