The document finds the characteristic polynomial, eigenvalues, and eigenvectors of the matrix A = [[0, 0, -2], [1, 2, 1], [1, 0, 3]]. It determines the characteristic polynomial is (k-1)(k-2)^2 = 0, with eigenvalues of 1, 2, 2. It then finds the corresponding eigenvectors of (1, -1/2, -1/2)^T, (1, 0, -1)^T, (0, 1, 0)^T. It concludes that since there are 3 independent eigenvectors, the matrix A can be diagonalized by the invertible matrix P containing the eigenvectors as columns.

1. Justify your answer
Find the characteristic polynomial, eigenvalues, and eigenvectors and, if possible an invertible
matrix P such that P-1 AP = D. A =
Solution
A=
0_0_-2
1_2_1
1_0_3
now
det of
A-kI=k^3-5k^2+8k-4
hencek^3-5k^2+8k-4 is characterstic polynomial
now for
A-kI=0
we have
k^3-5k^2+8k-4 =0
=(k-1)(k-2)^2=0
hence
k=1,2,2

2. for
k=1
we have
A-kI=
-1_0_-2
1_1_1
1_0_2
hence for
(A-I)X=0
we have
-x1-2x3=0
x1+x2+x3=0
x1+2x3=0
hence
x1=-2x3
x2=-x1-x3=2x3-x3=x3
hence
for
x1=1 we have
x2=x3=-1/2
so our eigen vector is
(1,-1/2,-1/2)t
t here means transpose

3. for
k=2
we get
A-kI=
-2_0_-2
1_0_1
1_0_1
hence
x1+x3=0
for
x1=1,x2=0
we have
x3=-1
now
for
x1=0,x2=1 we have
x3=0
so our eigen vector
are
(1,0,-1)t,(0,1,0)t
now as we have 3 independent eigen vectors
we can diagonaize this matrix
now write these eigen vectors coulmn wise

4. that would be P
hence p=
1_1_0
-0.5_0_1
0.5_-1_0