IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - 2014
\[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\-1 & 0 & 1 & 3\\0 & 1 & -2 & 1\\5 & 4 & -2 & -3\end{array}\right] \]
A = matrix(c(1,2,3,4, -1,0,1,3, 0,1,-2,1, 5,4,-2,-3), nrow=4, ncol=4, byrow=TRUE)
paste0("A=")## [1] "A="A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1print(paste("rank of matrix A is ",rankMatrix(A)))## [1] "rank of matrix A is 4"qr(A)$rank## [1] 4The rank of a matrix is defined as the number of linearly independent rows or linearly independent columns in a matrix A https://stattrek.com/matrix-algebra/matrix-rank.aspx
The process for computing the rank involves translating a matrix into echelon form. Putting this matrix into echelon form will be a tedious process but we will go through each and every step.
(R2 <- R1+R2) Add first row to second row: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 2 & 4 & 7\\0 & 1 & -2 & 1\\5 & 4 & -2 & -3\end{array}\right] \] (R4 <- -5R1+R4) Add -5 times row 1 to row 4: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 2 & 4 & 7\\0 & 1 & -2 & 1\\0 & -6 & -17 & -23\end{array}\right] \] (R2 <- R2/2) Divide row 2 by 2: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 1 & -2 & 1\\0 & -6 & -17 & -23\end{array}\right] \] (R3 <- R3-R2) Subtract third row from second row: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & -4 & -5/2\\0 & -6 & -17 & -23\end{array}\right] \] (R4 <- 6R2+R4) Add 6 times row 2 to row 4: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & -4 & -5/2\\0 & 0 & -5 & -2\end{array}\right] \] (R3 <- R3/-4) Divide row 3 by -4: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & 1 & 5/8\\0 & 0 & -5 & -2\end{array}\right] \] (R4 <- 5R3+R4) Add 5 times row 3 to row 4: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & 1 & 5/8\\0 & 0 & 0 & 9/8\end{array}\right] \] (R4 <- R4/(9/8)) Divide row 4 by 9/8: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & 1 & 5/8\\0 & 0 & 0 & 1\end{array}\right] \] (R3 <- R4/(-5/8)+R3) Add -5/8 row 4 to row 3: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 7/2\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] (R2 <- R2+(-7/2)R3) Add -7/2 row 4 to row 2: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 4\\0 & 1 & 2 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] (R1 <- -4R1+R4) Add -4 times row 1 to row 4: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 0\\0 & 1 & 2 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] (R2 <- R2+(-2)R3) Add -2 times row 3 plus row 2: \[ A = \left[\begin{array}{rrr}1 & 2 & 3 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] (R1 <- R1+(-3)R3) Add -3 times row 3 plus row 1: \[ A = \left[\begin{array}{rrr}1 & 2 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] (R1 <- R1+(-2)R1) Add -2 times row 2 plus row 1: \[ A = \left[\begin{array}{rrr}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right] \] The rank of matrix A is 4 since there are 4 non zero linearly independent rows
Please typeset your response using LaTeX mode in RStudio. If you do it in paper, please either scan or take a picture of the work and submit it. Please ensure that your image is legible and that your submissions are named using your rst initial, last name, assignment and problem set within the assignment. E.g. LFulton_Assignment2_PS1.png
For an mxn matrix where m > n, let B be a row-equivalent matrix in reduced row-echelon form.
Since m(rows) > n(columns), maximum rank of the matrix is n (number of columns)
Minimum rank of a non-zero matrisx is 1.
The maximum rank would be equal to n since the maximum rank is the lesser of either the number of rows or the number of columns. Since in this case n is less than m, n is the maximum rank. The minimum rank of any non-zero matrix is 1.
According to Theorem CRN(Computing Rank and Nullity), Let r denote the number of pivot columns (or the number of nonzero rows),then r(A) = r.
For matrix B, the left most nonzero entry of each nonzero row is equal to 1 and it is the only nonzero entry in its column. The pivot columns could form a indentity matrix if we rearrange matrix B. If this indentity matrix is in size kxk, then it is obvious that the maximum of k is equal to m or n which ever is smaller. Here m > n, so we know r < n. For nonzero matrix, the minimum number of pivot columns would be 1. So r >= 1. The minimum rank could be 1.
\[ B = \left[\begin{array}{rrr}1 & 2 & 1\\3 & 6 & 3\\2 & 4 & 2\end{array}\right] \]
B <- matrix(c(1,2,1, 3,6,3, 2,4,2), nrow=3, ncol=3, byrow=TRUE)
paste0("B=")## [1] "B="B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2rref(B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0print(paste("rank of matrix A is ",rankMatrix(B)))## [1] "rank of matrix A is 1"qr(B)$rank## [1] 1(R2 <- -3R1+R2) Add first row to second row: \[ B = \left[\begin{array}{rrr}1 & 2 & 1\\0 & 0 & 0\\2 & 4 & 2\end{array}\right] \] (R3 <- -2R1+R3) Add first row to second row: \[ B = \left[\begin{array}{rrr}1 & 2 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right] \] Since there is only one non-zero row, the rank of B is 1.
