1. λ1=1, λ2=3; P=[1121], D=[1003]
2. λ1=0, λ2=2; P=[1132], D=[0002]
3. λ1=2, λ2=3; P=[1132], D=[2003]
4. λ1=1, λ2=2; P=[1143], D=[1002]
5. λ1=1, λ2=3; P=[1143], D=[1003]
6. λ1=1, λ2=2; P=[2334], D=[1002]
7. λ1=1, λ2=2; P=[2152], D=[1002]
8. λ1=1, λ2=2; P=[3253], D=[1002]
9. The double eigenvalue λ1=λ2=1 has only the single associated eigenvector v1=(2, 1), so the matrix A is not diagonalizable.
10. The double eigenvalue λ1=λ2=2 has only the single associated eigenvector v1=(1, 1), so the matrix A is not diagonalizable.
11. The double eigenvalue λ1=λ2=2 has only the single associated eigenvector v1=(−1, 3), so the matrix A is not diagonalizable.
12. The double eigenvalue λ1=λ2=−1 has only the single associated eigenvector v1=(−3, 4), so the matrix A is not diagonalizable.
13. λ1=1, λ2=λ3=2; P=⎡⎣⎢100001310⎤⎦⎥, D=⎡⎣⎢100020002⎤⎦⎥
14. λ1=λ2=0, λ3=1;P=⎡⎣⎢−102111110⎤⎦⎥, D=⎡⎣⎢000000001⎤⎦⎥
15. λ1=0, λ2=λ3=1; P=⎡⎣⎢110−102320⎤⎦⎥, D=⎡⎣⎢000010001⎤⎦⎥
16. λ1= λ2=1, λ3=3; P=⎡⎣⎢001110−102⎤⎦⎥, D=⎡⎣⎢100010003⎤⎦⎥
17. λ1=−1, λ2=1, λ3=2; P=⎡⎣⎢110−102111⎤⎦⎥, D=⎡⎣⎢−100010002⎤⎦⎥
18. λ1=1, λ2=2, λ3=3; P=⎡⎣⎢110−102111⎤⎦⎥, D=⎡⎣⎢100020003⎤⎦⎥
19. λ1=1, λ2=2, λ3=3; P=⎡⎣⎢111110−102⎤⎦⎥, D=⎡⎣⎢100020003⎤⎦⎥
20. λ1=2, λ2=5, λ3=6; P=⎡⎣⎢1030−130−25⎤⎦⎥, D=⎡⎣⎢200050006⎤⎦⎥
21. The triple eigenvalue λ1=λ2=λ3=1 has only the two associated eigenvectors v1=(0, 0, 1) and v2=(1, 1, 0), so the matrix A is not diagonalizable.
22. The triple eigenvalue λ1=λ2=λ3=1 has only the single associated eigenvector v1=(1, 1, 1), so the matrix A is not diagonalizable.
23. The eigenvalues λ1=λ2=1 and λ3=2 have only the two associated eigenvectors v1=(1, 1, 1) and v3=(1, 1, 0), so the matrix A is not diagonalizable.
24. The eigenvalues λ1=1 and λ2=λ3=2 have only the two associated eigenvectors v1=(1, 1, 0) and v2=(1, 1, 1), so the matrix A is not diagonalizable.
25. λ1=λ2=−1, λ3=λ4=1; P=⎡⎣⎢⎢⎢⎢0001111001001000⎤⎦⎥⎥⎥⎥, D=⎡⎣⎢⎢⎢⎢−10000−10000100001⎤⎦⎥⎥⎥⎥
26. λ1=λ2=λ3=1, λ4=2; P=⎡⎣⎢⎢⎢⎢0000010010001011⎤⎦⎥⎥⎥⎥, D=⎡⎣⎢⎢⎢⎢1000010000100002⎤⎦⎥⎥⎥⎥
27. The eigenvalues λ1=λ2=λ3=1 and λ4=2 have only the two associated eigenvectors v1=(1, 0, 0, 0) and v4=(1, 1, 1, 1), so the matrix A is not diagonalizable.
28. The eigenvalues λ1=λ2=1 and λ3=λ4=2 have only the two associated eigenvectors v1=(1, 0, 0, 0) and v3=(1, 1, 1, 0), so the matrix A is not diagonalizable.