Problem set 1

1. What is the rank of the matrix A? A =

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

library(pracma)
A1 = matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow = 4, byrow = T)
rref(A1)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1


This is the \(I_4/\) matrix - rank is therefore 4. Each row and column are linearly independent.

??? ??? ??? ??? 2. 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 is n. If the matrix has fewer than n linear independent columns, the rank will be < n.

3. What is the rank of matrix B? B =

\[ \begin{bmatrix} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2\\ \end{bmatrix} \] 

library(Matrix)
## 
## Attaching package: 'Matrix'
## The following objects are masked from 'package:pracma':
## 
##     expm, lu, tril, triu
A3 = matrix(c(1,2,1,3,6,3,2,4,2), nrow = 3, byrow = T)
#reduced row echelon form
rref(A3)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0
#using rankMatrix function from Matrix package
rankMatrix(A3)[1]
## [1] 1


Matrix has one (non-zero) linearly independent row, so it’s rank is 1. Verified by rankMatrix function.

???
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 =???
\[ \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{bmatrix} \]
Step 1. Get determinant.

A2 = matrix(c(1,2,3,0,4,5,0,0,6), nrow = 3, byrow = T) 
A2
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6


From this week’s lecture:

To find eigenvalues and eigenvectors, we take the determinant of a special matrix and set it to zero:
det(A - \(\lambda\)I) = 0

This gives rise to a polynomial equation in terms of ?? called the characteristic polynomial.

To solve, using x in place of lamda:
\[ det( \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{bmatrix} - \begin{bmatrix} x & 0 & 0\\ 0 & x & 0\\ 0 & 0 & x\\ \end{bmatrix} ) = 0 \]
\[ det( \begin{bmatrix} 1 - x & 2 & 3\\ 0 & 4 - x & 5\\ 0 & 0 & 6 - x\\ \end{bmatrix} ) = 0 \]
\[ (1 - x) * det( \begin{bmatrix} 4 - x & 5\\ 0 & 6 - x\\ \end{bmatrix} ) - 2 * det( \begin{bmatrix} 0 & 5\\ 0 & 6 - x\\ \end{bmatrix} ) + 3 * det( \begin{bmatrix} 0 & 1 - x\\ 0 & 2\\ \end{bmatrix} ) \]
(1 - x)(x^2 - 10x + 24) - 2(0) + 3(0) x^2 - 10x + 24 - x^3 + 10x^2 - 24x Characteristic polynomial: -x^3 + 11x^2 - 34x + 24
Factored: (-x+1)(x-6)(x-4)
So our eigenvalues are 1, 6, and 4.
To get eigenvectors, we plug these values into the lambda/x equations above.
\[ ( \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}) * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5\\ \end{bmatrix} * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = 0 \]
Let’s use r to get the rref:

B1 = matrix(c(0,2,3,0,3,5,0,0,5), nrow = 3, byrow = T)
rref(B1)
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0


$x_2 = 0 $/ $x_3 = 0 $/ So \(x_2 = x_3 = 0\)/ and \(x_1\)/ is a free variable, which we’ll make 1.
\[ e_1 = \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} \]
Repeat for eigenvalue 4. \[ ( \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4\\ \end{bmatrix}) * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2\\ \end{bmatrix} * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = 0 \]
Let’s use r to get the rref:

B2 = matrix(c(-3,2,3,0,0,5,0,0,2), nrow = 3, byrow = T)
rref(B2)
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0


$x_1 - (2/3)*x_2 = 0 $/ $x_3 = 0 $/ So \(x_3 = 0\)/ and \(x_1 = 2/3\)/ and \(x_2 = 1\)/.
\[ e_4 = \begin{bmatrix} 2/3\\ 1\\ 0\\ \end{bmatrix} \]
Repeat for eigenvalue 6. \[ ( \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{bmatrix} - \begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6\\ \end{bmatrix}) * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = \begin{bmatrix} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0\\ \end{bmatrix} * \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} = 0 \]
Let’s use r to get the rref:

B3 = matrix(c(-5,2,3,0,-2,5,0,0,0), nrow = 3, byrow = T)
rref(B3)
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0


$x_1 - 1.6x_3 = 0 $/ $x_2 - 2.5x_3 = 0 $/ So \(x_1 = 1.6\)/ and \(x_2 = 2.5\)/ and \(x_3 = 1\)/.
\[ e_6 = \begin{bmatrix} 1.6\\ 2.5\\ 1\\ \end{bmatrix} \]
Our eigenvectors are: \[ \begin{bmatrix} 1.6\\ 2.5\\ 1\\ \end{bmatrix} , \begin{bmatrix} 2/3\\ 1\\ 0\\ \end{bmatrix} , \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} \]
Checking work in R.

#characteristic polynomial
charpoly(A2)
## [1]   1 -11  34 -24
eA2 = eigen(A2)
#eigenvalues
eA2$values
## [1] 6 4 1
#eigenvector
eA2$vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0



Please show your work using an R-markdown document. Please name your assignment submission with your ???rst initial and last name.