-
Eigenvectors for the matrix :
- , eigenvalue
- , eigenvalue
Details
-
From the definition of the eigenvector corresponding to the eigenvalue we haveThen:Equation has a nonzero solution if and only if
Details
-
For every we find its own vectors:
-
So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:
( 1 0.618 0 1 0.618 0 ) × ( − 1 ) ~ R 2 − 1 ⋅ R 1 → R 2 Subtract 1 1
from row2
:a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = 0.618 − 1 1 ⋅ ( 0.618 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 ( 1 0.618 0 0 0 0 ) { x 1 + 0.618 ⋅ x 2 = 0 (1) -
x 1 = − 0.618 ⋅ x 2
Answer:-
x 1 = − 0.618 ⋅ x 2 -
x 2 = x 2
General Solution :X = ( − 0.618 ⋅ x 2 x 2 ) The solution set:{ x 2 ⋅ ( − 0.618 1 ) } Letx 2 = 1 v 1 = ( − 0.618 1 ) -
-
λ 2 = 3.62 A − λ 2 I = ( − 0.618 1 1 − 1.62 ) A v = λ v ( A − λ I ) ⋅ v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:( − 0.618 1 0 1 − 1.62 0 ) × ( − 1.62 ) ~ R 1 ∕ ( − 0.618 ) → R 1 Divide row 1
by− 0.618 a 1 , 1 = − 0.618 ⋅ ( 1 − 0.618 ) = 1 a 1 , 2 = 1 ⋅ ( 1 − 0.618 ) = − 1.62 a 1 , 3 = 0 ⋅ ( 1 − 0.618 ) = 0 ( 1 − 1.62 0 1 − 1.62 0 ) × ( − 1 ) ~ R 2 − 1 ⋅ R 1 → R 2 Subtract 1 1
from row2
:a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = − 1.62 − 1 1 ⋅ ( − 1.62 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 ( 1 − 1.62 0 0 0 0 ) { x 1 − 1.62 ⋅ x 2 = 0 (1) -
x 1 = 1.62 ⋅ x 2
Answer:-
x 1 = 1.62 ⋅ x 2 -
x 2 = x 2
General Solution :X = ( 1.62 ⋅ x 2 x 2 ) The solution set:{ x 2 ⋅ ( 1.62 1 ) } Letx 2 = 1 v 2 = ( 1.62 1 ) -
-
-
A = ( 3 1 1 2 ) Eigenvectors for the matrix
A -
v = ( − 5 + 1 2 1 ) λ 1 = − 5 + 5 2 -
v = ( 5 + 1 2 1 ) λ 2 = 5 + 5 2
Details
-
From the definition of the eigenvector
v λ A v = λ v Then:A v − λ v = ( A − λ I ) ⋅ v = 0 Equation has a nonzero solution if and only ifdet ( A − λ I ) = 0 det ( A − λ I ) = | 3 − λ 1 1 2 − λ | = λ 2 − 5 λ + 5 = (?) ( λ + 5 − 5 2 ) ⋅ ( λ − 5 + 5 2 ) = 0 Details
-
λ 1 = − 5 + 5 2 -
λ 2 = 5 + 5 2
-
-
For every
λ -
λ 1 = − 5 + 5 2 A − λ 1 I = ( 5 + 1 2 1 1 5 − 1 2 ) A v = λ v ( A − λ I ) ⋅ v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:( 5 + 1 2 1 0 1 5 − 1 2 0 ) × ( 5 − 1 2 ) ~ R 1 ∕ ( 5 + 1 2 ) → R 1 Divide row 1
by5 + 1 2 a 1 , 1 = 5 + 1 2 ⋅ ( 1 5 + 1 2 ) = 1 a 1 , 2 = 1 ⋅ ( 1 5 + 1 2 ) = 5 − 1 2 a 1 , 3 = 0 ⋅ ( 1 5 + 1 2 ) = 0 ( 1 5 − 1 2 0 1 5 − 1 2 0 ) × ( − 1 ) ~ R 2 − 1 ⋅ R 1 → R 2 Subtract 1 1
from row2
:a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = 5 − 1 2 − 1 1 ⋅ ( 5 − 1 2 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 ( 1 5 − 1 2 0 0 0 0 ) { x 1 + 5 − 1 2 ⋅ x 2 = 0 (1) -
x 1 = − 5 + 1 2 ⋅ x 2
Answer:-
x 1 = − 5 + 1 2 ⋅ x 2 -
x 2 = x 2
General Solution :X = ( − 5 + 1 2 ⋅ x 2 x 2 ) The solution set:{ x 2 ⋅ ( − 5 + 1 2 1 ) } Letx 2 = 1 v 1 = ( − 5 + 1 2 1 ) -
-
λ 2 = 5 + 5 2 A − λ 2 I = ( − 5 + 1 2 1 1 − 5 − 1 2 ) A v = λ v ( A − λ I ) ⋅ v = 0 So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:( − 5 + 1 2 1 0 1 − 5 − 1 2 0 ) × ( − 5 − 1 2 ) ~ R 1 ∕ ( − 5 + 1 2 ) → R 1 Divide row 1
by− 5 + 1 2 a 1 , 1 = − 5 + 1 2 ⋅ ( 1 − 5 + 1 2 ) = 1 a 1 , 2 = 1 ⋅ ( 1 − 5 + 1 2 ) = − 5 − 1 2 a 1 , 3 = 0 ⋅ ( 1 − 5 + 1 2 ) = 0 ( 1 − 5 − 1 2 0 1 − 5 − 1 2 0 ) × ( − 1 ) ~ R 2 − 1 ⋅ R 1 → R 2 Subtract 1 1
from row2
:a 2 , 1 = 1 − 1 1 ⋅ 1 = 0 a 2 , 2 = − 5 − 1 2 − 1 1 ⋅ ( − 5 − 1 2 ) = 0 a 2 , 3 = 0 − 1 1 ⋅ 0 = 0 ( 1 − 5 − 1 2 0 0 0 0 ) { x 1 − 5 + 1 2 ⋅ x 2 = 0 (1) -
x 1 = 5 + 1 2 ⋅ x 2
Answer:-
x 1 = 5 + 1 2 ⋅ x 2 -
x 2 = x 2
General Solution :X = ( 5 + 1 2 ⋅ x 2 x 2 ) The solution set:{ x 2 ⋅ ( 5 + 1 2 1 ) } Letx 2 = 1 v 2 = ( 5 + 1 2 1 ) -
-
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