Homework3

Problem Set 1

1: What is the rank of matrix A?

4

## declare matrix
A = matrix( c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,3), nrow = 4, ncol=4, byrow = TRUE)

qr(A)$rank

## [1] 4

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

In an m x n matrix with m > n, the maximum rank is n,this is because the matrix is bounded by the smallest dimension. The minimum rank, assuming non-zero values, is 1, this is because at least one row or column must be linearly independent, making the matrix non-singular.

3: What is the rank of matrix B?

The rank of B is 1

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

qr(B)$rank

## [1] 1

Problem set 2

1: 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.

using the determinant formula

det(A - λI) = 0

substituting A - λI into the determinant, we get:

det( (1-λ 2 3 0 4-λ 5 0 0 6-λ) ) = 0

Expanding the determinant, we get

(1-λ)((4-λ)(6-λ)-0×5) - 2((6-λ)×0-0×5) + 3(0×0-0×(4-λ))

which simplifies to:

(1-λ)((24-10λ+λ^2)-0) - 0 + 0

(24-10λ+λ^2-24λ+10λ2-λ^3) + 0 + 0

-λ^3 + 11λ^2 - 34λ + 24 = 0

This equation represents the characteristic polynomial of matrix A. now we solve this polynomial equation to find the eigenvalues.

To find the eigenvalues, we need to solve the characteristic polynomial:

-λ^3 + 11λ^2 - 34λ + 24 = 0

we have to solve this cubic equation for λ we see that λ = 6 is a root because (6-6) = 0. We can then use division factorize the polynomial.

Given the characteristic polynomial:

-λ^3 + 11λ^2 - 34λ + 24 = 0

We have already found one root, λ = 6. let’s factorize the polynomial using this root:

-(λ - 6)((λ^2 - 5λ + 4)) = 0

Now, we have a quadratic equation:

λ^2 - 5λ + 4 = 0

This quadratic equation can be factored into:

(λ - 4)(λ - 1) = 0

So, the remaining roots are λ = 4 and λ = 1.

Therefore, the eigenvalues of matrix A are λ = 6, λ = 4, and λ = 1.

# To find the corresponding eigenvectors, we need to substitute each eigenvalue back into the equation (A - λI)x = 0 and solve for the vector x.

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

# Define the eigenvalues
eigenvalues <- c(6, 4, 1)

# Calculate eigenvectors corresponding to eigenvalues
eigenvectors <- sapply(eigenvalues, function(lambda) {
  eigen_result <- eigen(A - lambda * diag(3))
  eigen_result$vectors[, which(eigen_result$values == 0)]
})

# Display eigenvectors
print("Eigenvectors:")

## [1] "Eigenvectors:"

print(eigenvectors)

##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0