1. ${\lambda }_1=1,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{rr}1& 2\\ 1& 1\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}1& 0\\ 0& 3\end{array}\right]$

2. ${\lambda }_1=0,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}1& 3\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}0& 0\\ 0& 2\end{array}\right]$

3. ${\lambda }_1=2,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{ll}1& 3\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}2& 0\\ 0& 3\end{array}\right]$

4. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}1& 4\\ 1& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

5. ${\lambda }_1=1,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{rr}1& 4\\ 1& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 3\end{array}\right]$

6. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}2& 3\\ 3& 4\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

7. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}2& 5\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

8. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}3& 5\\ 2& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

9. The double eigenvalue ${\lambda }_1={\lambda }_2=1$ has only the single associated eigenvector ${\mathbf{v}}_1=(2,1)$, so the matrix A is not diagonalizable.

10. The double eigenvalue ${\lambda }_1={\lambda }_2=2$ has only the single associated eigenvector ${\mathbf{v}}_1=(1,1)$, so the matrix A is not diagonalizable.

11. The double eigenvalue ${\lambda }_1={\lambda }_2=2$ has only the single associated eigenvector ${\mathbf{v}}_1=(-1,3)$, so the matrix A is not diagonalizable.

12. The double eigenvalue ${\lambda }_1={\lambda }_2=-1$ has only the single associated eigenvector ${\mathbf{v}}_1=(-3,4)$, so the matrix A is not diagonalizable.

13. ${\lambda }_1=1,{\lambda }_2={\lambda }_3=2;\mathbf{P}=\left[\begin{array}{lll}1& 0& 3\\ 0& 0& 1\\ 0& 1& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{lll}1& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$

14. ${\lambda }_1={\lambda }_2=0,{\lambda }_3=1;\mathbf{P}=\left[\begin{array}{rrr}-1& 1& 1\\ 0& 1& 1\\ 2& 1& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{lll}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$

15. ${\lambda }_1=0,{\lambda }_2={\lambda }_3=1;\mathbf{P}=\left[\begin{array}{rrr}1& -1& 3\\ 1& 0& 2\\ 0& 2& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{lll}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

16. ${\lambda }_1={\lambda }_2=1,{\lambda }_3=3;\mathbf{P}=\left[\begin{array}{rrr}0& 1& -1\\ 0& 1& 0\\ 1& 0& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 3\end{array}\right]$

17. ${\lambda }_1=-1,{\lambda }_2=1,{\lambda }_3=2;\mathbf{P}=\left[\begin{array}{rrr}1& -1& 1\\ 1& 0& 1\\ 0& 2& 1\end{array}\right],\mathbf{D}=\left[\begin{array}{rrr}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right]$

18. ${\lambda }_1=1,{\lambda }_2=2,{\lambda }_3=3;\mathbf{P}=\left[\begin{array}{rrr}1& -1& 1\\ 1& 0& 1\\ 0& 2& 1\end{array}\right],\mathbf{D}=\left[\begin{array}{lll}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right]$

19. ${\lambda }_1=1,{\lambda }_2=2,{\lambda }_3=3;\mathbf{P}=\left[\begin{array}{rrr}1& 1& -1\\ 1& 1& 0\\ 1& 0& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{rrr}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right]$

20. ${\lambda }_1=2,{\lambda }_2=5,{\lambda }_3=6;\mathbf{P}=\left[\begin{array}{rrr}1& 0& 0\\ 0& -1& -2\\ 3& 3& 5\end{array}\right],\mathbf{D}=\left[\begin{array}{rrr}2& 0& 0\\ 0& 5& 0\\ 0& 0& 6\end{array}\right]$

21. The triple eigenvalue ${\lambda }_1={\lambda }_2={\lambda }_3=1$ has only the two associated eigenvectors ${\mathbf{v}}_1=(0,0,1)$ and ${\mathbf{v}}_2=(1,1,0)$, so the matrix A is not diagonalizable.

22. The triple eigenvalue ${\lambda }_1={\lambda }_2={\lambda }_3=1$ has only the single associated eigenvector ${\mathbf{v}}_1=(1,1,1)$, so the matrix A is not diagonalizable.

23. The eigenvalues ${\lambda }_1={\lambda }_2=1$ and ${\lambda }_3=2$ have only the two associated eigenvectors ${\mathbf{v}}_1=(1,1,1)$ and ${\mathbf{v}}_3=(1,1,0)$, so the matrix A is not diagonalizable.

24. The eigenvalues ${\lambda }_1=1$ and ${\lambda }_2={\lambda }_3=2$ have only the two associated eigenvectors ${\mathbf{v}}_1=(1,1,0)$ and ${\mathbf{v}}_2=(1,1,1)$, so the matrix A is not diagonalizable.

25. ${\lambda }_1={\lambda }_2=-1,{\lambda }_3={\lambda }_4=1;\mathbf{P}=\left[\begin{array}{rrrr}0& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{rrrr}-1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

26. ${\lambda }_1={\lambda }_2={\lambda }_3=1,{\lambda }_4=2;\mathbf{P}=\left[\begin{array}{rrrr}0& 0& 1& 1\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 1\end{array}\right],\mathbf{D}=\left[\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 2\end{array}\right]$

27. The eigenvalues ${\lambda }_1={\lambda }_2={\lambda }_3=1$ and ${\lambda }_4=2$ have only the two associated eigenvectors ${\mathbf{v}}_1=(1,0,0,0)$ and ${\mathbf{v}}_4=(1,1,1,1)$, so the matrix A is not diagonalizable.

28. The eigenvalues ${\lambda }_1={\lambda }_2=1$ and ${\lambda }_3={\lambda }_4=2$ have only the two associated eigenvectors ${\mathbf{v}}_1=(1,0,0,0)$ and ${\mathbf{v}}_3=(1,1,1,0)$, so the matrix A is not diagonalizable.