1. Pamela Leutwyler

2. Find the eigenvalues and eigenvectors next

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10. next characteristic polynomial

11. next characteristic polynomial

12. next potential rational roots:1,-1,3,-3,9,-9 synthetic division:

13. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 715 9

14. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9

15. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9 1

16. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9 1 1 8

17. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9 1 8 1 8 23

18. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9 1 8 23 1 8 31

19. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: 1 1 715 9 1 8 23 1 8 31 This is not zero. 1 is not a root.

20. next potential rational roots:1,-1,3,-3,9,-9 synthetic division: -3 1 715 9

21. next synthetic division: -3 1 715 9 1 potential rational roots:1,-1,3,-3,9,-9

22. next synthetic division: -3 1 715 9 -3 1 4 potential rational roots:1,-1,3,-3,9,-9

23. next synthetic division: -3 1 715 9 -3-12 1 4 3 potential rational roots:1,-1,3,-3,9,-9

24. next synthetic division: potential rational roots:1,-1,3,-3,9,-9 -3 1 715 9 -3-12 -9 1 4 3 0

25. next synthetic division: potential rational roots:1,-1,3,-3,9,-9 -3 1 715 9 -3-12 -9 1 4 3 0 This is zero. -3 is a root.

26. next synthetic division: potential rational roots:1,-1,3,-3,9,-9 -3 1 715 9 -3-12 -9 1 4 3 0

27. next synthetic division: potential rational roots:1,-1,3,-3,9,-9 -3 1 715 9 -3-12 -9 1 4 3 0

28. next synthetic division: -3 1 715 9 -3-12 -9 1 4 3 0 The eigenvalues are: -3, -3, -1

29. next The eigenvalues are: -3, -3, -1 To find an eigenvector belonging to the repeated root –3, consider the null space of the matrix –3I - A

30. next The eigenvalues are: -3, -3, -1 To find an eigenvector belonging to the repeated root –3, consider the null space of the matrix –3I - A The 2 dimensional null space of this matrix has basis =

31. next The eigenvalues are: -3, -3, -1 To find an eigenvector belonging to the repeated root –1, consider the null space of the matrix –1I - A The null space of this matrix has basis =

32. next The eigenvalues are: -3, -3, -1 The eigenvectors are:

33. next The eigenvalues are: -3, -3, -1 The eigenvectors are:

34. The eigenvalues are: -3, -3, -1 The eigenvectors are: A P P –1 diagonal matrix that is similar to A