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عدد التعديلات للمستخدم (user_editcount) | 0 |
اسم حساب المستخدم (user_name) | 'إسلام بهاء' |
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عنوان الصفحة (بدون نطاق) (page_title) | 'خوارزمية القيمه الخاصه' |
عنوان الصفحة الكامل (page_prefixedtitle) | 'خوارزمية القيمه الخاصه' |
آخر عشرة مساهمين في الصفحة (page_recent_contributors) | '' |
فعل (action) | 'edit' |
ملخص التعديل/السبب (summary) | '' |
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نص الويكي القديم للصفحة، قبل التعديل (old_wikitext) | '' |
نص الويكي الجديد للصفحة، بعد التعديل (new_wikitext) | 'Eigenvalues and Eigenvectors
We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!
Definitions
Let A be an nn matrix. The number is an eigenvalue of A if there exists a non-zero vector v such that
Av=v
In this case, vector v is called an eigenvector of A corresponding to .
Computing Eigenvalues and Eigenvectors
We can rewrite the condition Av=v as
(A−I)v=0
where I is the nn identity matrix. Now, in order for a non-zero vector v to satisfy this equation, A−I must not be invertible.
Otherwise, if A−I has an inverse,
(A−I)−1(A−I)v v = = (A−I)−10 0
But we are looking for a non-zero vector v .
That is, the determinant of A−I must equal 0. We call p()=det(A−I) the characteristic polynomial of A. The eigenvalues of A are simply the roots of the characteristic polynomial of A.
Example
Let A=2 −1 −4 −1 . Then
p() = = = = det2− −1 −4 −1− (2−)(−1−)−(−4)(−1) 2−−6 (−3)(+2)
Thus, 1=3 and 2=−2 are the eigenvalues of A.
To find eigenvectors v= v1 v2 vn corresponding to an eigenvalue , we simply solve the system of linear equations given by
(A−I)v=0
Example
The matrix A=2 −1 −4 −1 of the previous example has eigenvalues 1=3 and 2=−2. Let's find the eigenvectors corresponding to 1=3. Let v=v2v1 . Then (A−3I)v=0 gives us
2−3 −1 −4 −1−3 v1 v2 =0 0
from which we obtain the duplicate equations
−v1−4v2 −v1−4v2 = = 0 0
If we let v2=t, then v1=−4t. All eigenvectors corresponding to 1=3 are multiples of 1−4 and thus the eigenspace corresponding to 1=3 is given by the span of 1−4 . That is, 1−4 is a basis of the eigenspace corresponding to 1=3.
Repeating this process with 2=−2, we find that
4v1−4V2 −v1+v2 = = 0 0
If we let v2=t then v1=t as well. Thus, an eigenvector corresponding to 2=−2 is 11 and the eigenspace corresponding to 2=−2 is given by the span of 11 . 11 is a basis for the eigenspace corresponding to 2=−2.
In the following example, we see a two-dimensional eigenspace.
Example
Let A= 5 4 −4 8 1 −4 16 8 −11 . Then p()=det 5− 4 −4 8 1− −4 16 8 −11− =(−1)(+3)2 after some algebra! Thus, 1=1 and 2=−3 are the eigenvalues of A. Eigenvectors v= v1 v2 v3 corresponding to 1=1 must satisfy
4v1 4v1 −4v1 + − 8v2 4v2 + + − 16v3 8v3 12v3 = = = 0 0 0
Letting v3=t, we find from the second equation that v1=−2t, and then v2=−t. All eigenvectors corresponding to 1=1 are multiples of −2 −1 1 , and so the eigenspace corresponding to 1=1 is given by the span of −2 −1 1 . −2 −1 1 is a basis for the eigenspace corresponding to 1=1.
Eigenvectors corresponding to 2=−3 must satisfy
8v1 4v1 −4v1 + + − 8v2 4v2 4v2 + + − 16v3 8v3 8v3 = = = 0 0 0
The equations here are just multiples of each other! If we let v3=t and v2=s, then v1=−s−2t. Eigenvectors corresponding to 2=−3 have the form
−1 1 0 s+ −2 0 1 t
Thus, the eigenspace corresponding to 2=−3 is two-dimensional and is spanned by −1 1 0 and −2 0 1 . −1 1 0 −2 0 1 is a basis for the eigenspace corresponding to 2=−3.
Notes
Eigenvalues and eigenvectors can be complex-valued as well as real-valued.
The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue.
The techniques used here are practical for 22 and 33 matrices. Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods.
Key Concepts
Let A be an nn matrix. The eigenvalues of A are the roots of the characteristic polynomial
p()=det(A−I)
For each eigenvalue , we find eigenvectors v= v1 v2 vn by solving the linear system
(A−I)v=0
The set of all vectors v satisfying Av=v is called the eigenspace of A corresponding to .' |
فرق موحد للتغييرات المصنوعة بواسطة التعديل (edit_diff) | '@@ -1 +1,73 @@
+Eigenvalues and Eigenvectors
+We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!
