Assignment 3

Fundamentals of Computational Mathematics

CUNY MSDS DATA 605, Fall 2018

Rose Koh

09/13/2018

A <- matrix(c(1,2,3,4,
              1,0,1,3,
              0,1,-2,1,
              5,4,-2,-3), 4, byrow=T)
# manually
round(A %*% solve(A))

##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

# using function
qr(A)$rank

## [1] 4

This matrix has 4 pivots. It is invertible matrix The rank is the same as dimension

If m is grater than n, the maximum rank is n. Assuming that the matrix has at least one non-xero element, its minimum rank should be 1.

B <- matrix(c(1,2,1,3,6,3,2,4,2), 3, byrow=T)
qr(B)$rank

## [1] 1

The matrix B’s rows and columns are dependent, thus our rank is 1.

Based on the following formula,

\({Ax}= \lambda \,x\)

\({det}\left(A-\lambda \,I\right)\,=\,0\)

we find characteristic polynomial and solve for the roots. Subtract unknown lambda frrom the diagonal terms of matrix A to compute determinant.

\({det}\left(\begin{bmatrix}1-\lambda &2&3\\0&4-\lambda &5\\0&0&6-\lambda \end{bmatrix}\right)=0\)

Use algebra and simplify.

\(\left(1-\lambda\right) [\left(4-\lambda \right)\left(6-\lambda \right)-5\times 0] + 2[0 \times \left(6-\lambda \right)-5\times 0] + 3[0\times0 - 0\left(6-\lambda \right)]=0\)

Solve for the roots by setting the equation equal to zero.

\(\left(1-\lambda\right)\left(4-\lambda \right)\left(6-\lambda \right)=0\)

Characteristic Polynomial: \(-\lambda^3 + 11\lambda^2 - 34\lambda + 24=0\)

The roots are 6, 4 and 1. The roots of the polynomial are the eigenvalues of the system.

\(\lambda_1=6, \lambda_2=4, \lambda_3=1\)

\(\lambda_1=6\)

Find vectors x that satisfies the expression when lambda is 6.

\(\left(A-\lambda \,I\right)x\,=\,0\)

\(\begin{bmatrix}1-6 &2&3\\0&4-6 &5\\0&0&6-6 \end{bmatrix}\)

\(\begin{bmatrix}-5 &2&3\\0&-2 &5\\0&0&0 \end{bmatrix}\)

The singular matrix suggest why the lambda is the eigenvalue. Now we solve \(Ax = 0\)

\(\begin{bmatrix}-5&2&3\\0&-2&5\\0&0&0\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=0\)

The system of equations:

\(-5x + 2y + 3z = 0\)

\(-2 +5z = 0\)

If the \(Y = 1\), then \(-2 + 5z = 0\) thus \(z = \frac {2}{5}\)

Substitute Y and Z:

\(-5x + 2y + 3\frac {2}{5} = 0\)

\(-5x = \frac{-16}{5}\)

\(x = \frac{16}{25}\)

\(E_{\lambda_1=6}=c\begin{bmatrix}\frac {16}{25} \\ 1 \\ \frac {2}{5}\end{bmatrix}\)

\(\lambda_2=4\)

Find vectors x that satisfies the expression when lambda is 4.

\(\left(A-\lambda \,I\right)x\,=\,0\)

\(\begin{bmatrix}1-4 &2&3\\0&4-4 &5\\0&0&6-4 \end{bmatrix}\)

\(\begin{bmatrix}-3 &2&3\\0&0 &5\\0&0&-2 \end{bmatrix}\)

The singular matrix suggest why the lambda is the eigenvalue. Now we solve \(Ax = 0\)

\(\begin{bmatrix}-3 &2&3\\0&0 &5\\0&0&-2 \end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=0\)

The system of equations:

\(-3x + 2y + z = 0\)

\(5z = 0\)

\(2z = 0\)

If the \(Y = 1\), then \(X = \frac {2}{3}\)

\(E_{\lambda_2=4}=c\begin{bmatrix} \frac {2}{3} \\ 1 \\ 0 \end{bmatrix}\)

\(\lambda_3=1\)

Find vectors x that satisfies the expression when lambda is 1.

\(\left(A-\lambda \,I\right)x\,=\,0\)

\(\begin{bmatrix}1-1 &2&3\\0&4-1 &5\\0&0&6-1 \end{bmatrix}\)

\(\begin{bmatrix}0&2&3\\0&3&5\\0&0&5\end{bmatrix}\)

The singular matrix suggest why the lambda is the eigenvalue. Now we solve \(Ax = 0\)

\(\begin{bmatrix}0&2&3\\0&3&5\\0&0&5\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=0\)

The system of equations:

\(2y + 3z = 0\)

\(3y + 5z = 0\)

\(5z = 0\)

So the only non-zero solution is X = 0, Z = 0.

\(E_{\lambda_3=1}=c\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\)

# Check with the function
A <- matrix(c(1,2,3,0,4,5,0,0,6), 3, byrow=T)
A

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

#Eigenvalues
eigen(A)$values

## [1] 6 4 1

#Eigenvectors
eigen(A)$vectors

##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0