\[A = \begin{bmatrix}1&2&3&4\\-1&0&1&3\\0&1&-2&1\\5&4&-2&-3\end{bmatrix}\] The rank of a mxn matrix A is defined as the dimension of the column space of A, or \(r(A) = dim(C(A))\). Using reduced row echelon form, we can determine the dimensions and null spaces. Using the matlib function, echelon, I calculated the reduced rows of Matrix A below:
# Define Matrix A
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol = 4)
# Reduce rows
echelon(A, verbose = T)##
## Initial matrix:
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3
##
## row: 1
##
## exchange rows 1 and 4
## [,1] [,2] [,3] [,4]
## [1,] 5 4 -2 -3
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 1 2 3 4
##
## multiply row 1 by 0.2
## [,1] [,2] [,3] [,4]
## [1,] 1 0.8 -0.4 -0.6
## [2,] -1 0.0 1.0 3.0
## [3,] 0 1.0 -2.0 1.0
## [4,] 1 2.0 3.0 4.0
##
## multiply row 1 by 1 and add to row 2
## [,1] [,2] [,3] [,4]
## [1,] 1 0.8 -0.4 -0.6
## [2,] 0 0.8 0.6 2.4
## [3,] 0 1.0 -2.0 1.0
## [4,] 1 2.0 3.0 4.0
##
## subtract row 1 from row 4
## [,1] [,2] [,3] [,4]
## [1,] 1 0.8 -0.4 -0.6
## [2,] 0 0.8 0.6 2.4
## [3,] 0 1.0 -2.0 1.0
## [4,] 0 1.2 3.4 4.6
##
## row: 2
##
## exchange rows 2 and 4
## [,1] [,2] [,3] [,4]
## [1,] 1 0.8 -0.4 -0.6
## [2,] 0 1.2 3.4 4.6
## [3,] 0 1.0 -2.0 1.0
## [4,] 0 0.8 0.6 2.4
##
## multiply row 2 by 0.8333333
## [,1] [,2] [,3] [,4]
## [1,] 1 0.8 -0.400000 -0.600000
## [2,] 0 1.0 2.833333 3.833333
## [3,] 0 1.0 -2.000000 1.000000
## [4,] 0 0.8 0.600000 2.400000
##
## multiply row 2 by 0.8 and subtract from row 1
## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 -2.666667 -3.666667
## [2,] 0 1.0 2.833333 3.833333
## [3,] 0 1.0 -2.000000 1.000000
## [4,] 0 0.8 0.600000 2.400000
##
## subtract row 2 from row 3
## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 -2.666667 -3.666667
## [2,] 0 1.0 2.833333 3.833333
## [3,] 0 0.0 -4.833333 -2.833333
## [4,] 0 0.8 0.600000 2.400000
##
## multiply row 2 by 0.8 and subtract from row 4
## [,1] [,2] [,3] [,4]
## [1,] 1 0 -2.666667 -3.6666667
## [2,] 0 1 2.833333 3.8333333
## [3,] 0 0 -4.833333 -2.8333333
## [4,] 0 0 -1.666667 -0.6666667
##
## row: 3
##
## multiply row 3 by -0.2068966
## [,1] [,2] [,3] [,4]
## [1,] 1 0 -2.666667 -3.6666667
## [2,] 0 1 2.833333 3.8333333
## [3,] 0 0 1.000000 0.5862069
## [4,] 0 0 -1.666667 -0.6666667
##
## multiply row 3 by 2.666667 and add to row 1
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.000000 -2.1034483
## [2,] 0 1 2.833333 3.8333333
## [3,] 0 0 1.000000 0.5862069
## [4,] 0 0 -1.666667 -0.6666667
##
## multiply row 3 by 2.833333 and subtract from row 2
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.000000 -2.1034483
## [2,] 0 1 0.000000 2.1724138
## [3,] 0 0 1.000000 0.5862069
## [4,] 0 0 -1.666667 -0.6666667
##
## multiply row 3 by 1.666667 and add to row 4
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 -2.1034483
## [2,] 0 1 0 2.1724138
## [3,] 0 0 1 0.5862069
## [4,] 0 0 0 0.3103448
##
## row: 4
##
## multiply row 4 by 3.222222
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 -2.1034483
## [2,] 0 1 0 2.1724138
## [3,] 0 0 1 0.5862069
## [4,] 0 0 0 1.0000000
##
## multiply row 4 by 2.103448 and add to row 1
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0.0000000
## [2,] 0 1 0 2.1724138
## [3,] 0 0 1 0.5862069
## [4,] 0 0 0 1.0000000
##
## multiply row 4 by 2.172414 and subtract from row 2
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0.0000000
## [2,] 0 1 0 0.0000000
## [3,] 0 0 1 0.5862069
## [4,] 0 0 0 1.0000000
##
## multiply row 4 by 0.5862069 and subtract from row 3
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1The column index dimensions of the reduced Matrix A can be shown as \(D = {1,2,3,4}\). Each index represents an individual vector, meaning that \(r(A) = 4\). This can also be calculated using the R function found in the matlib package:
# Calculate rank of matrix
rankA <- R(A)
rankA## [1] 4The maximum rank of the matrix is n. The minimum rank of all non-zero matrices is 1.
