Problem Set 1

library("pracma")
## Warning: package 'pracma' was built under R version 3.4.1
  1. What is the rank of the matrix A?
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), nrow = 4,ncol = 4)
A
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3

The rank of A will be equal to the number of non-zero rows in reduced row echalo form.

rref(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

Since the result of putting A into reduced row echalon form is the Identity Matrix, A is rank 4.

  1. (i)Given an \(mxn\) matrix where \(m>n\), what can be the maximum rank?

Since rank is the number of non-zero rows in row echelon form, the maximum rank in these circumstances is \(n\), as you need a leading entry of 1 in each column in row echelon form, and \(n\) is the number of colunms.

B <- matrix(c(2,4,6,3,6,7,5,7,5,1,2,3), nrow = 4, ncol = 3, byrow = TRUE)
B
##      [,1] [,2] [,3]
## [1,]    2    4    6
## [2,]    3    6    7
## [3,]    5    7    5
## [4,]    1    2    3
rref(B)
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1
## [4,]    0    0    0
  1. The minimum rank, assuming that the matrix is non-zero.

The minimum rank in these circumstances is 1. In the trival case there is only 1 non-zero row. It could also be that all the rows differ by a multiplicative factor.

C <- matrix(c(2,4,6,4,8,12,6,12,18,1,2,3), nrow = 4, ncol = 3, byrow = TRUE)
C
##      [,1] [,2] [,3]
## [1,]    2    4    6
## [2,]    4    8   12
## [3,]    6   12   18
## [4,]    1    2    3
rref(C)
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    0    0
## [3,]    0    0    0
## [4,]    0    0    0
  1. What is the rank of matrix B?

It seems similar to my example C above, all of the row differ only by a multiplying ,so it should be 1.

B <- matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3,ncol = 3)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rref(B)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Which it is.

Problem Set 2

Compute the Eigenvalues and Eigenvectors of matrix A. You’ll need to show your work. You’ll need to wrtie the characteristic polynomial and show your solution.

The fact that this is an upper triangular matrix makes solving the determinant simple. It is the product of the diagonal terms.

\[det(A) = 24\] To varify using r:

A <- matrix(c(1,0,0,2,4,0,3,5,6), nrow = 3, ncol = 3)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
det(A)
## [1] 24

This means that the characteristic polynomial is

\[(1-\lambda)(4-\lambda)(6-\lambda) = 0\] Calculated from:

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

\[ \begin{vmatrix} (1-\lambda) && 2 && 3 \\ 0 &&(4-\lambda) && 5 \\ 0 && 0 && (6-\lambda) \end{vmatrix} \]

The roots are simply \(1,4,6\)

As for the Eigenvectors we have to solve systems of equations such that:

\[ AX_{n} = \lambda_{n}X_{n} \\ Ax_{n} - \lambda_{n}X_{n} = 0 \\ (A - \lambda_{n}I)X_{n} = 0 \]

We can treat this as a system of equations and find the reduced row echalon form.

For \(\lambda_{1} = 1\):

A1 <- matrix(c(0,0,0,2,3,0,3,5,5), nrow =3, ncol = 3)
A1
##      [,1] [,2] [,3]
## [1,]    0    2    3
## [2,]    0    3    5
## [3,]    0    0    5
rref(A1)
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

Looking at the rref(A1), \(x_{3} = x_{2} = 0\) and \(x_{1}\) is free and \(X_{1}\) for \(\lambda_{1}\) is:

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

For \(\lambda_{2} = 4\):

A2 <- matrix(c(-3,0,0,2,0,0,3,5,2), nrow =3, ncol = 3)
A2
##      [,1] [,2] [,3]
## [1,]   -3    2    3
## [2,]    0    0    5
## [3,]    0    0    2
rref(A2)
##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

Looking at the rref(A2), \(x_{3} = 0\) and \(x_{2}\) is free and \(x_{1} - \frac{2}{3}x_{2} = 0\) \(X_{2}\) for \(\lambda_{2}\) is:

\[ X_{2}= \begin{bmatrix} 1\\ \frac{3}{2}\\ 0 \end{bmatrix} \]

For \(\lambda_{3} = 6\):

A3 <- matrix(c(-5,0,0,2,-2,0,3,5,0), nrow =3, ncol = 3)
A3
##      [,1] [,2] [,3]
## [1,]   -5    2    3
## [2,]    0   -2    5
## [3,]    0    0    0
rref(A3)
##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

Looking at the rref(A3), and \(x_{3}\) is free and \(x_{1} - \frac{8}{5}x_{3} = 0\) and \(x_{2} - \frac{5}{2}x_{3} = 0\) , so \(X_{3}\) for \(\lambda_{3}\) is:

\[ X_{3} = \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix} \]

We can check our work using r’s eigen function.

eigen(A)
## eigen() decomposition
## $values
## [1] 6 4 1
## 
## $vectors
##           [,1]      [,2] [,3]
## [1,] 0.5108407 0.5547002    1
## [2,] 0.7981886 0.8320503    0
## [3,] 0.3192754 0.0000000    0
0.5108407/0.3192754
## [1] 1.6
0.7981886/0.3192754
## [1] 2.5
0.8320503/0.5547002
## [1] 1.5

Instead of declaring the free parameters 1 like I did r used different numbers but the proportions are correct. I included the arithmatic to show that.

Let matrix R be composed of the eigenvectors:

R <- matrix(c(1,0,0,1,1.5,0,1.6,2.5,1), nrow = 3, ncol = 3)
R
##      [,1] [,2] [,3]
## [1,]    1  1.0  1.6
## [2,]    0  1.5  2.5
## [3,]    0  0.0  1.0

If I did the eigenvectors correctly \(R^{-1}AR\) should return a diagonal matrix with the eigenvalues as the non-zero elements.

A <- solve(R)%*%A%*%R
A
##      [,1]         [,2]          [,3]
## [1,]    1 2.220446e-16  6.661338e-16
## [2,]    0 4.000000e+00 -8.881784e-16
## [3,]    0 0.000000e+00  6.000000e+00

There seems to be some rounding errors or r’s part but it is otherwise correct as \(10^{-16}\) are very very small numbers.