Problem set 1

1-What is the rank of the matrix A?

\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]

As we know that the Nonsingular Matrix Equivalence Theorem, often denoted as NME8, states that for a square matrix \(A\)

the following statements are equivalent:

1- \(A\) is nonsingular (invertible).

2- rank \((A) = n\) , where n is the number of rows (or columns) of \(A\)

3- The determinant of A is nonzero \((det(A ≠ 0)\)

4- The columns (or rows) of are linearly independent

5- The reduced row echelon form of \(A\) is the identity matrix.

# Define 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)

# Compute the determinant of A
det_A <- det(A)

# Compute the rank of A
rank_A <- Matrix::rankMatrix(A)

# Check if A is nonsingular (invertible) using NME8
if (det_A != 0) {
  cat("Matrix A is nonsingular.\n")
  cat("Rank of A:", rank_A, "\n")
} else {
  cat("Matrix A is singular.\n")
}
## Matrix A is nonsingular.
## Rank of A: 4

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?

For an \(m × n\) matrix where \(m > n\) , the maximum rank that can be achieved is \(n\), which is the smaller dimension. This is because the rank of a matrix is bounded by the number of columns or rows, whichever is smaller. Since in this case,\(n\) is smaller, the maximum rank cannot exceed n.

On the other hand, the minimum rank of the matrix, assuming it is non-zero, is 1. This is because a non-zero matrix will always have at least one linearly independent row or column, which implies that the rank cannot be zero, So,Maximum rank: \(n\) and Minimum rank: \(1\)

# Example of an m x n matrix where m > n
m <- 4  # number of rows
n <- 2  # number of columns

# Create a random non-zero matrix of size m x n
set.seed(123)  # for reproducibility
example_matrix <- matrix(runif(m * n, min = -5, max = 5), nrow = m)

# Display the matrix
print(example_matrix)
##            [,1]       [,2]
## [1,] -2.1242248  4.4046728
## [2,]  2.8830514 -4.5444350
## [3,] -0.9102308  0.2810549
## [4,]  3.8301740  3.9241904
# Compute the rank of the matrix
rank_example <- Matrix::rankMatrix(example_matrix)
print(paste("Rank of the matrix:", rank_example))
## [1] "Rank of the matrix: 2"

The maximum rank that can be achieved is 2 because \(n = 2\), which is the smaller dimension of the matrix. as computed, the rank of the matrix is indeed 2

This confirms the theoretical expectation that the maximum rank for this matrix, where \(m > n\)

3- What is the rank of matrix B?

\[\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\] 
# Define matrix B
B <- matrix(c(1, 2, 1,
              3, 6, 3,
              2, 4, 2), 
            nrow = 3, byrow = TRUE)

# Compute the rank of B
rank_B <- Matrix::rankMatrix(B)
print(paste("Rank of matrix B:", rank_B))
## [1] "Rank of matrix B: 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.

\(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\)

Solution :

A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = TRUE)
print(A)
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

Since A is a triangular matrix, its eigenvalues are values on the diagonal, so

\(λ1 = 1\), \(λ2 = 4\), and λ3 = 6

Double-check eigenvalues in R

eigen(A)$values
## [1] 6 4 1

The characteristic polynomial is \(p_A(\lambda) = (1 - \lambda)(4 - \lambda)(6 - \lambda)\) or

\(p_A(\lambda) = 24 - 34\lambda + 11\lambda^2 - \lambda^3\).

If \(\lambda = 1\), then \(A - \lambda I_3\) is row-reduced to:

rref(A - 1 * diag(3))
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\[ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \]

This gives us the *eigenspace8 \(E_{\lambda=1} = \langle \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \rangle\).

If \(\lambda = 4\), then \(A - \lambda I_3\) is row-reduced to:

rref(A - 4 * diag(3))
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

\[ \begin{pmatrix} 1 & -0.6666667 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \]

This gives us the eigenspace \(E_{\lambda=4} = \langle \begin{pmatrix} 1 \\ 1.5 \\ 0 \end{pmatrix} \rangle\).

If \(\lambda = 6\), then \(A - \lambda I_3\) is row-reduced to:

rref(A - 6 * diag(3))
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[ \begin{pmatrix} 1 & 0 & -1.6 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \\ \end{pmatrix} \]

This gives us the eigenspace \(E_{\lambda=6} = \langle \begin{pmatrix} 1.6 \\ 2.5 \\ 1 \end{pmatrix} \rangle\).