Rank-one Modification of Symmetric Eigenproblem Zack 11/8/2013
Eigenproblem β’ π΅ = π πΈπ π β’ Q is an orthonormal matrix, π π π = π π π = π½ β’ D is diagonal matrix, π 1 β€ π 2 β€ β― β€ π π
Initial modification of eigenproblem π΅ = π΅ + ππ£π£ π β’ β’ π΅ = π πΈπ π is known. β’ Compute eigenvalue decomposition of π΅ : π΅ = π πΈ π
π΅ = π΅ + ππ£π£ π = π πΈ + ππ¨π¨ π π π β’ β’ Define C = πΈ + ππ¨π¨ π , problem comes to eigenvalue decomposition of matrix C
β’ The eigenvalues of the matrix C satisfy the equation: det πΈ + ππ¨π¨ π β Ξ»π½ = 0 β’ det πΈ + ππ¨π¨ π β Ξ»π½ = det πΈ β Ξ»π½ det π½ + π πΈ β Ξ»π½ β1 π¨π¨ π π π 2 π π = ( π π β π )(1 + Ο π π β π ) π=1 π=1
π 2 π π 1 + Ο π π β π = 0 π=1
π 2 π π 1 + Ο π π β π = 0 π=1 β’ Define π = π π + ππ’ , π π = (π π β π π )/π π 2 π π π₯ π π’ = 1 + π π β π’ = 0 π’ β (0, π π+1 ) π=1
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).
Step1: choose an approximate function , for example: π’β0 π π’ = ββ lim π π ο π π’ = π + π’ + lim π’βπ π π’ = +β πβπ’ Step2: iteration 1. Choose π’ 0 β (0, Ξ΄) 2. solve π 0 ,π 0 , π 0 , with π π’ 0 = π₯ π π’ 0 , πβ² π’ 0 = π₯ π β² π’ 0 , πβ²β² π’ 0 = π₯ π β²β² π’ 0 β¦ 3. solve π π’ 1 = 0
π π β’ π π’ = π + πβπ’ ο Graggβs method π’ + Converges from any point in 0, π with a cubic order of convergence. 2 π π π β’ π π’ = π β πβπ’ ο fixed weight 1 method (FW1) π’ + 2 π π π+1 πβπ’ ο fixed weight 2 method (FW2) π π’ = π + π’ + Converges from any point in 0, π with quadratic order of convergence
β’ Divide π₯ π π’ into 2 part: π π 2 2 π π π π π₯ π π’ = 1 + π π β π’ + π π β π’ π=1 π=π+1 = 1 + π π’ + π(π’)
π 2 π π π π’ = π π β π’ π=1 π 2 π π π π’ = π π β π’ π=π+1 π 1 < β― < π π = 0 < π π+1 < β― < π π
β’ 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.
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.
Starting points π 2 π π 1 + π β π’ = 0 π π=1 π 2 2 2 π π 1 β π π π π+1 π’ + π π+1 β π’ + π β π’ = 0 π π=1 πβ π,π+1 π 2 2 2 π π 1 β π π π π+1 + + β 0 π’ 0 π π+1 β π’ 0 π π β π π+1 π=1 πβ π,π+1
Calculating the eigenvector β’ Assume π π is π th eigenvector of matrix A π π β π΅ π π π π = 0 β’ if π¨ = π π π and π¦ π = π π π π π π π½ + ππ¨π¨ π π¦ π = 0 πΈ β
Calculating the eigenvector β’ Theorem. If A is invertable and π β 0 , the following statements are equivalent: π΅ + ππ£π€ π π¦ = π And 1 π¦ = π΅ β1 π β ππ΅ β1 π£ , ππ = π€ π π΅ β1 π , where π = ( π + π€ π π΅ β1 π£) π π π½) β1 π¨ β’ Therefore, π¦ π = βπ(πΈ β
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