Problem Set 1

Problem 1

To determine the rank of matrix A, we could put it into rref and count the number of pivot columns or take the nullspace of the matrix.

Start with the matrix:

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

library(pracma)
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3) ,4 ,4)
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
nullspace(A)
## NULL

A row reduces to the identity matrix, and, as expected its nullspace is empty.
The rank is therefore 4.

Problem 2

The rank of the column space equals the rank of the rowspace. The rank is equal to the number of linearly independent rows or columns. Since A has n columns and m > n, A can have a rank of, at most, n.

Since A is not the zero matrix, it must have at least one column. The rank must be at least 1.

Problem 3

For the matrix: \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 4 & 5 \\ 2 & 4 & 2 \end{bmatrix}\)

we will use the same technique as problem 1.

A <- matrix(c(1,0,2,2,4,4,1,5,2),3,3)

rref(A)
##      [,1] [,2]  [,3]
## [1,]    1    0 -1.50
## [2,]    0    1  1.25
## [3,]    0    0  0.00
nullspace(A)
##            [,1]
## [1,]  0.6837635
## [2,] -0.5698029
## [3,]  0.4558423

The matrix has two pivots in rref, and, as expected, a nullspace with dimension 1.

The rank of A is therefore 2.

Problem Set 2

Problem 1

A is an upper triangular matrix, so its determinant will be the product of its diagonal entries. We need to solve:

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

This is our characteristic polynomial:

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

The eigenvalues are: \(\lambda=(1,4,6)\)

To get the eigenvectors, we compute the nullspace of \(A-\lambda I\)

For \(\lambda=-1\)

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

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

By inspection, we can see that any vector with the second and third entries being 0 will work as a solution.

the eigenvector is \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)

For \(\lambda = 4\):

we must find the nullspace of \(\begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix}\)

divide row 1 by -3: \(\begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix}\)

divide row 2 by 5 and zero out rows 1 and 3: \(\begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\)
The eigenvector is \(\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}\)

For \(\lambda=6\):

We find the nullspace of \(\begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix}\)

divide the first row by -5: \(\begin{bmatrix} 1 & -2/5 & -3/5 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix}\)

divide the second row by -2: \(\begin{bmatrix} 1 & -2/5 & -3/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\)

Add 2/5 times the second row to the first row: \(\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\)

The eigenvector is \(\begin{bmatrix} 8/5 \\ 5/2 \\ 1 \end{bmatrix}\)

The set of eigenvectors is: \(\begin{bmatrix} 8/5 \\ 5/2 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}\)

Checking the work:

A <- matrix(c(1,0,0,2,4,0,3,5,6),3,3)
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

Our solutions are scalar multiples of these eigenvectors, and we could arrive at these by setting our’s to be length 1.