1. PROBLEM SET 1

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

A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 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   -3
A[2,] <- A[2,] - A[2,1]/A[1,1]*A[1,]
A[4,] <- A[4,] - A[4,1]/A[1,1]*A[1,]

A[3,] <- A[3,] - A[3,2]/A[2,2]*A[2,]
A[4,] <- A[4,] - A[4,2]/A[2,2]*A[2,]

A[4,] <- A[4,] - A[4,3]/A[3,3]*A[3,]

(A)
##      [,1] [,2] [,3]   [,4]
## [1,]    1    2    3  4.000
## [2,]    0    2    4  7.000
## [3,]    0    0   -4 -2.500
## [4,]    0    0    0  1.125
A[2,] <- A[2,]/A[2,2]
A[3,] <- A[3,]/A[3,3]
A[4,] <- A[4,]/A[4,4]

(A)
##      [,1] [,2] [,3]  [,4]
## [1,]    1    2    3 4.000
## [2,]    0    1    2 3.500
## [3,]    0    0    1 0.625
## [4,]    0    0    0 1.000
A[3,] <- A[3,] - A[3,4]*A[4,]
A[2,] <- A[2,] - A[2,4]*A[4,]
A[1,] <- A[1,] - A[1,4]*A[4,]

A[2,] <- A[2,] - A[2,3]*A[3,]
A[1,] <- A[1,] - A[1,3]*A[3,]

A[1,] <- A[1,] - A[1,2]*A[2,]

(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

Since \(A\) can be shown to be reduced to echelon form without any non-zero rows, the rank of the matrix is equal to the number of rows in \(A\), 4.

(2) Given an \(mxn\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming the matrix is non-zero?

Given that \(m > n\), the maximum rank would be equal to the number of columns, \(n\). If the matrix was non-zero. The minimum rank would be one, if it had at least one non-zero element.

(3) What is the rank of matrix B?

B <- matrix(c(1,2,1,3,6,3,2,4,2), nrow = 3, byrow = TRUE)
(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
B[2,] <- B[2,] - B[2,1]/B[1,1]*B[1,]
B[3,] <- B[3,] - B[3,1]/B[1,1]*B[1,]
(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Since the the matrix has been reduced to Echelon form with only one non-zero row, the rank of the matrix is 1.

2. 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.

The equation to find the eigenvalues will take the form \[Ax = \lambda x \rightarrow (A - \lambda I)x = 0\] Since \((A-\lambda I)\) must be singular, the determinant of \((A - \lambda I)\) must be 0.

The determinant of \((A- \lambda I)\) and the corresponding eiganvalues:

\[det(A-\lambda I) = det\left ( \begin{bmatrix} 1 -\lambda & 2 & 3\\ 0 & 4 -\lambda & 5\\ 0 & 0 & 6 -\lambda \end{bmatrix} \right )\]

\[= (1-\lambda)\begin{vmatrix} 4-\lambda & 5\\ 0 & 6 - \lambda \end{vmatrix} - 2\begin{vmatrix} 0 & 5\\ 0 & 6 - \lambda \end{vmatrix} + 3\begin{vmatrix} 0 & 4 - \lambda\\ 0 & 0 \end{vmatrix}\]

\[= (1-\lambda)(4-\lambda)(6-\lambda) -2(0) + 3(0)= 0\] \[\therefore \lambda_1 = 1; \lambda_2 = 4; \lambda_3 = 6\] It seems that for any upper matrix U, the eigenvalues would be the diagonal.

Solving for the corresponding eiganvectors:

\[E_{1}=N\left ( \begin{bmatrix} 1 -1 & 2 & 3\\ 0 & 4 -1 & 5\\ 0 & 0 & 6 -1 \end{bmatrix} \right )=N\left ( \begin{bmatrix} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5 \end{bmatrix} \right )=N\left ( \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \right ) \\ E_1 =\left \{ \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =t\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\right \}\]

\[E_{4}=N\left ( \begin{bmatrix} 1 -4 & 2 & 3\\ 0 & 4 -4 & 5\\ 0 & 0 & 6 -4 \end{bmatrix} \right )=N\left ( \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2 \end{bmatrix} \right )=N\left ( \begin{bmatrix} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \right ) \\ E_4 =\left \{ \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=t\begin{bmatrix} x_1 = \frac{2}{3}x_2\\ x_2 = 1\\ 0 \end{bmatrix} =t\begin{bmatrix} \frac{2}{3}\\ 1\\ 0 \end{bmatrix} \right \}\]

\[E_{6}=N\left ( \begin{bmatrix} 1 -6 & 2 & 3\\ 0 & 4 -6 & 5\\ 0 & 0 & 6 -6 \end{bmatrix} \right )=N\left ( \begin{bmatrix} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{bmatrix} \right )=N\left ( \begin{bmatrix} 1 & 0 & -\frac{8}{5}\\ 0 & 1 & -\frac{5}{2}\\ 0 & 0 & 0 \end{bmatrix} \right ) \\ E_6 =\left \{ \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =t \begin{bmatrix} x_1 = \frac{8}{5}x_3 \\ x_2 = \frac{5}{2}x_3 \\ x_3 = 1 \end{bmatrix}=t \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix}\right \}\]