+Definitions
+
+
+Let A be an nn matrix. The number is an eigenvalue of A if there exists a non-zero vector v such that
+Av=v
+In this case, vector v is called an eigenvector of A corresponding to .
+
+
+
+
+Computing Eigenvalues and Eigenvectors
+
+We can rewrite the condition Av=v as
+(A−I)v=0
+where I is the nn identity matrix. Now, in order for a non-zero vector v to satisfy this equation, A−I must not be invertible.
+Otherwise, if A−I has an inverse,
+(A−I)−1(A−I)v v = = (A−I)−10 0
+But we are looking for a non-zero vector v .
+That is, the determinant of A−I must equal 0. We call p()=det(A−I) the characteristic polynomial of A. The eigenvalues of A are simply the roots of the characteristic polynomial of A.
+
+Example
+
+Let A=2 −1 −4 −1 . Then
+p() = = = = det2− −1 −4 −1− (2−)(−1−)−(−4)(−1) 2−−6 (−3)(+2)
+Thus, 1=3 and 2=−2 are the eigenvalues of A.
+
+To find eigenvectors v= v1 v2 vn corresponding to an eigenvalue , we simply solve the system of linear equations given by
+(A−I)v=0
+
+Example
+
+The matrix A=2 −1 −4 −1 of the previous example has eigenvalues 1=3 and 2=−2. Let's find the eigenvectors corresponding to 1=3. Let v=v2v1 . Then (A−3I)v=0 gives us
+2−3 −1 −4 −1−3 v1 v2 =0 0
+from which we obtain the duplicate equations
+−v1−4v2 −v1−4v2 = = 0 0
+If we let v2=t, then v1=−4t. All eigenvectors corresponding to 1=3 are multiples of 1−4 and thus the eigenspace corresponding to 1=3 is given by the span of 1−4 . That is, 1−4 is a basis of the eigenspace corresponding to 1=3.
+
+Repeating this process with 2=−2, we find that
+4v1−4V2 −v1+v2 = = 0 0
+If we let v2=t then v1=t as well. Thus, an eigenvector corresponding to 2=−2 is 11 and the eigenspace corresponding to 2=−2 is given by the span of 11 . 11 is a basis for the eigenspace corresponding to 2=−2.
+
+In the following example, we see a two-dimensional eigenspace.
+
+Example
+
+Let A= 5 4 −4 8 1 −4 16 8 −11 . Then p()=det 5− 4 −4 8 1− −4 16 8 −11− =(−1)(+3)2 after some algebra! Thus, 1=1 and 2=−3 are the eigenvalues of A. Eigenvectors v= v1 v2 v3 corresponding to 1=1 must satisfy
+
+4v1 4v1 −4v1 + − 8v2 4v2 + + − 16v3 8v3 12v3 = = = 0 0 0
+Letting v3=t, we find from the second equation that v1=−2t, and then v2=−t. All eigenvectors corresponding to 1=1 are multiples of −2 −1 1 , and so the eigenspace corresponding to 1=1 is given by the span of −2 −1 1 . −2 −1 1 is a basis for the eigenspace corresponding to 1=1.
+
+Eigenvectors corresponding to 2=−3 must satisfy
+
+8v1 4v1 −4v1 + + − 8v2 4v2 4v2 + + − 16v3 8v3 8v3 = = = 0 0 0
+The equations here are just multiples of each other! If we let v3=t and v2=s, then v1=−s−2t. Eigenvectors corresponding to 2=−3 have the form
+ −1 1 0 s+ −2 0 1 t
+Thus, the eigenspace corresponding to 2=−3 is two-dimensional and is spanned by −1 1 0 and −2 0 1 . −1 1 0 −2 0 1 is a basis for the eigenspace corresponding to 2=−3.
+
+Notes
+Eigenvalues and eigenvectors can be complex-valued as well as real-valued.
+
+The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue.
+
+The techniques used here are practical for 22 and 33 matrices. Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods.
+Key Concepts
+
+Let A be an nn matrix. The eigenvalues of A are the roots of the characteristic polynomial
+p()=det(A−I)
+For each eigenvalue , we find eigenvectors v= v1 v2 vn by solving the linear system
+(A−I)v=0
+The set of all vectors v satisfying Av=v is called the eigenspace of A corresponding to .
