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  • Chapter 7 Bounds and estimates for elements of f (A)

    Gérard MEURANT

    January-February, 2012

  • 1 Introduction

    2 Analytic bounds

    3 Examples

    4 Numerical experiments

    5 Estimates of traces and determinants

  • Introduction

    We would like to compute bounds or estimates of

    [f (A)]i ,j = (e i )T f (A)e j

    I Analytic bounds

    I Numerical experiments for f (λ) = 1/λ, exp(λ), √

    λ

    I Estimates of trace(A−1) and det(A)

  • Analytic bounds Performing analytically one or two Lanczos iterations, we are able to obtain bounds for the entries of A−1

    Theorem Let A be a symmetric positive definite matrix. Let

    s2i = ∑ j 6=i

    a2ji , i = 1, . . . , n

    Using the Gauss, Gauss–Radau and Gauss–Lobatto rules∑ k 6=i

    ∑ l 6=i ak,iak,lal ,i

    ai ,i ∑

    k 6=i ∑

    l 6=i ak,iak,lal ,i − (∑

    k 6=i a 2 k,i

    )2 ≤ (A−1)i ,i ai ,i − b +

    s2i b

    a2i ,i − ai ,ib + s2i ≤ (A−1)i ,i ≤

    ai ,i − a + s2i a

    a2i ,i − ai ,ia + s2i

    (A−1)i ,i ≤ a + b − aii

    ab

  • Compute analytically α1, η1, α2, the inverse of

    J2 =

    ( α1 η1 η1 α2

    ) is

    J−12 = 1

    α1α2 − η21

    ( α2 −η1 −η1 α1

    ) For Gauss–Radau we have to modify the (2, 2) element of J2

  • Using the nonsymmetric Lanczos algorithm

    Theorem Let A be a symmetric positive definite matrix and

    ti = ∑ k 6=i

    ak,i (ak,i + ak,j)− ai ,j(ai ,j + ai ,i )

    For (A−1)i ,j + (A −1)i ,i we have the two following estimates

    ai ,i + ai ,j − a + tia (ai ,i + ai ,j)2 − a(ai ,i + ai ,j) + ti

    , ai ,i + ai ,j − b + tib

    (ai ,i + ai ,j)2 − b(ai ,i + ai ,j) + ti

    If ti ≥ 0, the first expression with a gives an upper bound and the second one with b a lower bound

  • Other functions

    We have to compute f (J) for

    J =

    ( α η η ξ

    )

    Proposition

    Let δ = (α− ξ)2 + 4η2

    γ = exp

    ( 1

    2 (α + ξ −

    √ δ)

    ) , ω = exp

    ( 1

    2 (α + ξ +

    √ δ)

    ) The (1, 1) element of the exponential of J is

    1

    2

    [ γ + ω +

    ω − γ√ δ

    (α− ξ) ]

  • Theorem Let

    λ+ = 1

    2 (α + ξ +

    √ δ), λ− =

    1

    2 (α + ξ −

    √ δ)

    The (1, 1) element of f (J) is

    1

    2 √

    δ

    [ (α− ξ)(f (λ+)− f (λ−)) +

    √ δ(f (λ+) + f (λ−))

    ]

    We can obtain analytic bounds for the (i , i) element of f (A) for any function for which we can compute f (λ+) and f (λ−)

  • Examples

    Example F1 This is an example of dimension 10

    A = 1

    11

    

    10 9 8 7 6 5 4 3 2 1 9 18 16 14 12 10 8 6 4 2 8 16 24 21 18 15 12 9 6 3 7 14 21 28 24 20 16 12 8 4 6 12 18 24 30 25 20 15 10 5 5 10 15 20 25 30 24 18 12 6 4 8 12 16 20 24 28 21 14 7 3 6 9 12 15 18 21 24 16 8 2 4 6 8 10 12 14 16 18 9 1 2 3 4 5 6 7 8 9 10

    

  • The inverse of A is a tridiagonal matrix

    A−1 =

     2 −1 −1 2 −1

    . . . . . .

