Problem set 1

1.1 What is the rank of the matrix A?

Answer: 4. The number of non-zero rows in the matrix reduced row echelon form \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]

A1 <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), nrow=4, ncol=4)
A1
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
rref(A1)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1
rankMatrix(A1)
## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16

1.2 Given an mxn matrix where m > n, what can be the maximum rank? The mini- mum rank, assuming that the matrix is non-zero? Answer: The maximum rank can be equal or less than n, while the minimum rank for non-zero matrix is 1

1.3 What is the rank of the matrix B? Answer: 1. The number of non-zero row in the matrix reduced row echelon form

\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right] \]

A3 <- matrix(c(1, 3, 2, 2, 6, 4, 1, 3, 2), nrow=3, ncol=3)
A3
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rref(A3)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0
rankMatrix(A3)
## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16

Problem set 2

Compute the eigenvalues and eigenvectors of the matrix A. \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] \] \[Av = \lambda v\] \[0 = \lambda I_nv - Av\] \[0 = (\lambda I_n - A)v\] \[\lambda I_n - A = \ \lambda \left[\begin{array} {rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] -\left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] =\left[\begin{array} {rrr} \lambda-1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda-6 \end{array}\right] \]

\[ det(\lambda I_n - A) = (\lambda-1)[(\lambda-4)(\lambda-6)] = 0 \\ (\lambda^2 -4\lambda -\lambda + 4)(\lambda-6) = 0 \\ (\lambda^2 -5\lambda + 4)(\lambda-6) = 0 \\ \lambda^3 -5\lambda^2 + 4\lambda -6\lambda^2 + 30\lambda -24 = 0 \\ \lambda^3 -11\lambda^2 + 34\lambda -24 = 0 \\ \] Eigenvalues:

\[ \lambda = 1, \ \lambda = 4, \ \lambda = 6 \] Find Eigenvectors

\(\lambda = 1\)

\[0 = (\lambda I_n - A)v =\left[\begin{array} {rrr} \lambda-1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda-6 \end{array}\right] =\left[\begin{array} {rrr} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{array}\right] =\left[\begin{array} {rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \\ v = \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right] = |v| = \left[\begin{array} {rrr} 1 / \sqrt{(1^2)} \\ 0 / \sqrt{(1^2)} \\ 0 / \sqrt{(1^2)} \end{array}\right] = \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right] \]

\(\lambda = 4\) \[0 = (\lambda I_n - A)v =\left[\begin{array} {rrr} \lambda-1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda-6 \end{array}\right] =\left[\begin{array} {rrr} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{array}\right] =\left[\begin{array} {rrr} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \\ v = \left[\begin{array} {rrr} 2/3 \\ 1 \\ 0 \end{array}\right] = |v| = \left[\begin{array} {rrr} (2/3) / \sqrt{(1^2) + (2/3)^2} \\ 1 / \sqrt{(1^2) + (2/3)^2} \\ 0 / \sqrt{(1^2) + (2/3)^2 } \end{array}\right] = \left[\begin{array} {rrr} 0.5547 \\ 0.83205 \\ 0 \end{array}\right] \]

\(\lambda = 6\) \[0 = (\lambda I_n - A)v =\left[\begin{array} {rrr} \lambda-1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda-6 \end{array}\right] =\left[\begin{array} {rrr} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{array}\right] =\left[\begin{array} {rrr} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array}\right] \\ v = \left[\begin{array} {rrr} 1.6 \\ 2.5 \\ 1 \end{array}\right] = |v| = \left[\begin{array} {rrr} 1.6 / \sqrt{(1.6^2) + (2.5)^2 + (1)^2} \\ 2.5 / \sqrt{(1.6^2) + (2.5)^2 + (1)^2} \\ 1 / \sqrt{(1.6^2) + (2.5)^2 + (1)^2} \end{array}\right] = \left[\begin{array} {rrr} 0.510840685 \\ 0.798188571 \\ 0.319275428 \end{array}\right] \]

A <- matrix(c(1,0,0,2,4,0,3,5,6), nrow=3, ncol=3)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
eigen(A)
## eigen() decomposition
## $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