Problem Set 1

  1. What is the rank of the matrix A?
\(A = \begin{bmatrix}1 & 2 & 3 & 4\\−1 & 0 & 1 & 3\\0 & 1 & −2 & 1\\5 & 4 & −2 & −3\end{bmatrix}\)

Answer:

$

\(\overset{Start...}{\rightarrow} \begin{bmatrix}1 & 2 & 3 & 4\\−1 & 0 & 1 & 3\\0 & 1 & −2 & 1\\5 & 4 & −2 & −3\end{bmatrix} \overset{R_{1} + R_{2}}{\rightarrow} \begin{bmatrix}1 & 2 & 3 & 4\\0 & 2 & 4 & 7\\0 & 1 & −2 & 1\\5 & 4 & −2 & −3\end{bmatrix} \overset{-R_{1} + R_{4}}{\rightarrow} \begin{bmatrix}1 & 2 & 3 & 4\\0 & 2 & 4 & 7\\0 & 1 & −2 & 1\\0 & -6 & -17 & -23\end{bmatrix} \overset{\frac{1}{2}R_{2}}{\rightarrow} \begin{bmatrix}1 & 2 & 3 & 4\\0 & 1 & 2 &\frac{7}{2}\\0 & 1 & −2 & 1\\0 & -6 & -17 & -23\end{bmatrix}\)

\(\overset{-2R_{2}+R_{1}}{\rightarrow} \begin{bmatrix}1 & 0 & -1 & -3\\0 & 1 & 2 &\frac{7}{2}\\0 & 1 & −2 & 1\\0 & -6 & -17 & -23\end{bmatrix} \overset{-R_{2}+R_{3}}{\rightarrow} \begin{bmatrix}1 & 0 & -1 & -3\\0 & 1 & 2 &\frac{7}{2}\\0 & 0 & -4 & -\frac{5}{2}\\0 & -6 & -17 & -23\end{bmatrix} \overset{6R_{2}+R_{4}}{\rightarrow} \begin{bmatrix}1 & 0 & -1 & -3\\0 & 1 & 2 &\frac{7}{2}\\0 & 0 & -4 & -\frac{5}{2}\\0 & 0 & -5 & -2\end{bmatrix} \overset{-\frac{1}{4}R_{3}}{\rightarrow} \begin{bmatrix}1 & 0 & -1 & -3\\0 & 1 & 2 &\frac{7}{2}\\0 & 0 & 1 & \frac{5}{8}\\0 & 0 & -5 & -2\end{bmatrix}\)

\(\overset{R_{3}+R_{1}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 2 &\frac{7}{2}\\0 & 0 & 1 & \frac{5}{8}\\0 & 0 & -5 & -2\end{bmatrix} \overset{-2R_{3}+R_{2}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 0 &\frac{9}{4}\\0 & 0 & 1 & \frac{5}{8}\\0 & 0 & -5 & -2\end{bmatrix} \overset{5R_{3}+R_{4}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 0 &\frac{9}{4}\\0 & 0 & 1 & \frac{5}{8}\\0 & 0 & 0 & \frac{29}{8}\end{bmatrix} \overset{\frac{8}{29}R_{4}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 0 &\frac{9}{4}\\0 & 0 & 1 & \frac{5}{8}\\0 & 0 & 0 & 1\end{bmatrix}\)

\(\overset{-\frac{8}{5}R_{4}+R_{3}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 0 &\frac{9}{4}\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix} \overset{-\frac{4}{9}R_{4}+R_{2}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & -\frac{19}{8}\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix} \overset{\frac{8}{19}R_{4}+R_{1}}{\rightarrow} \begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}\)

For the four pivot columns, we get r(A) = 4.

