Solution rendered in RPubs here.

Problem Set 1
  1. What is the rank of the matrix \(A\)? \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\-1 & 0 & 1 & 3 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix}\] \[ A = \begin{bmatrix}-1 & -2 & -3 & -4 \\-1 & 0 & 1 & 3 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix}\] \[ A = \begin{bmatrix}5 & 10 & 15 & 20 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 1 & -2 & 1 \\0 & -6 & -17 & -23\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 1 & 2 & 7/2 \\0 & 1 & -2 & 1 \\0 & -6 & -17 & -23\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 1 & 2 & 7/2 \\0 & 0 & -4 & -5/2 \\0 & -6 & -17 & -23\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & -6 & -12 & -21 \\0 & 0 & -4 & -5/2 \\0 & -6 & -17 & -23\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 1 & 2 & 7/2 \\0 & 0 & -4 & -5/2 \\0 & 0 & -5 & -2\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & -4 & -5/2 \\0 & 0 & -5 & -2\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & 1 & 5/8 \\0 & 0 & -5 & -2\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & 5 & -25/8 \\0 & 0 & -5 & -2\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & 1 & 5/8 \\0 & 0 & 0 & 9/8\end{bmatrix}\] \[ A = \begin{bmatrix}1 & 2 & 3 & 4 \\0 & 2 & 4 & 7 \\0 & 0 & -4 & -5/2 \\0 & 0 & 0 & 9/8\end{bmatrix}\] \[ Rank\,\,=\,4 \]
library(matrixcalc)
# check with R
m <- matrix(data = c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, ncol = 4, byrow = TRUE)
m
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
matrix.rank(m)
## [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?

Maximum rank = \(m\) (rows)
Minimum rank = 1 (all other rows could be linearly dependent)

  1. What is the rank of matrix \(B\)? \[ B = \begin{bmatrix}1 & 2 & 1 \\3 & 6 & 3 \\2 & 4 & 2 \end{bmatrix}\]
    R2 is a multiple of R1. R3 is a multiple of R1. They are linearly dependent. \[ B = \begin{bmatrix}3 & 6 & 3 \\3 & 6 & 3 \\2 & 4 & 2 \end{bmatrix}\] \[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\2 & 4 & 2 \end{bmatrix}\] \[ B = \begin{bmatrix}2 & 4 & 2 \\0 & 0 & 0 \\2 & 4 & 2 \end{bmatrix}\] \[ B = \begin{bmatrix}1 & 2 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}\] \[ Rank\,\,=\,1 \]
# check with R
m <- matrix(data = c(1,2,1,3,6,3,2,4,2), nrow = 3, ncol = 3, byrow = TRUE)
m
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
matrix.rank(m)
## [1] 1
Problem Set 2

Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to write out the characteristic polynomial and show your solution.
Eigenvalues:
\[ A = \begin{bmatrix}1 & 2 & 3 \\0 & 4 & 5 \\0 & 0 & 6 \end{bmatrix}\] \[ \lambda\,I_3 = \begin{bmatrix}\lambda & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda \end{bmatrix}\] \[ det(A-\lambda\,I_n)=0\] \[ det\,\begin{bmatrix}1-\lambda & 2 & 3 \\0 & 4-\lambda & 5 \\0 & 0 & 6-\lambda \end{bmatrix} = 0\] \[(1-\lambda)(4-\lambda)(6-\lambda)=0\] \[ Eigenvalues\,of\,A:\] \[\lambda=1,\, \lambda=4,\, \lambda=6\]

Eigenvectors:
\[\lambda=1\] \[ \begin{bmatrix}1-\lambda & 2 & 3 \\0 & 4-\lambda & 5 \\0 & 0 & 6-\lambda \end{bmatrix}\]

\[ \begin{bmatrix}0 & 2 & 3 \\0 & 3 & 5\\0 & 0 & 5\end{bmatrix}\,\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\]

\[The\,first\,pivot\,is\,0.\,x_1 = free.\,Let\,the\,value=1.\]

\[3 x_2 + 5 x_3 = 0 \,and\, 5 x_3 = 0\]

\[x_{\lambda=1}\,=\begin{bmatrix}1 \\0 \\0\end{bmatrix}\]

\[\lambda=4\] \[ \begin{bmatrix}-3 & 2 & 3 \\0 & 0 & 5\\0 & 0 & 2\end{bmatrix}\,\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\] \[Second\,pivot\,is\, 0.\,x_2=free.\,Let\,the\,value=1.\] \[-3x_1+2x_2 +3x_3 = 0\,and\, 2x_3 = 0\] \[x_3=0,\,x_2=1\,and\,x_1=2/3\] \[x_{\lambda=4}\,=\begin{bmatrix}2/3\\1 \\0\end{bmatrix}\]

\[\lambda=6\] \[ \begin{bmatrix}-5 & 2 & 3\\0 & -2 & 5\\0 & 0 & 0\end{bmatrix}\,\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=0\] \[Third\,pivot\,is\, 0.\,x_3=free.\,Let\,the\,value=1.\] \[-5x_1 +2x_2 +3x_3 = 0 and, -2x_1+5x_3 = 0\] \[x_3 = 1,\,x_2 = 5/2,\,and\,x_1=8/5\]

\[x_{\lambda=6}\,=\begin{bmatrix}8/5 \\5/2 \\1\end{bmatrix}\]