(1) What is the rank of the 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))
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
print(det(A))
## [1] -9

The determinant of A is non-zero therefore, the rank of matrix A is 4 ()

print(Rank(A))
## [1] 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?

Answer:

Since the rank is the number of all non-zero row, the rank has to be no greater than the minimum value of either m or n.

(3) What is the rank of matrix B?

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

From the above matrix, Row 2 and Row 3 are multiples of Row 1 in varying degrees. Therefore,

R1 <- 
R2 <- 
R3 <- 

a <- B[1, ]-(1/3)%*%B[2, ]
b <- B[3, ]-(2/3)%*%B[2, ]

matrix(c(a,b,B[2, ]), nrow = 3, byrow = T)
##      [,1] [,2] [,3]
## [1,]    0    0    0
## [2,]    0    0    0
## [3,]    3    6    3

This implies that the rank of matirx B 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.

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

\[ Ax = \lambda x \] \[ Ax - \lambda x = 0 \]

\[ det \left( A - \lambda I \right)x = 0\]

\[\begin{equation*} \mathbf{} A-\lambda I = \left[\begin{matrix} 1 - \lambda & 2 & 4\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda \end{matrix}\right] \end{equation*} \]

The determinant of the matrix will be the product of the diagonal elements. Therefore,

\[det\left( A-\lambda I\right) = \left(1-\lambda\right)\left(4-\lambda \right)\left(6-\lambda \right) = 0 \]

\[ \left(4-\lambda - 4\lambda + \lambda^{2}\right)\left(6-\lambda \right) = 0\]

\[ \left(24-6\lambda - 24\lambda + 6\lambda^{2}-4\lambda + \lambda^{2} + 4\lambda^{2} - \lambda^3\right) = 0\]

The characteristic polynomial is:

\[-\lambda^3 + 11\lambda^2 - 34\lambda + 24 = 0\]

The eigen values are: 1, 4, and 6

At \(\lambda_1 = 1\), the corresponding eigen vectors are:

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 1.0000\\ 0.0000\\ 0.0000 \end{matrix}\right] \end{equation*} \] At \(\lambda_2 = 4\), the corresponding eigen vectors are:

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 1.6000\\ 2.5000\\ 1.0000 \end{matrix}\right] \end{equation*} \]

At \(\lambda_3 = 6\), the corresponding eigen vectors are:

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 1.6667\\ 1.0000\\ 0.0000 \end{matrix}\right] \end{equation*} \]