Problem 1
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Solution 1
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Problem 2
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Solution 2
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Problem 3
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Problem 3a
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Define the concept of steepest descent d k {\displaystyle d_{k}\!\,} and show how to compute the optimal stepsize α k {\displaystyle \alpha _{k}\!\,}
Descent Direction
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∇ Q ( x k ) ⋅ d k < 0 {\displaystyle \nabla Q(x_{k})\cdot d_{k}<0\!\,}
Optimal step size
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Choose α k {\displaystyle \alpha _{k}\!\,} such that Q ( x k + α k d k ) {\displaystyle Q(x_{k}+\alpha _{k}d_{k})\!\,} is minimized i.e.
α k : min α k Q ( x k + α k d k ) {\displaystyle \alpha _{k}:\min _{\alpha _{k}}Q(x_{k}+\alpha _{k}d_{k})\!\,}
Q ( x k + α k d k ) = ⟨ x k + α k d k , A x k + α k A d k ⟩ − ⟨ x k + α k d k , b ⟩ {\displaystyle Q(x_{k}+\alpha _{k}d_{k})=\langle x_{k}+\alpha _{k}d_{k},Ax_{k}+\alpha _{k}Ad_{k}\rangle -\langle x_{k}+\alpha _{k}d_{k},b\rangle \!\,}
d d α Q ( x k + α k d k ) = ⟨ A d k , x k ⟩ + α k ⟨ A d k , d k ⟩ − ⟨ d k , b ⟩ {\displaystyle {\frac {d}{d\alpha }}Q(x_{k}+\alpha _{k}d_{k})=\langle Ad_{k},x_{k}\rangle +\alpha _{k}\langle Ad_{k},d_{k}\rangle -\langle d_{k},b\rangle \!\,}
Setting the above expression equal to zero gives the optimal α k {\displaystyle \alpha _{k}\!\,} :
α k = ⟨ d k , b − A x k ⟩ ⟨ A d k , d k ⟩ {\displaystyle \alpha _{k}={\frac {\langle d_{k},b-Ax_{k}\rangle }{\langle Ad_{k},d_{k}\rangle }}\!\,}
Note that since A {\displaystyle A\!\,} is symmetric
⟨ d k , A x k ⟩ = ⟨ A d k , x k ⟩ {\displaystyle \langle d_{k},Ax_{k}\rangle =\langle Ad_{k},x_{k}\rangle \!\,}
Problem 3b
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Formulate the steepest descent (or gradient method) method and write a pseudocode which implements it.
Solution 3b
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Note that d k = − ∇ Q ( x k ) = b − A x k = r k {\displaystyle d_{k}=-\nabla Q(x_{k})=b-Ax_{k}=r_{k}\!\,} . Then the minimal α k {\displaystyle \alpha _{k}\!\,} is given by ⟨ r k , r k ⟩ ⟨ r k , r k ⟩ A {\displaystyle {\frac {\langle r_{k},r_{k}\rangle }{\langle r_{k},r_{k}\rangle _{A}}}\!\,}
Given x 0 ∈ R n {\displaystyle x_{0}\in R^{n}\!\,}
For k = 0 , 1 , 2 , … {\displaystyle k=0,1,2,\ldots \!\,}
r k = b − A x k If r k = 0 , then quit d k = r k α k = ⟨ r k , r k ⟩ ⟨ r k , r k ⟩ A x k + 1 = x k + α k d k {\displaystyle {\begin{aligned}r_{k}&=b-Ax_{k}\\&{\mbox{If }}r_{k}=0,{\mbox{ then quit }}\\d_{k}&=r_{k}\\\alpha _{k}&={\frac {\langle r_{k},r_{k}\rangle }{\langle r_{k},r_{k}\rangle _{A}}}\\x_{k+1}&=x_{k}+\alpha _{k}d_{k}\end{aligned}}}
Problem 3c
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Solution 3
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Since B {\displaystyle B\!\,} is symmetric, positive definite, B = E T E {\displaystyle B=E^{T}E\!\,} where E {\displaystyle E\!\,} is upper triangular (Cholesky Factorization).
Then B − 1 = E − 1 E − T {\displaystyle B^{-1}=E^{-1}E^{-T}\!\,}
Hence,
B − 1 A x = B − 1 b E − 1 E − T A x = E − 1 E − T b E − T A x = E − T b E − T A E − 1 ⏟ A ~ E x ⏟ x ~ = E − T b ⏟ b ~ {\displaystyle {\begin{aligned}B^{-1}Ax&=B^{-1}b\\E^{-1}E^{-T}Ax&=E^{-1}E^{-T}b\\E^{-T}Ax&=E^{-T}b\\\underbrace {E^{-T}AE^{-1}} _{\tilde {A}}\underbrace {Ex} _{\tilde {x}}&=\underbrace {E^{-T}b} _{\tilde {b}}\end{aligned}}\!\,}
A ~ {\displaystyle {\tilde {A}}\!\,} is symmetric:
( E − T A E − 1 ) T = E − T A E − 1 {\displaystyle (E^{-T}AE^{-1})^{T}=E^{-T}AE^{-1}\!\,} since A {\displaystyle A\!\,} symmetric
A ~ {\displaystyle {\tilde {A}}\!\,} is positive definite:
x T E − T A E − 1 x = ( E − 1 x ) T A ( E − 1 x ) > 0 {\displaystyle x^{T}E^{-T}AE^{-1}x=(E^{-1}x)^{T}A(E^{-1}x)>0\!\,} since A {\displaystyle A\!\,} positive definite
Pseudocode
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Given x 0 ∈ R n {\displaystyle x_{0}\in R^{n}\!\,}
For k = 0 , 1 , 2 , … {\displaystyle k=0,1,2,\ldots \!\,}
x k ~ = E x k r k = b ~ − A ~ x k ~ If r k = 0 , then quit d k = r k α k = ⟨ r k , r k ⟩ ⟨ r k , r k ⟩ A ~ x k + 1 = x k + α k d k {\displaystyle {\begin{aligned}{\tilde {x_{k}}}&=Ex_{k}\\r_{k}&={\tilde {b}}-{\tilde {A}}{\tilde {x_{k}}}\\&{\mbox{If }}r_{k}=0,{\mbox{ then quit }}\\d_{k}&=r_{k}\\\alpha _{k}&={\frac {\langle r_{k},r_{k}\rangle }{\langle r_{k},r_{k}\rangle _{\tilde {A}}}}\\x_{k+1}&=x_{k}+\alpha _{k}d_{k}\end{aligned}}}