  1. Given an mxn matrix where m > n, what can be the maximum rank? The mini- mum rank, assuming that the matrix is non-zero?
Answer: The maximum rank is the lower of the two matrix dimensions so:\
\(r_{max}(M) = m\)
\(r_{min}(M) = 0\)
  1. What is the rank of matrix B?
\(B = \begin{bmatrix}1 & 2 & 1\\3 & 6 &3\\2 & 4 & 2\end{bmatrix}\)

\(\begin{bmatrix}1 & 2 & 1\\3 & 6 & 3\\2 & 4 & 2\end{bmatrix} \overset{-3R_{1} + R_{2}}{\rightarrow} \begin{bmatrix}1 & 2 & 1\\0 & 0 & 0\\2 & 4 & 2\end{bmatrix} \overset{-2R_{1} + R_{3}}{\rightarrow} \begin{bmatrix}1 & 2 & 1\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}\)

So, r(B) = 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.

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

\(\begin{bmatrix}1 & 2 & 3\\0 & 4 & 5\\0 & 0 &6\end{bmatrix} \overset{(\frac{R_{2}}{4})}{\rightarrow} \begin{bmatrix}1 & 2 & 3\\0 & 1 & \frac{5}{4}\\0 & 0 &6\end{bmatrix} \overset{-2R_{2}+R_{1}}{\rightarrow} \begin{bmatrix}1 & 0 & \frac{1}{2}\\0 & 1 & \frac{5}{4}\\0 & 0 &6\end{bmatrix} \overset{(\frac{R_{3}}{6})}{\rightarrow} \begin{bmatrix}1 & 0 & \frac{1}{2}\\0 & 1 & \frac{5}{4}\\0 & 0 & 1\end{bmatrix} \overset{RREF}{\rightarrow} \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\)

To find the eigenvalues…

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

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

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

The characteristic polynomial is: \((\lambda-1)(\lambda-4)(\lambda-6) - 0 - 0\)

The eigenvalues are \(\lambda-1 = (1, 4, 6)\)

For \(\lambda = 1\)

\(E_{1} = N\left( \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} - \begin{bmatrix}1 & 2 & 3\\0 & 4 & 5\\0 & 0 &6\end{bmatrix} \right) = N\left( \begin{bmatrix}0 & -2 & -3\\0 & -3 & -5\\0 & 0 & -5\end{bmatrix} \right)\)

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

\(-2v_{2}-3v_{3} = 0\)

\(-3v_{2}-5v_{3} = 0\)

\(-5v_{3} = 0\)

We see that \(v_{3} = 0\) and \(v_{2} = 0\) and \(v_{1} = 1\).

So \(E_{1} = \left\langle \begin{bmatrix}1\\0\\0\end{bmatrix} \right\rangle\)

For \(\lambda = 4\)

\(E_{4} = N\left( \begin{bmatrix}41 & 0 & 0\\0 & 4 & 0\\0 & 0 & 4\end{bmatrix} - \begin{bmatrix}1 & 2 & 3\\0 & 4 & 5\\0 & 0 &6\end{bmatrix} \right) = N\left( \begin{bmatrix}3 & -2 & -3\\0 & 0 & -5\\0 & 0 & -2\end{bmatrix} \right)\)

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

So, \(v_{3} = 0\) and

\(3v_{1}-2v_{2} = 0\)

\(v_{1} = \frac{2}{3}v_{2}\)

And we can assume \(v{2} = 1\) so

\(E_{4} = \left\langle \begin{bmatrix}\frac{2}{3}\\1\\0\end{bmatrix} \right\rangle\)

For \(\lambda = 6\)

\(E_{6} = N\left( \begin{bmatrix}6 & 0 & 0\\0 & 6 & 0\\0 & 0 & 6\end{bmatrix} - \begin{bmatrix}1 & 2 & 3\\0 & 4 & 5\\0 & 0 &6\end{bmatrix} \right) = N\left( \begin{bmatrix}5 & -2 & -3\\0 & 2 & -5\\0 & 0 & 0\end{bmatrix} \right)\)

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

\(5v_{1} - 2v_{2} - 3v_{3} = 0\)

\(2v_{2} - 5v_{3} = 0\)

\(v_{3} = 1\)

So. \(2v_{2} - 5 = 0\)

\(v_{2}= \frac{5}{2}\)

\(5v_{1} - 5 - 3 = 0\) and \(v_{1} = \frac{8}{5}\)

\(E_{6} = \left\langle \begin{bmatrix}\frac{8}{5}\\\frac{5}{2}\\1\end{bmatrix} \right\rangle\)