Problem set 1:

1. What is the rank of the matrix \(A\)?

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

To determine the rank of the given matrix \(A\), we performed a series of row operations to reduce \(A\) to its row echelon form (REF).

The row operations applied to Matrix \(A\) are:

  1. \(R_2 = R_2 + R_1\)
  2. \(R_4 = R_4 - 5R_1\)
  3. \(R_3 = R_3 + 0.5R_2\)
  4. \(R_2 = \frac{1}{2}R_2\)
  5. \(R_4 = R_4 + 6R_2\)
  6. \(R_3 = -\frac{1}{4}R_3\)
  7. \(R_4 = R_4 + 5R_3\)

These operations reduce Matrix \(A\) to its row echelon form (REF), allowing us to determine its rank.

\[ A_{\text{REF}} = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3.5 \\ 0 & 0 & 1 & 0.625 \\ 0 & 0 & 0 & 1.125 \end{bmatrix} \]

Each row in \(A_{\text{REF}}\) contains a leading 1 with zeros in all entries below, indicating that all rows are linearly independent. Therefore, the rank of matrix \(A\) is equal to the number of non-zero rows in its row echelon form, which is 4.

2. Given an \(m \times n\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

The maximum rank is \(n\) and the minimum rank is 1.

Proof Summary:
  • Maximum rank \(n\): Limited by the number of columns, as you cannot have more linearly independent vectors than there are dimensions in the space spanned by the columns.
  • Minimum rank 1: A single non-zero element in the matrix implies at least one linearly independent vector, ensuring the rank is at least 1 for any non-zero matrix.

3. What is the rank of matrix \(B\)?

\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \] To find the rank of matrix \(B\), we can perform row reduction (Gaussian elimination) to bring it to its row echelon form (REF) or reduced row echelon form (RREF), and then count the number of non-zero rows.

I performed the following operations

  1. \(R_2 = R_2 - 3R_1\) results in \(R_2 = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}\)

  2. \(R_3 = R_3 - 2R_1\) results in \(R_3 = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}\)

Transformed \(B\) :

\[ B_{\text{REF}} = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

In the row echelon form of \(B\), there is only one non-zero row, which indicates that there is only one linearly independent row in matrix \(B\).

Therefore, the rank of matrix \(B\) is \(1\).

Problem set 2:

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. First, we find the eigenvalues by solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix, and \(\lambda\) is the eigenvalues.

  2. Second, we find the eigenvectors associated with each eigenvalue by solving \((A - \lambda I)x = 0\) for each eigenvalue \(\lambda\), where \(x\) represents the eigenvector.

Step 1: Finding the Eigenvalues
We can get the characteristic polynomial from \(\det(A - \lambda I) = 0\):

\[ A - \lambda I = \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} \]

The determinant of a triangular matrix is the product of its diagonal entries, so:

\[ \det(A - \lambda I) = (1-\lambda)(4-\lambda)(6-\lambda) = 0 \]

Solving the above gives us the eigenvalues:

\[ \lambda_1 = 1, \quad \lambda_2 = 4, \quad \lambda_3 = 6 \]

Step 2: Finding the Eigenvectors
We solve \((A - \lambda I)x = 0\) for each eigenvalue:

For \(\lambda_1 = 1\):

  1. \[ A - 1I = \begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix} \]

  2. We solve \((A - 1I)x = 0\) for \(x\):

  • Since the first row becomes \(0x_1 + 2x_2 + 3x_3 = 0\), and considering the triangular form of the matrix, we can choose \(x_2 = 0\) and \(x_3 = 0\) to simplify the solution, leading to \(x_1\) as a free variable. A simple choice is \(x_1 = 1\).
  • Thus, an eigenvector for \(\lambda_1 = 1\) is \(x_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\).

For \(\lambda_2 = 4\):

  1. \[ A - 4I = \begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix} \]

  2. We solve \((A - 4I)x = 0\) for \(x\):

  • The second row suggests \(5x_3 = 0\), leading to \(x_3 = 0\).
  • From the first row, \(-3x_1 + 2x_2 + 3x_3 = 0\), with \(x_3 = 0\), we can choose \(x_1 = 2\) and \(x_2 = 3\) to satisfy the equation.
  • Thus, an eigenvector for \(\lambda_2 = 4\) is \(x_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}\).

For \(\lambda_3 = 6\):

  1. \[ A - 6I = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix} \]

  2. We solve \((A - 6I)x = 0\) for \(x\):

  • The third row is all zeros, which does not provide a constraint on \(x_3\), so \(x_3\) can be a free variable. Choosing \(x_3 = 2\) for simplicity.
  • From the second row, \(-2x_2 + 5x_3 = 0\). Substituting \(x_3 = 2\) gives \(x_2 = 5\).
  • From the first row, \(-5x_1 + 2x_2 + 3x_3 = 0\). Substituting \(x_2 = 5\) and \(x_3 = 2\) gives \(x_1 = 3\).
  • Thus, an eigenvector for \(\lambda_3 = 6\) is \(x_3 = \begin{bmatrix} 3 \\ 5 \\ 2 \end{bmatrix}\).