Exercise 7.1.2

Compute the characteristic polynomial to find eigenvalues as we did before. For each $\lambda$, find a basis for $K_{\lambda }$ consisting of a union of disjoint cycles by computing bases for the null space of ${(A-\mathit{\lambda I})}^p$ for each $p$. Write down the matrix $S$ whose columns consist of these cycles of generalized eigenvectors. Then we will get the Jordan canonical form $J=S^{-1}\mathit{AS}$. When the matrix is diagonalizable, the Jordan canonical form should be the diagonal matrix similar to $A$. For more detail, please see the examples in the textbook. On the other hand, these results were computed by Wolfram Alpha. For example, the final question needs the command below.

JordanDecomposition[{{2,1,0,0},{0,2,1,0},{0,0,3,0},{0,1,-1,3}}]

1.

$$S=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),J=\left(\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right).$$

2.

$$S=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right),J=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill \end{array}\right).$$

3.

$$S=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right),J=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right).$$

4.

$$S=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right),J=\left(\begin{array}{cccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right).$$