Problem Set 1 1. What is the rank of Matrix \(A\)?

a <- matrix(c(1, -1, 0, 5, 2, 0, 1, 4, 3, 1, -2, -2, 4, 3, 1, -3), nrow = 4, ncol = 4, byrow = F)
a
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3

Got a little help figuring out ranking matrix from this site: https://stackoverflow.com/questions/10881392/rank-of-a-matrix-in-r

#install the matrix package in R
qr(a)$rank
## [1] 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?

Since the rank can never be larger than the number of rows, and it can never be larger than the number of columns either, the maximum rank for the matrix is \(n\)

The minimum rank, assuming that the matric is non-zero is 1.

  1. What is the rank of matrix \(B\)?
b <- matrix (c(1, 3, 2, 2, 6, 4, 1, 3, 2), ncol = 3, nrow = 3, byrow = F)
b
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
qr(b)$rank
## [1] 1

Problem Set 2 Compute the eigenvalues and eigenvectors of the matrix A

\[\mathbf{A} = \left[\begin{array} {rr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right]\]

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

we will need to find the eigenvalues for 1, 4, 6, assuming that \(Av = \lambda v\), then $$ is called the eigenvalue associated with the eigenvector of \(v\) of \(A\)

\[\mathbf{A} = \left[\begin{array} {rr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] - \mathbf{\lambda} \left[\begin{array} {rr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\]

equals \[\mathbf{A} = \left[\begin{array} {rr} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{array}\right]\]

=\[(1-\lambda)det\left[\begin{array} {rr} 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{array}\right]-2 * det\left[\begin{array} {rr} 0 & 5 \\ 0 & 6-\lambda \end{array}\right]+3det\left[\begin{array} {rr} 0 & 4-\lambda \\ 0 & 0 \end{array}\right]\]

=\[det\left[\begin{array} {rr} 4-\lambda & 5 \\ 0 & 6-\lambda \end{array}\right] = \lambda^2 - 10\lambda + 24\]

= \[(4 - \lambda)(6 - \lambda) - 5 * 0\] = \[(4 - \lambda)(6 - \lambda) - 5 * 0 = \lambda^2 - 10\lambda + 24\]

=\[det\left[\begin{array} {rr} 0 & 5 \\ 0 & 6-\lambda \end{array}\right]\]

= \(0 * (6 - \lambda) - 5 * 0\)

= \(0\)

=\[det\left[\begin{array} {rr} 0 & 4-\lambda \\ 0 & 0 \end{array}\right]\]

= \(0*0 - (4 - \lambda) * 0\)

= \(0\)

This equals: \(-\lambda^3 + 11\lambda^2 - 34\lambda + 24 + 0 + 0\) or \(-\lambda^3 + 11\lambda^2 - 34\lambda + 24\)

Solving \(-\lambda^3 + 11\lambda^2 - 34\lambda + 24\)

factor \(-(\lambda - 1)(\lambda - 4)(\lambda - 6) = 0\)

\(0 = (\lambda - 1)\) \(\lambda = 1\)

\(0 = (\lambda - 4)\) \(\lambda = 4\)

\(0 = (\lambda - 6)\) \(\lambda = 6\)

Eigenvalues are 1, 4, 6

Now to check with R

ev <- eigen(A)
ev
## eigen() decomposition
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0