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),4,4)
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
qr(A)$rank
## [1] 4

Since each row is linear independent from other rows, the rank of the matrix is 4.

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

The maximum rank can be equal or less than n, the minimum rank of non-zero matrix is 1.

What is the rank of matrix B?

B = matrix(c(1, 3, 2, 2, 6, 4, 1, 3, 2),3,3)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
qr(B)$rank
## [1] 1
2* B[1,] == B[3,]
## [1] TRUE TRUE TRUE
3* B[1,] == B[2,]
## [1] TRUE TRUE TRUE

Since each row is not linearly independent with other rows, the rank of B is 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.

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

\[ \lambda E-A= \left[\begin{array} {rrr} \lambda - 1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{array}\right] \\ det(\lambda E-A) = (\lambda -1 )(\lambda - 4)(\lambda -6 ) = 0 \\ \lambda^{3}-11\lambda^{2}+34\lambda - 24 = 0 \\ Eigenvalue: \lambda_1 = 6 , \lambda_2 = 4, \lambda_3 = 1 \\ \] Now we need to use the eigenvalue to calculate the eigenvectors:

\[ (\lambda_1 I-A)h_1 = 0 \\ \left[\begin{array} {rrr} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{array}\right]h_1 = \left[\begin{array} {rrr} 0 \\ 0 \\ 0\end{array}\right] \\ h_1 = \left[\begin{array} {rrr} 1.6 \\ 2.5 \\ 1\end{array}\right] \\ \] \[ (\lambda_2 I-A)h_2 = 0 \\ \left[\begin{array} {rrr} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{array}\right]h_2 = \left[\begin{array} {rrr} 0 \\ 0 \\ 0\end{array}\right] \\ h_2 = \left[\begin{array} {rrr} 1 \\ 1.5 \\ 0 \end{array}\right] \\ \]

\[ (\lambda_3 I-A)h_3 = 0 \\ \left[\begin{array} {rrr} 1 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{array}\right]h_2 = \left[\begin{array} {rrr} 0 \\ 0 \\ 0\end{array}\right] \\ h_3 = \left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array}\right] \\ \]