Rank and Eigen Vectors



Rank


What is the rank of the matrix A ?


\[A \ = \ \begin{bmatrix} 1&2&3&4\\ -1&0&1&3\\ 0&1&-2&1 \\5&4&-2&-3 \end{bmatrix}\]

If a square matrix has a determinant, then we immediately know that no row can be eliminated as being linearly dependent on any other row or linear combination of rows.


a<-matrix(c(1, 2, 3, 4, -1, 0, 1, 3,0, 1, -2, 1,5, 4, -2, -3), byrow = TRUE, ncol = 4)
det(a)
## [1] -9


So in this case, we know that A is full rank, its rank is 4.

Matrix::rankMatrix(a)[1]
## [1] 4


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


The maximum rank of m rows, is m. That would be when n=m, meaning you have enough variables and independent equations to find the intersection all n coefficients.


The minimum is 1, which means you only have 1 independent equation.


What is the rank of matrix B?


\[A \ = \ \begin{bmatrix} 1&2&1\\ 3&6&3\\ 2&4&2 \end{bmatrix}\]
One can see pretty easily that all 3 are dependent so the rank is 1 and the determinent is 0 

a<-matrix(c(1, 2, 1, 3, 6, 3, 2, 4,2), byrow = TRUE, ncol = 3)
det(a)
## [1] 0
Matrix::rankMatrix(a)[1]
## [1] 1




Eigen Values and Vectors


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.


The Eigen Values are valid for all vectors with the indicated ratio between \(X_1\) and \(X_2\) and \(X_3\)


Note : When \(\lambda \ = \ 1\) we deduce that \(X_2\) and a\(X_3\) are zero because there is no other solution.

\[X_1 \ + \ 2X_2 \ + \ + 3X_3 \ = X_1\] \[ 4X_2 \ + \ + 5X_3 \ = X_2\] \[ 6X_3 \ = X_3\]


Thus as a result we know that

\[X_1 \ = \ X_1\]


We can double check our work.


\[\lambda \ = 6\]


a<-matrix(c(1, 2, 3, 0, 4, 5, 0, 0,6), byrow = TRUE, ncol = 3)

b<-c(8/5, 5/2,1)
a %*% b
##      [,1]
## [1,]  9.6
## [2,] 15.0
## [3,]  6.0
6 * b   
## [1]  9.6 15.0  6.0


\[\lambda \ = 4\]


a<-matrix(c(1, 2, 3, 0, 4, 5, 0, 0,6), byrow = TRUE, ncol = 3)

b<-c(1, 3/2,0)
a %*% b
##      [,1]
## [1,]    4
## [2,]    6
## [3,]    0
4 * b   
## [1] 4 6 0


\[\lambda \ = 1\]


a<-matrix(c(1, 2, 3, 0, 4, 5, 0, 0,6), byrow = TRUE, ncol = 3)

b<-c(1, 0,0)
a %*% b
##      [,1]
## [1,]    1
## [2,]    0
## [3,]    0
1 * b   
## [1] 1 0 0