A−λI=⎣⎡1−λ−36−3−5−λ−6334−λ⎦⎤The characteristic equation
det(A−λI)=0
(1−λ)(−5−λ)(4−λ)−54+54
−18(−5−λ)+18(1−λ)−9(4−λ)=0
−λ3+30λ+52=0
−λ2(λ+2)+2λ(λ+2)+26(λ+2)=0
λ1=−2
λ2−2λ−26=0
(λ−1)2=27
λ2=1−33,λ3=1+33
λ1=−2,λ2=1−33,λ3=1+33 These are eigenvalues.
λ=−2
⎣⎡1−λ−36−3−5−λ−6334−λ⎦⎤=⎣⎡3−36−3−3−6336⎦⎤ R1=R1/3
⎣⎡1−36−1−3−6136⎦⎤ R2=R2+3R1
⎣⎡106−1−6−6166⎦⎤ R3=R3−6R1
⎣⎡100−1−60160⎦⎤ R2=R2/(−6)
⎣⎡100−1101−10⎦⎤ R1=R1+R2
⎣⎡1000100−10⎦⎤
If we take v3=t, then v1=0,v2=t,v3=t.
Thus the eigenvector is
v=⎣⎡011⎦⎤
λ=1−33
⎣⎡1−λ−36−3−5−λ−6334−λ⎦⎤
=⎣⎡33−36−3−6+33−6333+33⎦⎤
R1=3R1/9
⎣⎡1−36−3/3−6+33−63/333+33⎦⎤ R2=R2+3R1
⎣⎡106−3/3−6+23−63/33+33+33⎦⎤ R3=R3−6R1
⎣⎡100−3/3−6+23−6+233/33+33+3⎦⎤ R2=R2/(−6+23)
⎣⎡100−3/31−6+233/3−1−3/23+3⎦⎤ R1=R1+R2/3
⎣⎡10001−6+23−1/2−1−3/23+3⎦⎤R3=R3+(6−23)R2
⎣⎡100010−1/2−1−3/20⎦⎤
If we take u3=t, then u1=t/2,u2=(2+3)t/2,u3=t.
Thus the eigenvector is
u=⎣⎡1/2(2+3)/21⎦⎤
λ=1+33
⎣⎡1−λ−36−3−5−λ−6334−λ⎦⎤
=⎣⎡−33−36−3−6−33−6333−33⎦⎤R1=−3R1/9
⎣⎡1−363/3−6−33−6−3/333−33⎦⎤ R2=R2+3R1
⎣⎡1063/3−6−23−6−3/33−33−33⎦⎤ R3=R3−6R1
⎣⎡1003/3−6−23−6−23−3/33−33−3⎦⎤ R2=R2/(−6−23)
⎣⎡1003/31−6−23−3/3−1+3/23−3⎦⎤ R1=R1−R2/3
⎣⎡10001−6−23−1/2−1+3/23−3⎦⎤R3=R3+(6+23)R2
⎣⎡100010−1/2−1+3/20⎦⎤
If we take w3=t, then w1=t/2,w2=(2−3)t/2,w3=t.
Thus the eigenvector is
w=⎣⎡1/2(2−3)/21⎦⎤
Form the matrix P, whose column i is eigenvector no. i:
P=⎣⎡0111/21+3/211/21−3/21⎦⎤ Form the matrix D whose element at row i, column i is eigenvalue no. i:
D=⎣⎡−20001−330001+33⎦⎤The matrices P and D are such that the initial matrix
A=⎣⎡1−36−3−5−6334⎦⎤=PDP−1
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