1. What is the rank of the matrix A?
\[ A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{pmatrix} \]
Matrix A is a 4x4 matrix, then the rank is 4.
2. Maximum and Minimum Rank of a Matrix
For a matrix of size \(m \times n\) with \(m > n\), the maximum possible rank is \(n\), which corresponds to the maximum number of linearly independent column vectors in the matrix. The minimum rank, assuming the matrix is non-zero, is 1, because even a single non-zero element in the matrix indicates a non-zero determinant for some \(1 \times 1\) submatrix.
3. What is the rank of Matrix B
\[ B = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{pmatrix} \]
Matrix B is a 3x3 matrix, then the rank is 3.
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix} \]
Computing Eigenvalues and Eigenvectors:
# Defining the matrix A
A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = TRUE)
# Calculating eigenvalues and eigenvectors
eigen_A <- eigen(A)
eigen_A$values## [1] 6 4 1eigen_A$vectors## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0Characteristic Polynomial of Matrix A
The eigenvalues are 1, 4, and 6. The characteristic polynomial \(p(\lambda)\) is given by the determinant of \(A - \lambda I\):
\[p(\lambda) = \det(A - \lambda I) = (\lambda - 1)(\lambda - 4)(\lambda - 6)\]
# Assuming matrix A is already defined as before
# Calculate the characteristic polynomial coefficients
char_poly <- function(lambda) {
det(matrix(c(1 - lambda, 2, 3,
0, 4 - lambda, 5,
0, 0, 6 - lambda), nrow = 3, byrow = TRUE))
}
# Display the characteristic polynomial for each eigenvalue
eigenvalues <- c(1, 4, 6)
sapply(eigenvalues, char_poly)## [1] 0 0 0My problem here so far, how to ask R to express the result as an expression of \(\lambda\).