    . . .

    −1 2 −1 −1 2

    

  • Example F3 This is an example proposed by Z. Strakoš. Let Λ be a diagonal matrix

    λi = λ1 +

    ( i − 1 n − 1

    ) (λn − λ1)ρn−i , i = 1, . . . , n

    Let Q be the orthogonal matrix of the eigenvectors of the tridiagonal matrix (−1, 2, −1). Then the matrix is

    A = QTΛQ

    We will use λ1 = 0.1, λn = 100 and ρ = 0.9

  • Example F4 The matrix is arising from the 5–point finite difference approximation of the Poisson equation in a unit square with an m ×m mesh This gives a linear system Ax = c of order m2

    A =

     T −I −I T −I

    . . . . . .

    . . .

    −I T −I −I T

    

  • Each block is of order m and

    T =

     4 −1 −1 4 −1

    . . . . . .

    . . .

    −1 4 −1 −1 4

    

  • Diagonal elements

    Example F1, GL, A−15,5 = 2

    rule Nit=1 2 3 4 5 6 7

    G 0.3667 1.3896 1.7875 1.9404 1.9929 1.9993 2

    G–R bL 1.3430 1.7627 1.9376 1.9926 1.9993 2.0000 2

    G–R bU 3.0330 2.2931 2.1264 2.0171 2.0020 2.0001 2

    G–L 3.1341 2.3211 2.1356 2.0178 2.0021 2.0001 2

  • Example F3, GL, n = 100, A−150,50 = 4.2717

    Nit G G–R bL G–R bU G–L

    10 2.7850 3.0008 5.1427 5.1664

    20 4.0464 4.0505 4.4262 4.4643

    30 4.2545 4.2553 4.2883 4.2897

    40 4.2704 4.2704 4.2728 4.2733

    50 4.2716 4.2716 4.2718 4.2718

    60 4.2717 4.2717 4.2717 4.2717

  • Example F4, GL, n = 900, A−1150,150 = 0.3602

    Nit G G–R bL G–R bU G–L

    10 0.3578 0.3581 0.3777 0.3822

    20 0.3599 0.3599 0.3608 0.3609

    30 0.3601 0.3601 0.3602 0.3602

    40 0.3602 0.3602 0.3602 0.3602

  • Non–diagonal elements with the nonsymmetric Lanczos algorithm

    Example F1, GNS, A−12,2 + A −1 2,1 = 1

    rule Nit=1 2 4 5 6 7

    G 0.4074 0.6494 0.9512 0.9998 1.0004 1

    G–R bL 0.6181 0.8268 0.9998 1.0004 1.0001 1

    G–R bU 2.6483 1.4324 1.0035 1.0012 0.9994 1

    G–L 3.2207 1.4932 1.0036 1.0012 0.9993 0.9994

  • Example F3, GNS, n = 100, A−150,50 + A −1 50,49 = 1.4394

    Nit G G–R bL G–R bU G–L

    10 0.8795 0.9429 2.2057 2.2327

    20 1.3344 1.3362 1.5535 1.5839

    30 1.4301 1.4308 1.4510 1.4516

    40 1.4386 1.4387 1.4404 1.4404

    50 1.4394 1.4394 1.4395 1.4395

    60 1.4394 1.4394 1.4394 1.4394

  • Example F4, GNS, n = 900, A−1150,150 + A −1 150,50 = 0.3665

    Nit G G–R bL G–R bU G–L

    10 0.3611 0.3615 0.3917 0.3979

    20 0.3656 0.3657 0.3678 0.3680

    30 0.3663 0.3664 0.3666 0.3666

    40 0.3665 0.3665 0.3665 0.3665

  • Non–diagonal elements with the block Lanczos algorithm

    Let (J−1k )1,1 the 2× 2 (1, 1) block of the inverse of Jk with

    Jk =

     Ω1 Γ

    T 1

    Γ1 Ω2 Γ T 2

    . . . . . .

