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Rank-one Modification of Symmetric Eigenproblem
Zack 11/8/2013
Eigenproblem
- π΅ = π πΈπ π
- Q is an orthonormal matrix, π π π = π ππ = π½
- D is diagonal matrix, π1 β€ π2 β€ β― β€ ππ
Rank-one Modification of Symmetric Eigenproblem Zack 11/8/2013 - - PowerPoint PPT Presentation
Rank-one Modification of Symmetric Eigenproblem Zack 11/8/2013 - - PowerPoint PPT Presentation
Rank-one Modification of Symmetric Eigenproblem Zack 11/8/2013 Eigenproblem = Q is an orthonormal matrix, = = D is diagonal matrix, 1 2
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Initial modification of eigenproblem
- π΅ = π΅ + ππ£π£π
- π΅ = π πΈπ π is known.
- Compute eigenvalue decomposition of
π΅: π΅ = π πΈ π
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- π΅ = π΅ + ππ£π£π = π πΈ + ππ¨π¨π π π
- Define C = πΈ + ππ¨π¨π, problem comes to eigenvalue decomposition
- f matrix C
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- The eigenvalues of the matrix C satisfy the equation:
det πΈ + ππ¨π¨π β Ξ»π½ = 0
- det πΈ + ππ¨π¨π β Ξ»π½ = det πΈ β Ξ»π½ det π½ + π πΈ β Ξ»π½ β1π¨π¨π
= (
π=1 π
ππ β π )(1 + Ο
π=1 π
ππ
2
ππ β π )
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1 + Ο
π=1 π
ππ
2
ππ β π = 0
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1 + Ο
π=1 π
ππ
2
ππ β π = 0
- Define π = ππ + ππ’, ππ = (ππ β ππ)/π
π₯π π’ = 1 +
π=1 π
ππ
2
ππ β π’ = 0 π’ β (0, ππ+1)
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Newtonβs method would be an obvious choice for finding the solution. However, Newtonβs method is based on a local linear approximation to the function. Since our functions are rational functions, it seems more natural to develop a method based on a local approximation via simple rational function f(t).
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Step1: choose an approximate function, for example: lim
π’β0 π π’ = ββ
lim
π’βπ π π’ = +β
ο π π’ = π +
π π’ + π πβπ’
Step2: iteration
- 1. Choose π’0 β (0, Ξ΄)
- 2. solve π0,π0, π0 , with π π’0 = π₯π π’0 , πβ² π’0 = π₯πβ² π’0 ,
πβ²β² π’0 = π₯πβ²β² π’0 β¦
- 3. solve π π’1 = 0
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- π π’ = π +
π π’ + π πβπ’ ο Graggβs method
Converges from any point in 0, π with a cubic order of convergence.
- π π’ = π β
ππ
2
π’ + π πβπ’ ο fixed weight 1 method (FW1)
π π’ = π +
π π’ + ππ+1
2
πβπ’ ο fixed weight 2 method (FW2)
Converges from any point in 0, π with quadratic order of convergence
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- Divide π₯π π’ into 2 part:
π₯π π’ = 1 +
π=1 π
ππ
2
ππ β π’ +
π=π+1 π
ππ
2
ππ β π’ = 1 + π π’ + π(π’)
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π π’ =
π=1 π
ππ
2
π
π β π’
π π’ =
π=π+1 π
ππ
2
π
π β π’
π1 < β― < ππ = 0 < ππ+1 < β― < ππ
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- choose an approximate function for π π’ ,π π’
π π’ ο π + ππ’β1 π π’ ο π + π(π β π’)β1 This method is named βthe middle wayβ. For this method, convergence cannot be guaranteed unless the starting point lies close enough to the
- root. In case of convergence, the order is quadratic.
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βapproaching from the leftβ (BNS1) π π’ ο π(π β π’)β1 π π’ ο π + π(π β π’)β1 Converge from any point in (0, π’β] and the order of convergence is quadratic. βapproaching from the rightβ (BNS2) π π’ ο π + ππ’β1 π π’ ο π(π β π’)β1 Converge from any point in [π’β,π) and the order of convergence is quadratic.
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Starting points
1 +
π=1 π
ππ
2
π
π β π’ = 0
1 β ππ
2
π’ + ππ+1
2
ππ+1 β π’ +
π=1 πβ π,π+1 π
ππ
2
π
π β π’ = 0
1 β ππ
2
π’0 + ππ+1
2
ππ+1 β π’0 +
π=1 πβ π,π+1 π
ππ
2
π
π β ππ+1
β 0
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Calculating the eigenvector
- Assume ππ is πth eigenvector of matrix A
π΅ ππ β ππ ππ = 0
- if π¨ = π ππ and π¦π = π π
ππ πΈ β πππ½ + ππ¨π¨π π¦π = 0
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Calculating the eigenvector
- Theorem. If A is invertable and π β 0, the following statements are
equivalent: π΅ + ππ£π€π π¦ = π And π¦ = π΅β1π β ππ΅β1π£, ππ = π€ππ΅β1π, where π = (
1 π + π€ππ΅β1π£)
- Therefore, π¦π = βπ(πΈ β
πππ½)β1π¨
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