Problem Set 1

  1. What is the rank of the matrix A?

\[A = \begin{bmatrix}1&2&3&4\\-1&0&1&3\\0&1&-2&1\\5&4&-2&-3\\\end{bmatrix}\] The rank of an invertible matrix is equal to its dimension. So let’s check whether this matrix is invertible by calculating it’s determinant.

A<-matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3),nrow=4,byrow=TRUE)
det.matrixA<-det(A)
det.matrixA
## [1] -9

Since the determinant is not 0, the matrix is invertible. Therefore it’s rank is equal to it’s dimension. r=4

This can be cross-checked using a couple of R functions

library(matrixcalc)
matrix.rank(A)
## [1] 4
rank.A<- qr(A)$rank
rank.A
## [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?

The rank of a matrix is equal to the minimum of the number of rows and the number of columns. So in this case, the maximum rank = n. If the matrix is non-zero, it means there is at least one element that is non-zero, and there is at least one element in its column space. Therefore the minimum rank can be 1.

  1. What is the rank of matrix B? \[B = \begin{bmatrix}1&2&1\\3&6&3\\2&4&2\\\end{bmatrix}\] The rank of a matrix is equal to the number of pivots. A visual inspection of the above matrix shows that columns 2 and 3 are multiples of column 1. Similarly rows 2 and 3 are multiples of row 1. So there is linear dependence amongst the rows and columns. When this matrix is row-reduced, 2 of the rows will consist of only 0s. Only the first row will be non-zero, with a pivot element. Therefore it’s rank = 1. This can be cross-checked using R functions as shown below:
B<-matrix(c(1,2,1,3,6,3,2,4,2),nrow=3,byrow=TRUE)
matrix.rank(B)
## [1] 1
rank.B<- qr(B)$rank
rank.B
## [1] 1

Problem Set 2

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.

\[A = \begin{bmatrix}1&2&3\\0&4&5\\0&0&6\\\end{bmatrix}\] To find the eigenvalues and eigenvectors of this matrix, we set:

\[det(A-\lambda I)=0\]

\[ det\Bigg\{\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\\\end{bmatrix}\ - \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\\end{bmatrix}\Bigg\}=0 \\ \\ = det\Bigg\{\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\\\end{bmatrix}\ - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\\\end{bmatrix}\Bigg\}=0 \\ \\ = det\Bigg\{\begin{bmatrix}1-\lambda&2&3\\0&4-\lambda&5\\0&0&6-\lambda\\\end{bmatrix}\Bigg\}=0 \\ \\ = \Bigg\{\big(1-\lambda) det \begin{bmatrix}4-\lambda&5\\0&6-\lambda\\\end{bmatrix}\Bigg\}=0 \\ \\ = \big(1-\lambda)[(4-\lambda)(6-\lambda) - 5.0]=0\\ \\ = \big(1-\lambda)(4-\lambda)(6-\lambda)=0\\ \\ = \lambda^{3}-11\lambda^{2} +34\lambda-24=0 \]

This is the characteristic polynomial equation, which has 3 roots which then yield 3 eigenvalues: 1,4 and 6 in ascending order

Substituting this back in the matrix one by one will gives us the eigenvectors as shown below:

\[\begin{bmatrix}1-1&2&3\\0&4-1&5\\0&0&6-1\\\end{bmatrix} \begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\\\end{bmatrix}\\ \\ = \begin{bmatrix}0&2&3\\0&3&5\\0&0&5\\\end{bmatrix} \begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\\\end{bmatrix}\\\]

The third row (5z = 0) gives us z = 0 which then substituting back in the second row (3y+0=0) gives us y = 0 Since x is unconstrained by the 3 rows, we get the following eigenvector corresponding to eigenvalue = 1

\[\begin{bmatrix}1\\0\\0\\\end{bmatrix}\]

Following a similar process of back-substitution for the other two eigenvalues 4 and 6 yields their correspondng eigenvectors as well. Below, we cross-check the eigenvalues using R functions, and we also extract the eigenvectors:

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

evalues.A <- eigen(A)$values
evalues.A
## [1] 6 4 1
evectors.A <- eigen(A)$vectors
evectors.A
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0