    . . .

    Γk−2 Ωk−1 Γ T k−1

    Γk−1 Ωk

     ∆1 = Ω1, ∆i = Ωi − Γi−1Ω−1i−1Γ

    T i−1, i = 2, . . . , k

  • Ck = ∆ −1 1 Γ

    T 1 ∆

    −1 2 Γ

    T 2 · · ·∆−1k−1Γ

    T k−1∆

    −1 k Γ

    T k

    (J−1k+1)1,1 = (J −1 k )1,1 + Ck∆

    −1 k+1C

    T k

    Going from step k to step k + 1 we compute Ck+1 incrementally Note that we can reuse Ck∆

    −1 k+1 to compute Ck+1

  • Example F3, GB, n = 100, A−12,1 = −3.2002

    Nit G G–R bL G–R bU G–L

    2 -3.0808 -3.0948 -3.9996 -4.1691

    3 -3.1274 -3.1431 -3.5655 -3.6910

    4 -3.2204 -3.2187 -3.2637 -3.5216

    5 -3.2015 -3.2001 -3.1974 -3.2473

    6 -3.1969 -3.1966 -3.1964 -3.1969

    7 -3.1970 -3.1972 -3.1995 -3.1994

    8 -3.1993 -3.1995 -3.2008 -3.1999

    9 -3.2001 -3.2001 -3.2005 -3.2008

    10 -3.2002 -3.2002 -3.2002 -3.2004

    We see that we obtain good approximations but not always bounds As a bonus we also obtain estimates of A−11,1 and A

    −1 2,2

  • Example F4, GB, n = 900, A−1400,100 = 0.0597

    Nit G G–R bL G–R bU G–L

    10 0.0172 0.0207 0.0632 0.0588

    20 0.0527 0.0532 0.0616 0.0621

    30 0.0590 0.0591 0.0597 0.0597

    40 0.0597 0.0597 0.0597 0.0597

    Note that for this problem the Gauss rule gives a lower bound, Gauss–Radau a lower and an upper bound

  • Dependence on the eigenvalue estimates

    We take Example F4 with m = 6 We look at the number of Lanczos iterations needed to obtain an upper bound for the element (18, 18) with four exact digits

    Example F4, GL, n = 36

    a=10−4 10−2 0.1 0.3 0.4 1 6

    15 13 11 11 8 8 9

    With the exact eigenvalue a = 0.3961 we need 9 Lanczos iterations Note that it works even when a > λmin

  • Bounds for the elements of the exponential

    Example F3, GL, n = 100, exp(A)50,50 = 5.3217 10 41. Results

    ×10−41

    Nit G G–R bL G–R bL G–L

    2 0.0000 0.0000 7.0288 8.8014

    3 0.0075 0.2008 5.6649 6.0776

    4 1.0322 2.5894 5.3731 5.4565

    5 3.9335 4.7779 5.3270 5.3385

    6 5.1340 5.2680 5.3235 5.3232

    7 5.3070 5.3178 5.3218 5.3219

    8 5.3203 5.3209 5.3218 5.3218

    9 5.3212 5.3213 5.3217 5.3217

    10 5.3215 5.3217 5.3217 5.3217

    11 5.3217 5.3217 5.3217 5.3217

    Convergence is faster than with A−1

  • Example F4, GNS, n = 900, exp(A)50,50 + exp(A)50,49 = 83.8391

    rule Nit=2 3 4 5 6 7

    G 63.4045 81.4124 83.6607 83.8318 83.8389 83.8391

    G–R bL 108.0918 86.3239 83.8796 83.8420 83.8392 83.8391

    G–R bU 76.1266 83.7668 83.7781 83.8383 83.8391 83.8391

    G–L 163.8043 90.9304 84.1878 83.8530 83.8395 83.8391

    Convergence is quite fast

  • Bounds for