Problem Set 1

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

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

The rank of a matrix is the number of pivots it has. Further, for rectangular matrices the rank has to be no greater than the smaller of the row or column dimension.

The Rank of matrix A is 4

(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 for an mxn matrix where m > n is n

The minimum rank for an mxn matrix where m > n and assuming that the matrix is non-zero is 1

(3) What is the rank of matrix B?

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

The Rank of matrix A 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.

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

\(\lambda\) is an eigenvalue of A iff, \(\det(\lambda I_n - A) = 0\)

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

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

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

\((\lambda - 1) (\lambda - 4) (\lambda - 6) + (-2 \times -5 \times 0) + (-3 \times 0 \times 0) - (-2 \times 0 \times (\lambda - 6)) - ((\lambda - 1) \times -5 \times 0) - (-3 \times (\lambda - 4) \times 0) = 0\)

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

This means our possible eigenvalues are \(\lambda = 1\), or \(\lambda = 4\), or \(\lambda = 6\)

To find the eigenvectors we need to return to

\(\det(\lambda I_n - A) = 0\) and remember this comes from

\((\lambda I_n - A) \vec{v} = \vec{0}\), which is the same as

\(\left[ \begin{array}{ccc} \lambda - 1 & -2 & -3 \\ 0 & \lambda -4 & -5 \\ 0 & 0 & \lambda -6 \end{array}\right] \vec{v} = \vec{0}\), and we need to solve for our eigenvalues, so

\(\lambda = 1, \left[ \begin{array}{ccc} (1 - 1) & -2 & -3 \\ 0 & (1 - 4) & -5 \\ 0 & 0 & (1 - 6) \end{array}\right] \vec{v} = \vec{0}\)

\(\left[ \begin{array}{ccc} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{array}\right] \vec{v} = \vec{0}\)

If we multiply row 1 by 3 and row 2 by 2 and then subtract row 2 from row 1, then row 1 = 0, 0, 1

If we subtract row 3 from from row 2, then row 2 = 0, -3, 0

If we divide row 3 by 5 and add to row 1, then row 1 = 0, 0, 0

\(\left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -5 \end{array}\right] \left[ \begin{array}{ccc} v_1 \\ v_2 \\ v_3 \end{array}\right] = \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right]\)

This does not give me any relation between \(v_1, v_2, v_3\). I think the eigenvectors would be would be this.

\(E_{\lambda=1} = span (\left[ \begin{array}{ccc} 0 \\ -3 \\ 0 \end{array}\right] \left[ \begin{array}{ccc} 0 \\ 0 \\ -5 \end{array}\right])\)

\(\lambda = 4, \left[ \begin{array}{ccc} (4 - 1) & -2 & -3 \\ 0 & (4 - 4) & -5 \\ 0 & 0 & (4 - 6) \end{array}\right] \vec{v} = \vec{0}\)

\(\left[ \begin{array}{ccc} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{array}\right] \vec{v} = \vec{0}\)

If we multiply row 2 by 2 and row 3 by 5 and subtract row 3 from the row 2, then row 2 = 0, 0, 0

If we multiply row 1 by 2 and row 3 by 3 and subtract row 3 from row 1, then row 1 = 6, -4, 0

If we divide all rows by 2, then row 1 = 3, -2, 0 and row 3 = 0, 0, -1

\(\left[ \begin{array}{ccc} 3 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right] \left[ \begin{array}{ccc} v_1 \\ v_2 \\ v_3 \end{array}\right] = \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right]\)

This gives us \(3v_1 = 2v_2\), which I think means this:

\(E_{\lambda=4} = span (\left[ \begin{array}{ccc} 3 \\ -2 \\ 0 \end{array}\right])\)

\(\lambda = 6, \left[ \begin{array}{ccc} (6 - 1) & -2 & -3 \\ 0 & (6 - 4) & -5 \\ 0 & 0 & (6 - 6) \end{array}\right] \vec{v} = \vec{0}\)

\(\left[ \begin{array}{ccc} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{array}\right] \vec{v} = \vec{0}\)

This gives us \(5v_1 = 8v_3\), which I think means this:

\(E_{\lambda=6} = span (\left[ \begin{array}{ccc} 5 \\ 0 \\ -8 \end{array}\right])\)