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### Transcript of Chapter 7 Bounds and estimates for elements of f(A)...

• 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