' |
حجم الصفحة الجديد (new_size) | 4158 |
حجم الصفحة القديم (old_size) | 0 |
الحجم المتغير في التعديل (edit_delta) | 4158 |
السطور المضافة في التعديل (added_lines) | [
0 => 'Eigenvalues and Eigenvectors',
1 => 'We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!',
2 => 'Definitions',
3 => false,
4 => false,
5 => 'Let A be an nn matrix. The number is an eigenvalue of A if there exists a non-zero vector v such that',
6 => 'Av=v ',
7 => 'In this case, vector v is called an eigenvector of A corresponding to . ',
8 => false,
9 => false,
10 => false,
11 => false,
12 => 'Computing Eigenvalues and Eigenvectors',
13 => false,
14 => 'We can rewrite the condition Av=v as',
15 => '(A−I)v=0 ',
16 => 'where I is the nn identity matrix. Now, in order for a non-zero vector v to satisfy this equation, A−I must not be invertible. ',
17 => 'Otherwise, if A−I has an inverse,',
18 => '(A−I)−1(A−I)v v = = (A−I)−10 0 ',
19 => 'But we are looking for a non-zero vector v .',
20 => 'That is, the determinant of A−I must equal 0. We call p()=det(A−I) the characteristic polynomial of A. The eigenvalues of A are simply the roots of the characteristic polynomial of A.',
21 => false,
22 => 'Example',
23 => false,
24 => 'Let A=2 −1 −4 −1 . Then',
25 => 'p() = = = = det2− −1 −4 −1− (2−)(−1−)−(−4)(−1) 2−−6 (−3)(+2) ',
26 => 'Thus, 1=3 and 2=−2 are the eigenvalues of A.',
27 => false,
28 => 'To find eigenvectors v= v1 v2 vn corresponding to an eigenvalue , we simply solve the system of linear equations given by',
29 => '(A−I)v=0 ',
30 => false,
31 => 'Example',
32 => false,
33 => 'The matrix A=2 −1 −4 −1 of the previous example has eigenvalues 1=3 and 2=−2. Let's find the eigenvectors corresponding to 1=3. Let v=v2v1 . Then (A−3I)v=0 gives us',
34 => '2−3 −1 −4 −1−3 v1 v2 =0 0 ',
35 => 'from which we obtain the duplicate equations',
36 => '−v1−4v2 −v1−4v2 = = 0 0 ',
37 => 'If we let v2=t, then v1=−4t. All eigenvectors corresponding to 1=3 are multiples of 1−4 and thus the eigenspace corresponding to 1=3 is given by the span of 1−4 . That is, 1−4 is a basis of the eigenspace corresponding to 1=3.',
38 => false,
39 => 'Repeating this process with 2=−2, we find that',
40 => '4v1−4V2 −v1+v2 = = 0 0 ',
41 => 'If we let v2=t then v1=t as well. Thus, an eigenvector corresponding to 2=−2 is 11 and the eigenspace corresponding to 2=−2 is given by the span of 11 . 11 is a basis for the eigenspace corresponding to 2=−2.',
42 => false,
43 => 'In the following example, we see a two-dimensional eigenspace.',
44 => false,
45 => 'Example',
46 => false,
47 => 'Let A= 5 4 −4 8 1 −4 16 8 −11 . Then p()=det 5− 4 −4 8 1− −4 16 8 −11− =(−1)(+3)2 after some algebra! Thus, 1=1 and 2=−3 are the eigenvalues of A. Eigenvectors v= v1 v2 v3 corresponding to 1=1 must satisfy',
48 => false,
49 => '4v1 4v1 −4v1 + − 8v2 4v2 + + − 16v3 8v3 12v3 = = = 0 0 0 ',
50 => 'Letting v3=t, we find from the second equation that v1=−2t, and then v2=−t. All eigenvectors corresponding to 1=1 are multiples of −2 −1 1 , and so the eigenspace corresponding to 1=1 is given by the span of −2 −1 1 . −2 −1 1 is a basis for the eigenspace corresponding to 1=1.',
51 => false,
52 => 'Eigenvectors corresponding to 2=−3 must satisfy',
53 => false,
54 => '8v1 4v1 −4v1 + + − 8v2 4v2 4v2 + + − 16v3 8v3 8v3 = = = 0 0 0 ',
55 => 'The equations here are just multiples of each other! If we let v3=t and v2=s, then v1=−s−2t. Eigenvectors corresponding to 2=−3 have the form',
56 => ' −1 1 0 s+ −2 0 1 t',
57 => 'Thus, the eigenspace corresponding to 2=−3 is two-dimensional and is spanned by −1 1 0 and −2 0 1 . −1 1 0 −2 0 1 is a basis for the eigenspace corresponding to 2=−3.',
58 => false,
59 => 'Notes',
60 => 'Eigenvalues and eigenvectors can be complex-valued as well as real-valued. ',
61 => false,
62 => 'The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. ',
63 => false,
64 => 'The techniques used here are practical for 22 and 33 matrices. Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods.',
65 => 'Key Concepts',
66 => false,
67 => 'Let A be an nn matrix. The eigenvalues of A are the roots of the characteristic polynomial',
68 => 'p()=det(A−I)',
69 => 'For each eigenvalue , we find eigenvectors v= v1 v2 vn by solving the linear system',
70 => '(A−I)v=0 ',
71 => 'The set of all vectors v satisfying Av=v is called the eigenspace of A corresponding to .'
] |
السطور المزالة في التعديل (removed_lines) | [] |
ما إذا كان التعديل قد تم عمله من خلال عقدة خروج تور (tor_exit_node) | 0 |
طابع زمن التغيير ليونكس (timestamp) | 1398203231 |