4 10/19

1. New: 9.1.4, 9.1.6 Old: 8.1.4, 8.1.6. Find a complete set of independent eigenvectors for each eigenvalue. Make sure to write the null space for any multiple eigenvalues. No Gerschgorin. State whether singular and/or defective.

2. New: 9.1.14 Old: 8.1.14 Find a complete set of independent eigenvectors for each eigenvalue. Make sure to write the null space for any multiple eigenvalues. No Gerschgorin. Explain whether singular and/or defective or not.

3. New: 9.1.4, 9.1.6 Old: 8.1.4, 8.1.6. Refer to your previous work on these matrices. Check that $E^{-1}AE$ is indeed $\Lambda$ for the eigenvalues and eigenvectors you found. If not, explain why not.

4. New: 9.2.11 Old: 8.2.13. First, ensure that the book knows what it is talking about by taking the matrix $A$ of 9.1.6/8.1.6,
   
   $$\left( \begin{array}{cc} 0 & 1 0 & 0 \end{array} \right),$$
   
   
   and showing that $A^2$ can be diagonalized. Then check that $A$ can indeed be diagonalized as the book says. If you find that the author knows what he is talking about, prove that the theorem is true for any arbitrary matrix $A$. If you do not believe that the author has a clue, then prove, for every matrix $A$, that if $A$ is diagonalizable, $A^2$ is diagonalizable. Hint: relate the eigenvectors and eigenvalues of $A^2$ to those of $A$.

5. New: 9.2.12 Old: 8.2.14. Show first that in the basis of the eigenvectors,
   
   $${A'}^k = \Lambda^k = \left( \begin{array}{cccc} \lam... ...vdots & \vdots & \vdots & \ddots \\ \end{array} \right)$$
   
   
   To do so, show first that this is true for $k=1$. Then show that if it is true for a value $k$, such as $k=1$, it is true for the next larger value of $k$:
   
   $$\left( \begin{array}{cccc} \lambda_1^k & 0 & 0 & \cdot... ...vdots & \vdots & \vdots & \ddots \\ \end{array} \right)$$
   
   
   That will imply the desired result for $k=2,3,4,\ldots$ in succession through recursion. Next show that since $A=EA'E^{-1}$
   
   $$A^k=EA'E^{-1} EA'E^{-1} EA'E^{-1} \ldots EA'E^{-1} = E{A'}^kE^{-1}$$
   
   

   Use the theorem to find a square root of the matrix of 9.2.5/8.2.5, i.e. $A^2$ is the matrix of question 9.2.5/8.2.5, and you must find $A$. Indicate $\sqrt{-1}=i$. Note: the eigenvalues and eigenvectors of the 9.2.5/8.2.5 matrix are:
   

   $$\lambda_1 = 0 \quad \vec e_1 = \left(\begin{array}{r} 0 ... ...ft(\begin{array}{r} 0 -3 2 \end{array}\right) \qquad$$
   
   

6. New: 9.1.4 Old: 8.1.4. Refer to your previous work on the matrix. For the given matrix solve the system $\dot{\vec x}=A\vec x$ using the same method of diagonalization as used in class. Accurately draw a comprehensive set of typical solution curves in the $x_1$, $x_2$ plane.