###(1)
A <- rbind(c( 1, 2, 3, 4),
c(-1, 0, 1, 3),
c( 0, 1,-2, 1),
c( 5, 4,-2, -3))
qr(A)$rank
## [1] 4
###(2)
# n is the maximum rank for an m x n matrix where m > n
# 1 is the minimum rank, assuming the matrix is non-zero
###(3)
B <- rbind(c( 1, 2, 1),
c( 3, 6, 3),
c( 2, 4, 2))
qr(B)$rank
## [1] 1
A <- rbind(c( 1, 2, 3),
c( 0, 4, 5),
c( 0, 0, 6))
\[A= \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix} \]
\[ A x = \lambda x \] \[ A x - \lambda x = 0 \] \[ (A - \lambda I) x = 0 \] \[ det(A - \lambda I) = 0 \] \[{A - \lambda I} = \left[\begin{array} {rrr} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{array}\right] \]
The determinant of this matrix is just the product of the diagonal because it is an upper triangular matrix. So…
\[ det(A - \lambda I) = (1 - \lambda)(4 - \lambda)(6 - \lambda) = 0 \] \[ (4 - \lambda - 4\lambda + \lambda^2)(6 - \lambda) = 0 \]
\[ 24 - 6\lambda - 24\lambda + 6\lambda^2 - 4\lambda + \lambda^2 + 4\lambda^2 - \lambda^3 = 0 \] So the characteristic polynomial is:
\[ - \lambda^3 + 11\lambda^2 - 34\lambda + 24 = 0 \] The eigenvalues are: 1, 4, and 6
Now, plug them in to find eigenvectors:
library(pracma)
# augment the matrices with 0 vector, perform elimination, and solve by setting free variable to 1
rref( cbind( A-1*diag(3), c(0,0,0)) )
## [,1] [,2] [,3] [,4]
## [1,] 0 1 0 0
## [2,] 0 0 1 0
## [3,] 0 0 0 0
rref( cbind( A-4*diag(3), c(0,0,0)) )
## [,1] [,2] [,3] [,4]
## [1,] 1 -0.6666667 0 0
## [2,] 0 0.0000000 1 0
## [3,] 0 0.0000000 0 0
rref( cbind( A-6*diag(3), c(0,0,0)) )
## [,1] [,2] [,3] [,4]
## [1,] 1 0 -1.6 0
## [2,] 0 1 -2.5 0
## [3,] 0 0 0.0 0
The eigenvectors are:
\[\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \] \[\begin{bmatrix} 0.67 \\ 1 \\ 0 \\ \end{bmatrix} \] \[\begin{bmatrix} 1.6 \\ 2.5 \\ 1 \\ \end{bmatrix} \]