Assignment 3

Problem Set 1

Problem 1

What is the rank of the matrix A? \[\begin{equation*} A = \mathbf{}\left[\begin{matrix} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3 \end{matrix}\right] \end{equation*}\]

The rank is 4 because there are 4 non-zero rows in the row reduced echelon form.

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

Problem 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 could be m. The minimum rank could be 1 since we’re assuming the matrix is non-zero.

Problem 3

What is the rank of matrix B? \[\begin{equation*}B = \mathbf{}\left[\begin{matrix} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2\\ \end{matrix}\right] \end{equation*}\]

The rank is 1 because there is 1 non-zero row in row reduced echelon form.

You can see that row 3 is a scalar multiple of row 1 (2*R1 = R3). You can also see that row 2 is a scalar multiple of row 1 (3*R1 = R2). That means that rows 1 & 3 are linearly depedent and rows 1 & 2 are linearly dependent. THat leaves us with one linearly indepedent row. That also shows us the rank is 1.

A <- matrix(c(1,2,1,3,6,3,2,4,2),nrow=3,byrow=TRUE)
A
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rref(A)
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0

Problem Set 2

Problem 1

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.

Eigenvalues By Hand:

\[\begin{equation*}A = \mathbf{}\left[\begin{matrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{matrix}\right] \end{equation*}\]

To find the eigenvalues we know the following is true:
Step 1 \[\begin{equation*}det( \mathbf{} \left[ \begin{matrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda\\ \end{matrix} \right] - \left[ \begin{matrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\\ \end{matrix} \right] ) = 0 \end{equation*}\]

Step 2 \[\begin{equation*}det( \mathbf{} \left[ \begin{matrix} \lambda - 1 & 2 & 3\\ 0 & \lambda - 4 & 5\\ 0 & 0 & \lambda - 6\\ \end{matrix} \right] ) = 0 \end{equation*}\]

Step 3 \[\begin{equation*}\lambda - 1 \mathbf{} \left[ \begin{matrix} \lambda - 4 & 5\\ 0 & \lambda - 6\\ \end{matrix} \right] - 2 \mathbf{} \left[ \begin{matrix} 0 & 5\\ 0 & \lambda - 6\\ \end{matrix} \right] + 3 \mathbf{} \left[ \begin{matrix} 0 & \lambda - 4 \\ 0 & 0\\ \end{matrix} \right] \end{equation*}\]

Step 4 \[\begin{equation*} (\lambda - 1)(\lambda - 4)(\lambda - 6) - (2*0*0) + (3*0*0) \end{equation*}\]

Step 5
\[\begin{equation*} (\lambda - 1)(\lambda - 4)(\lambda - 6) \end{equation*}\]

The eigenvalues are 1,4, and 6

Multiplying those together, we see that the characteristic polynomial is: \[\begin{equation*} (\lambda - 1)(\lambda - 4)(\lambda - 6) \end{equation*}\]

\[\begin{equation*} (\lambda^{2} - 1\lambda - 4\lambda + 4) (\lambda - 6) \end{equation*}\]

\[\begin{equation*} (\lambda^{3} - 5\lambda^{2} + 4\lambda) - (6\lambda^{2} - 30\lambda + 24) \end{equation*}\]

\[\begin{equation*} (\lambda^{3} - 11\lambda^{2} + 34\lambda - 24) \end{equation*}\]

Find the Eigenvectors:

Step 1 We know that I\(\lambda\) - A = 0, which gives us the following: \[\begin{equation*}det( \mathbf{} \left[ \begin{matrix} \lambda - 1 & 2 & 3\\ 0 & \lambda - 4 & 5\\ 0 & 0 & \lambda - 6\\ \end{matrix} \right] \end{equation*}\]

For \(\lambda\) = 1 we get:
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 0 & 2 & 3\\ 0 & -3 & 5\\ 0 & 0 & -5\\ \end{matrix} \right] \end{equation*}\]

For \(\lambda\) = 4 we get:
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & -2\\ \end{matrix} \right] \end{equation*}\]

For \(\lambda\) = 6 we get:
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 5 & 2 & 3\\ 0 & 2 & 5\\ 0 & 0 & 0\\ \end{matrix} \right] \end{equation*}\]

Step 2 In order to get the eigen vectors, each matrix needs to be in row reduced echelon form:
For \(\lambda\) = 1 the following matrix is rref:

lambda1 <- matrix(c(0,2,3,0,-3,5,0,0,-5),nrow=3,byrow=TRUE)
rref(lambda1)
##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

For \(\lambda\) = 4 the following matrix is rref:

lambda1 <- matrix(c(3,2,3,0,0,5,0,0,-2),nrow=3,byrow=TRUE)
rref(lambda1)
##      [,1]      [,2] [,3]
## [1,]    1 0.6666667    0
## [2,]    0 0.0000000    1
## [3,]    0 0.0000000    0

For \(\lambda\) = 6 we get:

lambda1 <- matrix(c(5,2,3,0,2,5,0,0,0),nrow=3,byrow=TRUE)
rref(lambda1)
##      [,1] [,2] [,3]
## [1,]    1    0 -0.4
## [2,]    0    1  2.5
## [3,]    0    0  0.0

Step 3 Let t be a constant, this means the eigenvector is:
\(\lambda\) = 1
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = \left[ \begin{matrix} 0\\ 0\\ 0\\ \end{matrix} \right] \end{equation*}\]

\[\begin{equation*} v2 = 0 \end{equation*}\] \[\begin{equation*} v3 = 0 \end{equation*}\]

\[\begin{equation*} \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = t \left[ \begin{matrix} 1\\ 0\\ 0\\ \end{matrix} \right] \end{equation*}\]

\(\lambda\) = 4
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 1& 2/3 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = \left[ \begin{matrix} 0\\ 0\\ 0\\ \end{matrix} \right] \end{equation*}\]

\[\begin{equation*} v1 + 2/3v2 = 0 \end{equation*}\] \[\begin{equation*} v3 = 0 \end{equation*}\]

\[\begin{equation*} \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = t \left[ \begin{matrix} -2/3\\ 1\\ 0\\ \end{matrix} \right] \end{equation*}\]

\(\lambda\) = 6
\[\begin{equation*} \mathbf{} \left[ \begin{matrix} 1& 0 & -0.4\\ 0 & 1 & 2.5\\ 0 & 0 & 0\\ \end{matrix} \right] \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = \left[ \begin{matrix} 0\\ 0\\ 0\\ \end{matrix} \right] \end{equation*}\]

\[\begin{equation*} v1 - 0.4v3 = 0 \end{equation*}\] \[\begin{equation*} v2 + 2.5v3 = 0 \end{equation*}\]

\[\begin{equation*} \left[ \begin{matrix} v1\\ v2\\ v3\\ \end{matrix} \right] = t \left[ \begin{matrix} 0.4\\ -2.5\\ 1\\ \end{matrix} \right] \end{equation*}\]

Using R:

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

The characteristic polynomial is 1\(\lambda\)3 - 11\(\lambda\)2 - 34\(\lambda\) - 24\(\lambda\).
The eigenvalues are \(\lambda\) = 6,4,and 1

charpoly <- charpoly(A,info=TRUE)
## Error term: 0
charpoly$cp
## [1]   1 -11  34 -24
roots(charpoly$cp)
## [1] 6 4 1
Resources: