** DATA_605_Assignment_3_Thonn - Eigenvalues & Eigenvectors **
#install.packages("matrixcalc")** Problem Set 1 **
1-1). What is the rank of matrix A?
\(A = \left[\begin{array}{rrrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array} \right]\)
library(matrixcalc)
#create Matrix M1
A <- matrix(data = c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, ncol = 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 -3matrix.rank(A)## [1] 4# Answer 1-1: the rank of A is 4
# [1] 41-2). Given an \(m x n\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero ? Maximum rank = \(m\) (rows) Minimum rank = 1 (all other rows could be linearly dependent)
Answer-1-2:
Rank:
The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we transform the matrix to its row echelon form and count the number of non-zero independent rows.
View an example \(m x n\) matrix \[\left[ {\begin{array}{rrrr} a & b & c\\ d & e & f\\ g & h & i\\ j & k & l\ \end{array} } \right]\]
View in reduced form: \[\left[ {\begin{array}{rrrr} x1 & 0 & 0\\ 0 & x2 & 0\\ 0 & 0 & x3\\ 0 & 0 & x4\\ \end{array} } \right]\]
Notice that the rows below the nth column become multiples of each other, and therefore can be eliminated.
Therefor the minimum rank of an m x n matrix that is non-zero is 1.
The minimum rank is 1
And the maximimum rank is m (where m is the number of rows)
1-3). What is the rank of matrix \(B\)?
Rank:
The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we transform the matrix to its row echelon form and count the number of non-zero independent rows.
\[ B = \begin{bmatrix}1 & 2 & 1 \\3 & 6 & 3 \\2 & 4 & 2 \end{bmatrix}\]
row 2 is a multiple of row 1.
\[ B = \begin{bmatrix}3 & 6 & 3 \\3 & 6 & 3 \\2 & 4 & 2 \end{bmatrix}\] subtract row 1 from row 2.
\[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\2 & 4 & 2 \end{bmatrix}\] row 3 is a multiple of row 1. \[ B = \begin{bmatrix}2 & 4 & 2 \\0 & 0 & 0 \\2 & 4 & 2 \end{bmatrix}\] subtract row 1 from row 3. \[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}\] There is one independent row in this matrix. \[ Rank =1 \]
1-3). What is the rank of matrix B?
B <- matrix(data = c(1,2,1,3,6,3,2,4,2), nrow = 3, ncol = 3, byrow = TRUE)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2matrix.rank(B)## [1] 1# Answer 1-3: the rank of A is 1
# [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.
Eigenvalues:
\[A = \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}\]
\[ \lambda,I_3 = \begin{bmatrix}\lambda & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda \end{bmatrix}\]
\[\lambda I_3 - A = \begin{pmatrix} \lambda -1 & -2 & -3\\ 0 & \lambda - 4 & -5 \\ 0 & 0 & \lambda -6 \end{pmatrix}\]
\[det ( \lambda I_3 -A ) = 0\] \[(\lambda-6)(\lambda-4)(\lambda-1)-(-2)(0) +-3(0)\] \[(\lambda-6)(\lambda-4)(\lambda-1) = 0\]
\[\lambda = 6 , \lambda=4 , \lambda = 1\]
characteristic polynomial = \[x^3 - 9x^2 +24x - 16\]
Eigenvectors:
1). lambda = 1 \[\lambda=1\]
\[ \begin{bmatrix}\lambda - 1 & 2 & 3 \\0 & \lambda - 4 & 5 \\0 & 0 & \lambda - 6 \end{bmatrix}\]
\[ \begin{bmatrix}0 & 2 & 3 \\0 & -3 & 5\\0 & 0 & -5\end{bmatrix},\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\]
\[first, pivot = 0,x_1 = free,with, value=1\]
\[-3 x_2 + 5 x_3 = 0 ,and, 5 x_3 = 0\]
\[x_{\lambda=1},=\begin{bmatrix}1 \\0 \\0\end{bmatrix}\]
2). lambda = 4 \[ \begin{bmatrix}\lambda - 1 & 2 & 3 \\0 & \lambda - 4 & 5 \\0 & 0 & \lambda - 6 \end{bmatrix}\]
\[\lambda=4\] \[ \begin{bmatrix}3 & 2 & 3 \\0 & 0 & 5\\0 & 0 & -2\end{bmatrix},\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\]
\[Second,pivot = 0,x_2=free, with value=1.\]
\[3x_1+2x_2 +3x_3 = 0,and, 2x_3 = 0\]
\[x_3=0,x_2=1,and,x_1=-2/3\]
\[x_{\lambda=4},=\begin{bmatrix}-2/3\\1 \\0\end{bmatrix}\]
3). lambda = 6
\[ \begin{bmatrix}\lambda - 1 & 2 & 3 \\0 & \lambda - 4 & 5 \\0 & 0 & \lambda - 6 \end{bmatrix}\]
\[\lambda=6\] \[ \begin{bmatrix}5 & 2 & 3\\0 & 2 & 5\\0 & 0 & 0\end{bmatrix},\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\]
\[Third,pivot = 0 ,x_3=free,with value=1.\]
\[5x_1 +2x_2 +3x_3 = 0 ,and, 2x_1+5x_3 = 0\]
\[x_3 = 1,x_2 = 2/5,and,x_1=-5/2\]
\[x_{\lambda=6},=\begin{bmatrix}-5/2 \\2/5 \\1\end{bmatrix}\]
A <- matrix(data = c(1,0,0,2,4,0,3,5,6), nrow = 3, ncol = 3, byrow = FALSE)
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6EV1 <- eigen(A)
EV1## $values
## [1] 6 4 1
##
## $vectors
## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0EV2 <- EV1$values
EV2## [1] 6 4 1END