1. Problem set 1

  1. What is the rank of the matrix A? \[ A = \left(\begin{array}{cc} 1 & 2 & 3 & 4\\ −1 & 0 & 1 & 3\\ 0 & 1 & −2 & 1\\ 5 & 4 & −2 & −3 \end{array}\right) \]

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

  3. What is the rank of matrix B? \[ B = \left(\begin{array}{cc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{array}\right) \]

2. 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. \[ A = \left(\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{array}\right) \]

Please show your work using an R-markdown document. Please name your assignment submission with your first initial and last name.

Solutions:

Problem set 1

\[ A = \left(\begin{array}{cc} 1 & 2 & 3 & 4\\ −1 & 0 & 1 & 3\\ 0 & 1 & −2 & 1\\ 5 & 4 & −2 & −3 \end{array}\right) = (r_{1} -> r_{3}) \left(\begin{array}{cc} −1 & 0 & 1 & 3\\ 0 & 1 & −2 & 1\\ 1 & 2 & 3 & 4\\ 5 & 4 & −2 & −3 \end{array}\right) \] =>

\[ A= (-r_{1}, r_{1}-r{3}, -5r_{1}+r{4}) \left(\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 1 & −2 & 1\\ 0 & 2 & 4 & 7\\ 0 & 0 & 3 & 12 \end{array}\right) = (-r_{1}, r_{1}-r{3}, -5r_{1}+r{4}) \left(\begin{array}{cc} 1 & 0 & -1 & -3\\ 0 & 1 & −2 & 1\\ 0 & 1 & 2 & 7/4\\ 0 & 0 & 1/4 & 1 \end{array}\right) \] Rank \(A = 4\)

  1. Given \(mxn\) matrix where \(m > n\)
  • maximum rank:

\(\begin{cases} row\cdot rank \leq m \\ column\space\cdot rank \leq n \end{cases}\)

  • minimum rank (matrix \(\neq\) 0):

\(\begin{cases} rank(A_{mxn}) \leq n \end{cases}\)

  1. \[ B = \left(\begin{array}{cc} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{array}\right) = \left(\begin{array}{cc} 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 2 & 1 \end{array}\right) = \left(\begin{array}{cc} 1 & 2 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right) \]

Rank \(B = 1\)

Problem set 2

  • Finding eigenvalues:

\(det(A-\lambda I)=0\), where I is the 3×3 identity matrix

\[ A-\lambda I = \left(\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \end{array}\right) - \left(\begin{array}{cc} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda \end{array}\right) = \left(\begin{array}{cc} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda \end{array}\right) \]

\[ det(A-\lambda I) = (1-\lambda) \left(\begin{array}{cc} 4-\lambda & 5\\ 0 & 6-\lambda \end{array}\right) -(-3) \left(\begin{array}{cc} 0 & 5\\ 0 & 6-\lambda \end{array}\right) +3 \left(\begin{array}{cc} 0 & 4-\lambda\\ 0 & 0 \end{array}\right) \] =>

\[ det(A-\lambda I) = (1-\lambda)((4-\lambda)(6-\lambda)-0)+3(0-0)+3(0-0) = (1-\lambda)(4-\lambda)(6-\lambda) \] To find solutions to det(A − λI) = 0: \((1-\lambda)(4-\lambda)(6-\lambda) = 0\) Therefore, these are the eigenvalues of A: \(\begin{cases} \lambda = 1 \\ \lambda = 4 \\ \lambda = 6 \end{cases}\)

  • Finding eigenvectors by Gaussian Elimination:

    1. For each eigenvalue, \(\begin{cases} (A-\lambda I) x = 0 \end{cases}\) (where x is the eigenvector associated with eigenvalue λ)
    1. Gaussian Elimination \(\begin{cases} (A-\lambda I : 0) \end{cases}\), solve the resulting linear system by back substitution


For \(\lambda = 1\)

\[ A-I = \left(\begin{array}{cc} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5 \end{array}\right) \]

\[ \left(\begin{array}{cc} 0 & 2 & 3 & 0\\ 0 & 3 & 5 & 0\\ 0 & 0 & 5 & 0 \end{array}\right) \] => \(\begin{cases} 2x_2+3x_3 = 0 \\ 3x_2+5x_3 = 0 \\ 5x_3 = 0 \end{cases}\)

\[ x = \left(\begin{array}{cc} 1\\ 0\\ 0 \end{array}\right) \]

For \(\lambda = 4\)

\[ A-4I = \left(\begin{array}{cc} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2 \end{array}\right) \]

\[ \left(\begin{array}{cc} -3 & 2 & 3 & 0\\ 0 & 0 & 5 & 0\\ 0 & 0 & 2 & 0 \end{array}\right) \] => \(\begin{cases} -3x_1+2x_2+3x_3 = 0 \\ 5x_3 = 0 \\ 2x_3 = 0 \end{cases}\)

\[ x = \left(\begin{array}{cc} x_1=2/3x_2\\ x_2\\ x_3 \end{array}\right) = \left(\begin{array}{cc} 2/3x_2\\ x_2\\ x_3 \end{array}\right) = x_2 \left(\begin{array}{cc} 2/3\\ 1\\ 0 \end{array}\right) +x_3 \left(\begin{array}{cc} 0\\ 0\\ 1 \end{array}\right) \]

For \(\lambda = 6\)

\[ A-6I = \left(\begin{array}{cc} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{array}\right) \]

\[ \left(\begin{array}{cc} -5 & 2 & 3 & 0\\ 0 & -2 & 5 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \] => \(\begin{cases} -5x_1+2x_2+3x_3 = 0 \\ -2x_2+5x_3 = 0 \end{cases}\)

\[ x = \left(\begin{array}{cc} x_1=11/10x_3\\ x_2=5/2x_3\\ x_3 \end{array}\right) = \left(\begin{array}{cc} 11/10x_3\\ 5/2x_3\\ x_3 \end{array}\right) = x_3 \left(\begin{array}{cc} 11/10\\ 5/2\\ 1 \end{array}\right) \]