\[B = \begin{bmatrix}1&2&1\\3&6&3\\2&4&2\end{bmatrix} \\\]
# Define Matrix A
B <- matrix(c(1,3,2,2,6,4,1,3,2), ncol = 3)
# Reduce rows
echelon(B, verbose = T)##
## Initial matrix:
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2
##
## row: 1
##
## exchange rows 1 and 2
## [,1] [,2] [,3]
## [1,] 3 6 3
## [2,] 1 2 1
## [3,] 2 4 2
##
## multiply row 1 by 0.3333333
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 1 2 1
## [3,] 2 4 2
##
## subtract row 1 from row 2
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 2 4 2
##
## multiply row 1 by 2 and subtract from row 3
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0
##
## row: 2In this example, you can clearly see that matrix B has 1 non-zero row, thus the rank of this Matrix is 1. This is further proved using the R function below:
# Calculate rank
rankB <- R(B)
rankB## [1] 1Compute 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.
\[A = \begin{bmatrix}1&2&3\\0&4&5\\0&0&6\end{bmatrix} \\\]
Because Matrix A is a upper-triangle matrix, we know that the eigenvalues of A are the diagonals. However, we can also calculate the values and vectors from the characteristic polynomial of Matrix A.
The eigenvalues of Matrix A can be solved using the function, \(f(\lambda)=det(\lambda I - A)=0\):
\(= \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}-\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\end{bmatrix}=0 \\ = \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{bmatrix}-\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\end{bmatrix}=0 \\ = \begin{bmatrix}\lambda -1&-2&-3\\0&\lambda-4&-5\\0&0&\lambda-6\end{bmatrix}=0\\= (\lambda-1)(\lambda-4)(\lambda-6)=0\)
We can use R to compute the characteristic polynomial using the charpoly function from the pracma package.
# Define matrix
A <- matrix(c(1,0,0,2,4,0,3,5,6), ncol= 3)
# Compute characteristic polynomial
p <- Poly(A)
p## [1] 1 -11 34 -24We can now determine characteristic polynomial is \(\lambda^3-11\lambda ^2+34 \lambda-24 = 0\) and can be solved using the following values:
\[\begin{matrix}\lambda_1=1\\\lambda_2=4\\\lambda_3=6\end{matrix}\] These solutions can be checked in R using the eigen function:
# Check eigenvalues
lam <- eigen(A)$values
lam## [1] 6 4 1To get the eigenvector, we substitue each \(\lambda\) value into the matrix:
\(det(\lambda I-A)=\begin{bmatrix}\lambda-1&-2&-3\\0&\lambda-4&-5\\0&0&\lambda-6\end{bmatrix}=0\)
Solve for \(\lambda=1\):
\[A-1\lambda =\begin{bmatrix}1-1&-2&-3\\0&1-4&-5\\0&0&1-6\end{bmatrix}=\begin{bmatrix}0&-2&-3\\0&-3&-5\\0&0&-5\end{bmatrix}=0\\\]
# Define Matrix when Lambda = 1
lam=1
lam1 <- matrix(c(lam-1,0,0,-2,lam-4,0,-3,-5,lam-6), ncol= 3)
# Calculate eigen space
lam1 <- echelon(lam1)
lam1 <- lam1[1:3,2]
# Show vector values for Lambda = 1
as.matrix(lam1)## [,1]
## [1,] 1
## [2,] 0
## [3,] 0Solve for \(\lambda=4\):
\[A-4\lambda =\begin{bmatrix}4-1&-2&-3\\0&4-4&-5\\0&0&4-6\end{bmatrix}=\begin{bmatrix}3&-2&-3\\0&0&-5\\0&0&-2\end{bmatrix}=0\\\]
# Define Matrix when Lambda = 4
lam=4
lam2 <- matrix(c(lam-1,0,0,-2,lam-4,0,-3,-5,lam-6), ncol= 3)
lam2 ## [,1] [,2] [,3]
## [1,] 3 -2 -3
## [2,] 0 0 -5
## [3,] 0 0 -2# Calculate eigen space
lam2 <- echelon(lam2)
lam2## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0lam2 <- lam2[1:3,2]
# Show vector values for Lambda = 4
as.matrix(lam2)## [,1]
## [1,] -0.6666667
## [2,] 0.0000000
## [3,] 0.0000000Solve for \(\lambda=6\):
\[A-6\lambda =\begin{bmatrix}6-1&-2&-3\\0&6-4&-5\\0&0&6-6\end{bmatrix}=\begin{bmatrix}5&-2&-3\\0&2&-5\\0&0&0\end{bmatrix}=0\\\]
# Define Matrix when Lambda = 6
lam=6
lam3 <- matrix(c(lam-1,0,0,-2,lam-4,0,-3,-5,lam-6), ncol= 3)
lam3 ## [,1] [,2] [,3]
## [1,] 5 -2 -3
## [2,] 0 2 -5
## [3,] 0 0 0# Calculate eigen space
lam3 <- echelon(lam3)
lam3## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0lam3 <- lam3[1:3,3]
# Show vector values for Lambda = 6
as.matrix(lam3)## [,1]
## [1,] -1.6
## [2,] -2.5
## [3,] 0.0Eigenvector:
v <- c(lam3, lam2, lam1)
matrix(v, ncol=3)## [,1] [,2] [,3]
## [1,] -1.6 -0.6666667 1
## [2,] -2.5 0.0000000 0
## [3,] 0.0 0.0000000 0