1. Problem set 1
  1. What is the rank of the matrix A? First, we reduce A to a reduced-row-echelon form.
A <-matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5, 4, -2, -3), nrow=4, byrow = TRUE)

rreform4 <- function(A) {
  A[2,] <- (((A[2,1])/A[1,1])*A[1,]-A[2,])
  A[3,] <- (((A[3,1])/A[1,1])*A[1,]-A[3,])
  A[4,] <- (((A[4,1])/A[1,1])*A[1,]-A[4,])
    A[3,] <- (((A[3,2])/A[2,2])*A[2,]-A[3,])
    A[4,] <- (((A[4,2])/A[2,2])*A[2,]-A[4,])
      A[4,] <- (((A[4,3])/A[3,3])*A[3,]-A[4,])
  
  return(A)
}

rreform4(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

Then, we see there are no 0 rows, so we conclude that rank(A)=4.

  1. Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

The maximum rank of a matrix, where m > n, is n. The minimum rank of a non-zero matrix is 1.

  1. What is the rank of matrix B? B = 1 2 1 3 6 3 2 4 2

We can see right away that row 2 and row 3 are multiples of row 1, so we know rank(B)=1.

We can check this using the reduced-row-echelon form of B.

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

rreform3 <- function(B) {
  B[2,] <- (((B[2,1])/B[1,1])*B[1,]-B[2,])
  B[3,] <- (((B[3,1])/B[1,1])*B[1,]-B[3,])
    B[3,] <- (((B[3,2])/B[2,2])*B[2,]-B[3,])
  
  return(B)
}

rreform3(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]  NaN  NaN  NaN

So we see that rank(B)=1.

  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.

First we need to find the eigen values. We will start by finding the solutions to the det(A-λI). I will store (A-λI) in matrix B.

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

B <- matrix(c(("1-λ"), 2, 2, 0, ("4-λ"), 5, 0, 0, ("6-λ")), 3, 3, byrow = TRUE)

Then, det(A-λI)=(1-λ)((4-λ)(6-λ)-(5*0)). When we solve for λ, we find our eigen values to be λ=1, 4, 6. To find our eigen vectors, we solve (A-λI)[x]=[0] for each eigenvalue we found, where [x]=c(x, y, z).

First, λ=1.

#Store A-λI in matrix B, sub in λ=1

B <- matrix(c(0, 2, 3, 0, 3, 5, 0, 0, 5), nrow=3, byrow = TRUE)

#So, (B)[x]=[0] gives us the system of equations:

# 0*x + 2y + 3z = 0
# 0*x + 3y + 5z = 0
# 0*x + 0y + 5z = 0

#So we know z=0, y=0, x=x. let x=1. Therefore, our eigenvector for λ=1 is c(1, 0, 0).

Then, λ=4.

#Store A-λI in matrix B, sub in λ=4

B <- matrix(c(-3, 2, 3, 0, 0, 5, 0, 0, 2), nrow=3, byrow = TRUE)

#So, (B)[x]=[0] gives us the system of equations:

# -3x + 2y + 3z = 0
# 0*x + 0*y + 5z = 0
# 0*x + 0*y + 2z = 0

#So we know z=0, y=y, And we solve for x.

# -3x + 2y + 3*0 = 0
# 2y = 3x
# x = (2/3)y

#So, x = (2/3)y. Now let y=1. Therefore, our eigenvector for λ=4 is c((2/3), 1, 0).

Then, λ=6.

#Store A-λI in matrix B, sub in λ=6

B <- matrix(c(-5, 2, 3, 0, -2, 5, 0, 0, 0), nrow=3, byrow = TRUE)

#So, (B)[x]=[0] gives us the system of equations:

# -5x + 2y + 3z = 0
# 0*x + -2*y + 5z = 0
# 0*x + 0*y + 0z = 0

#So we know z=z, And we solve for x and y. Let's start with y.

# 0*x + -2*y + 5z = 0
# 2y = 5z
# y = (5/2)z

#So, y = (5/2)z. Now we solve for x.

# -5x + 2y + 3z = 0
# 2y + 3z = 5x
# 2(5/2)z + 3z = 5x
# 8z = 5x
# x = (8/5)z

#So, x = (8/5)z. Let z=1. Therefore, our eigenvector for λ=6 is c((8/5), 5/2, 1).