Section 1

1. What is the rank of matrix A

  • Matrix A is an identity matrix in its row enduced form, therefore its rank is equal to its dimension which is 4
library(pracma)
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),ncol=4)
rref(A)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

2. Given an mxn matrix

  • The maximum rank is equivalent to the smaller of the number of rows/columns, therefore if \(M>N\) then the maximum rank is N
  • If a matrix has non zero elements, its minimum rank is 1

3. What is the rank of matrix B?

  • rank 1
b <- matrix(c(1,3,2,2,6,4,1,3,2),ncol=3)
rref(b)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Section 2 Compute eigen values

\[ det (\lambda I_{n} -A)=0\]

\[\begin{equation*} \mathbf{}\left[\begin{matrix} \lambda -1 & -2&-3\\ 0 & \lambda -4 &-5 \\ 0 &0& \lambda -6 \end{matrix}\right] \end{equation*}\]

\[\begin{equation*} \mathbf{A*det}\left[\begin{matrix} \lambda -4 & -5 &\\ 0 & \lambda -6\\ \end{matrix}\right]-\mathbf{B*det}\left[\begin{matrix} \ 0&-5\\ 0&\lambda -6\\ \end{matrix}\right]+\mathbf{C*det}\left[\begin{matrix} \ 0&\lambda-4\\ 0&0\\ \end{matrix}\right] \end{equation*}\]

\[ \lambda -1(\lambda -4*\lambda -6)-(-5*0)-((0*\lambda -6) -(-5*0))+((0*0)-(\lambda -4 *0) ) \]

  • Our characteristic polynomial is

\[ (\lambda -1)(\lambda^{2}-10\lambda +24)=0\]

  • our eigenvalues are \[ \lambda= \{4,6,1\} \]

  • To get vectors we now need to plug eigen values in to original matrix \[\lambda I_{n} -A\]

  • eigenvalue 4 yields

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 3 &-2&-3\\ 0 & 0 &-5 \\ 0 &0& -2\end{matrix}\right] \end{equation*}\]

  • Now solve row reduced form
C <- matrix(c(3,0,0,-2,0,0,-3,-5,-2),ncol=3)
rref(C)
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

\[ x-2/3y=0\] \[z=0\]

  • in terms of x our vector for eigenvalue \[\mathbf{E_4}= \left[\begin{array} {rrr} 1 \\ 3/2 \\ 0 \end{array}\right] \]

  • eigenvalue 6 yields

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 5 & -2&-3\\ 0 & 2 &-5 \\ 0 &0&0 \end{matrix}\right] \end{equation*}\]

  • Now solve row reduced form
C <- matrix(c(5,0,0,-2,2,0,-3,-5,0),ncol=3)
rref(C)
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[ x-(8/5z)=0\] \[ y-(5/2z)=0\]

    • in terms of z our vector for eigenvalue \[\mathbf{E_6}= \left[\begin{array} {rrr} 8/5 \\ 5/2 \\ 1 \end{array}\right] \]
  • eigenvalue 1 yields

\[\begin{equation*} \mathbf{}\left[\begin{matrix} 0 & -2&-3\\ 0 & -3 &-5 \\ 0 &0&-5 \end{matrix}\right] \end{equation*}\]

  • Now solve row reduced form
C <- matrix(c(0,0,0,-2,-3,0,-3,-5,-5),ncol=3)
rref(C)
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\[y=0\] \[z=0\]

\[\mathbf{E_1}= \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right] \]