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.
\[ A = \left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] \]
Please show your work using an R-markdown document. Please name your assignment submission with your rst initial and last name.
A <- matrix(c(1,2,3, 0,4,5, 0,0,6), nrow=3, ncol=3, byrow=TRUE)
eigen_a <- eigen(A)
paste("eigenValues includes:")## [1] "eigenValues includes:"(eigen_a$values)## [1] 6 4 1paste("eigenVectors includes:")## [1] "eigenVectors includes:"(eigen_a$vectors)## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0#characteristic polynomial
p <- charpoly(A, info = TRUE)## Error term: 0p$cp## [1] 1 -11 34 -24Therefore, the characteristic polynomial is \[ (\lambda^3 - 11\lambda^2 + 34\lambda - 24) = 0 \]
To find the eigenvalues of A, we must solve \[det( \lambda I-A) =0\] for \[\lambda\] where I is Identitity Matrix.
\[ Ax = \lambda x \] \[ Ax = \lambda Ix \] \[ Ax - \lambda Ix = 0 \]
\[ (A - \lambda I) x = 0 \] \[ det(A - \lambda I) = 0 \]
\[ I = \left[\begin{array}{rrr}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right] \]
\[det(A - \lambda I) =0\] \[det(\left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] - \lambda \left[\begin{array}{rrr}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]) = 0 \]
\[det(\left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] - \left[\begin{array}{rrr}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{array}\right]) = 0 \]
\[det(\left[\begin{array}{rrr}1-\lambda & 2 & 3\\0 & 4-\lambda & 5\\0 & 0 & 6-\lambda\end{array}\right]) = 0 \] \[ ((1-\lambda)(4-\lambda)(6-\lambda) + (2*5*0) + (3*0*0)) - (3(4-\lambda)*0 + (1-\lambda)*5*0 + (2*0*(6-\lambda))) = 0 \] \[ ((1-\lambda)(4-\lambda)(6-\lambda) = 0 \] \[ (4-5\lambda+\lambda^2)(6-\lambda) = 0 \] \[ (24-34\lambda+11\lambda^2-\lambda^3) = 0 \] \[ (\lambda^3 - 11\lambda^2 + 34\lambda - 24) = 0 \]
The eigenvalues are: 1, 4, and 6
\[ A - \lambda I = 0 \] \[ \lambda = 1 \]
\[ (\left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] - \left[\begin{array}{rrr}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]) * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \] \[ \left[\begin{array}{rrr}0 & 2 & 3\\0 & 3 & 5\\0 & 0 & 5\end{array}\right] * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \]
A1 = matrix(c(0,2,3,0,3,5,0,0,5), nrow = 3, byrow = T)
rref(A1)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0# augment the matrices with 0 vector, perform elimination, and solve by setting free variable to 1
rref( cbind( A-1*diag(3), c(0,0,0)) )## [,1] [,2] [,3] [,4]
## [1,] 0 1 0 0
## [2,] 0 0 1 0
## [3,] 0 0 0 0\[ e_1 = \left[\begin{array}{rrr}1\\0\\0\end{array}\right] \]
\[ \lambda = 4 \]
\[ (\left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] - \left[\begin{array}{rrr}4 & 0 & 0\\0 & 4 & 0\\0 & 0 & 4\end{array}\right]) * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \] \[ \left[\begin{array}{rrr}-3 & 2 & 3\\0 & 0 & 5\\0 & 0 & 2\end{array}\right] * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \]
A4 = matrix(c(0,2,3,0,3,5,0,0,5), nrow = 3, byrow = T)
rref(A4)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0# augment the matrices with 0 vector, perform elimination, and solve by setting free variable to 4
rref( cbind( A-4*diag(3), c(0,0,0)) )## [,1] [,2] [,3] [,4]
## [1,] 1 -0.6666667 0 0
## [2,] 0 0.0000000 1 0
## [3,] 0 0.0000000 0 0\[ e_4 = \left[\begin{array}{rrr}0.667\\1\\0\end{array}\right] \]
\[ \lambda = 6 \]
\[ (\left[\begin{array}{rrr}1 & 2 & 3\\0 & 4 & 5\\0 & 0 & 6\end{array}\right] - \left[\begin{array}{rrr}6 & 0 & 0\\0 & 6 & 0\\0 & 0 & 6\end{array}\right]) * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \] \[ \left[\begin{array}{rrr}-5 & 2 & 3\\0 & -2 & 5\\0 & 0 & 0\end{array}\right] * \left[\begin{array}{rrr}v_1\\v_2\\v_3\end{array}\right] = 0 \]
A6 = matrix(c(0,2,3,0,3,5,0,0,5), nrow = 3, byrow = T)
rref(A6)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0# augment the matrices with 0 vector, perform elimination, and solve by setting free variable to 6
rref( cbind( A-6*diag(3), c(0,0,0)) )## [,1] [,2] [,3] [,4]
## [1,] 1 0 -1.6 0
## [2,] 0 1 -2.5 0
## [3,] 0 0 0.0 0\[ e_6 = \left[\begin{array}{rrr}1.6\\2.5\\1\end{array}\right] \]
The eigenvectors are: