Lecture 15

Question 1

Inverse power iteration

Answer 1

A-1 x = 1/λ x
|1/λn| > ... > |1/λ1|
xk+1 = A-1 xk
xk → (1/λn)^k αn un, converges to un
λn = (unT A un) / (unT un)

Question 2

Inverse power iteration steps

Answer 2

1. factorize PLU=A
Ax{k+1}=xk so LUx{k+1}=P.T xk
2. Solve Ly=P.T xk
3. Solve U.x{k+1}=y
4. Normalize x{k+1}=x{k+1}/||x{k+1}||

Question 3

Cost of computing largest eigenvalue? Smallest?

Answer 3

O(n2) vs O(n3) – indeed factorization n3

Question 4

eigenvalue of (A+B/2)^-1?

Answer 4

2/(2λ1+λ2)

Question 5

Inverse power method shifted matrix

Answer 5

(A-σI)^-1 x = ƛx
xk+1=(A-σI)^-1 xk
converges to largest ƛ, smallest (λ-σ), that is λ closest to σ (not |λ|!!!)

Question 6

Inverse power method shifted matrix steps

Answer 6

1. Factorize B=(A-σI)=PLU
2. Solve Ly=P.T xk
3. Solve Uxk+1 = y
   O(n^3))

Question 7

Convergence inverse power method shifted matrix

Answer 7

||ek+1|| = |(λclosest - σ)/(λ2ndclosest - σ)| ||ek||

Question 8

Rayleigh quotient iteration

Answer 8

inverse power method shifted matrix xk+1=(A-σkI)^-1 xk

(A-σkI) with σk updated at each iteration s.t. σk=xkT.A.xk/xkT.xk, convergence close to cubic, but cost per it n^3