Problem Set 1

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

\[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\-1 & 0 & 1 & 3 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix} (R2 <- R1 +R2)\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\0 & -6 & -17 & -23\end{bmatrix} (R4 <- -5R1 +R4)\]

\[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\0 & 0 & -5 & -2\end{bmatrix} (R4 <- 3R2+R4)\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & -4 & -5/2 \\0 & 0 & -5 & -2\end{bmatrix} (R3 <- 1/2R2+R3)\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & -4 & -5/2 \\0 & 0 & 0 & 9/8\end{bmatrix} (R4 <- 5/4R2+R4)\]

\[ Rank\,\,=\,4 \] ### R solution

library(matrixcalc)

m <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, ncol = 4, byrow = TRUE)
m
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
matrix.rank(m)
## [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 is \(m\) (rows)
The Minimum rank is 1, the others rows may be linearly dependent.

3. What is the rank of matrix \(B\)?

\[ B = \begin{bmatrix}1 & 2 & 1 \\3 & 6 & 3 \\2 & 4 & 2 \end{bmatrix}\]

\[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\2 & 4 & 2 \end{bmatrix} (R2 <- -3R1+R2)\] \[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} (R3 <- -2R1+R3)\] \[ Rank\,\,=\,1 \]
R2 and R3 are linearly dependent.

# Test
m <- matrix(c(1,2,1,3,6,3,2,4,2), nrow = 3, ncol = 3, byrow = TRUE)
m
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
matrix.rank(m)
## [1] 1

Problem Set 2

Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to write out the characteristic polynomial and show your solution.
Eigenvalues:

characteristic polynomial:

\[ P(\lambda)=det(A-\lambda\,I_n)=0\] \[ A = \begin{bmatrix}1 & 2 & 3 \\0 & 4 & 5 \\0 & 0 & 6 \end{bmatrix};\,\, \lambda\,I_3 = \begin{bmatrix}\lambda & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda \end{bmatrix}\]

\[ det\,\begin{bmatrix}1-\lambda & 2 & 3 \\0 & 4-\lambda & 5 \\0 & 0 & 6-\lambda \end{bmatrix} = 0\] \[(1-\lambda)(4-\lambda)(6-\lambda)=0\] \[ \lambda ^{3}-11\lambda^{2}+34\lambda-24=0\] ##Eigenvalues of A: \[\lambda=1,\, \lambda=4,\, \lambda=6\]

Eigenvectors:

\(\lambda_1=6\)

\[(A- \lambda_1I)v=0\]

\[-5x + 2y + 3z = 0 \\ -2y + 5z = 0\] If we make \[Y=1\]: \[-2 + 5z = 0; z = 2/5\] Substituting Y and Z: \[-5x + 2 + 3(2/5) = 0 \\ -5x = -16/5 \\ x = 16/25\] \[E_{\lambda_1=6}=c\begin{bmatrix}16/25 \\ 1 \\ 2/5\end{bmatrix}\]

\(\lambda_2=4\)

\[(A- \lambda_2 I)v=0\]

\[-3x + 2y + z = 0\\ 5z = 0\\2z = 0\] If we set \[Y=1\] then \[X=2/3\] \[E_{\lambda_2=4}=c\begin{bmatrix}2/3 \\ 1 \\ 0\end{bmatrix}\]

\(\lambda_3=1\)

\[(A- \lambda_3 I)v=0\]

\[\begin{bmatrix}0&2&3\\0&3&5\\0&0&5\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=0\]

Equations:
\[2y + 3z = 0\\ 3y + 5z = 0\\5z = 0\]

The only non-zero solution is \[X=1\] and Y and Z = 0 \[E_{\lambda_3=1}=c\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\]

Test

library(pracma)
## Warning: package 'pracma' was built under R version 3.3.3
mat <- matrix(data = c(1,0,0,2,4,0,3,5,6), nrow = 3, ncol = 3, byrow = FALSE)
mat
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
charpoly(mat)
## [1]   1 -11  34 -24
eigen(mat)
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0
# eign vectors
eigen(mat)$values
## [1] 6 4 1
#for L = 6
eigen(mat)$vectors[,1]
## [1] 0.5108407 0.7981886 0.3192754
#for L = 4
eigen(mat)$vectors[,2]
## [1] 0.5547002 0.8320503 0.0000000
#for L = 1
eigen(mat)$vectors[,3]
## [1] 1 0 0