Homework3

Problem Set 1

Question 1: What is the rank of the matrix A?

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

Reduce to row echelon form:

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

\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\)

\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 8 & 5 \\ 5 & 4 & -2 & -3 \end{bmatrix}\)

\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 8 & 5 \\ 0 & -6 & -17 & -23 \end{bmatrix}\)

\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & -5 & -2 \end{bmatrix}\)

\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & 0 & 14/3 \end{bmatrix}\)

Since the matrix contains no rows of all zeros, the matrix has a rank equal to the number of rows, which is 4.

Answer: Rank = 4

Question 2: Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Answer: The maximum rank could be the total number of columns, which would be n. This is because the extra row(s) would be all zeros when reduced to row echelon form. The minimum rank could be 1 if the matrix is non-zero. If a matrix contains any numbers at all, its rank is always at least 1.

Question 3: What is the rank of matrix B?

B = \(\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\)

Reduce to row echelon form:

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

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

\(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)

Two rows are all zeros, so only 1 row is counted for the rank.

Answer: Rank = 1

Problem Set 2

Question 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.

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

1. Write the characteristic polynomial.

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

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

(4 - \(\lambda\) - \(\lambda\) + \(\lambda^2\))(6 - \(\lambda\)) = 0

24 - 6\(\lambda\) - 6\(\lambda\) + 6\(\lambda^2\) - 4\(\lambda\) + \(\lambda^2\) +\(\lambda^2\) - \(\lambda^3\) = 0

-\(\lambda^3\) + 8\(\lambda^2\) - 16\(\lambda\) + 24 = 0

Characteristic polynomial: -\(\lambda^3\) + 8\(\lambda^2\) - 16\(\lambda\) + 24

2. Find the eigenvalues.

We already know the factored version of the characteristic polynomial, so we will solve for the values of lambda.

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

\(\lambda\) = 1, 4, and 6

Eigenvalues: 1, 4, 6

3. Find the eigenvectors.

To find the eigenvectors, each eigenvalue must be used in the matrix.

First, we will use \(\lambda\) = 1.

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

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

We can row reduce this matrix.

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

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

\(\begin{bmatrix} 0 & 1 & 3/2 \\ 0 & 0 & 1/2 \\ 0 & 0 & 0 \end{bmatrix}\)

\(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1/2 \\ 0 & 0 & 0 \end{bmatrix}\)

\(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\)

Set up the equation to solve for x, y, and z.

\(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\) * \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

x = t

y = 0

z = 0

Eigenvector = \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)

Next, we will use \(\lambda\) = 4.

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

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

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

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

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

Set up the equation to solve for x, y, and z.

\(\begin{bmatrix} 1 & -2/3 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\) * \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

x - 2/3y - z = 0

z = 0

x - 2/3y = 0

x = 2/3y

Eigenvector = \(\begin{pmatrix} 2/3 \\ 1 \\ 0 \end{pmatrix}\)

Lastly, we will use \(\lambda\) = 6.

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

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

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

\(\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\)

Set up the equation to solve for x, y, and z.

\(\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}\) * \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

x - 8/5z = 0

y - 5/2z = 0

z = t

x = 8/5t

y = 5/2t

Eigenvector = \(\begin{pmatrix} 8/5 \\ 5/2 \\ 1 \end{pmatrix}\)

The eigenvectors are \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\), \(\begin{pmatrix} 2/3 \\ 1 \\ 0 \end{pmatrix}\), and \(\begin{pmatrix} 8/5 \\ 5/2 \\ 1 \end{pmatrix}\).