total least squares - hong kong baptist university · c.c. paige and z. strakosˇ , unifying least...

Total least squares

Gerard MEURANT

October, 2008

1 Introduction to total least squares

2 Approximation of the TLS secular equation

3 Numerical experiments

Introduction to total least squares

In least squares (LS) we have only a perturbation of the right handside whereas Total Least Squares (TLS) considers perturbations ofthe vector of observations c and of the m × n data matrix A

minimize ‖(E r

)‖F ,

E , r

subject to the constraint (A + E )x = c + r

This is finding the smallest perturbations E and r such that c + ris in the range of A + E

see Golub and Van Loan; Van Huffel and Vandewalle; Paige andStrakos

Theorem (Golub and Van Loan)

Let C =(A c

)and UTCV = Σ be its SVD

Assume that the singular values of C are such that

σ1 ≥ · · · ≥ σk > σk+1 = · · ·σn+1

Then the solution of the TLS problem is given by

min ‖(E r

)‖F = σn+1

andxTLS = − y

α

where the vector(y α

)Tof norm 1 with α 6= 0 is in the

subspace Sk spanned by the right singular vectors{vk+1, . . . , vn+1} of V . If there is no such vector with α 6= 0, theTLS problem has no solution

The right singular vectors v i are the eigenvectors of CTC and Sk isthe invariant subspace associated to the smallest eigenvalue σ2

n+1

The TLS solution xTLS solves the eigenvalue problem

CTC

(x−1

)= σ2

n+1

(x−1

)

TheoremIf σA

n > σn+1, then xTLS exists and is the unique solution of theTLS problem

xTLS = (ATA− σ2n+1I )

−1AT c

Moreover, σn+1 satisfies the following secular equation

σ2n+1

[1 +

n∑i=1

d2i

(σAi )2 − σ2

n+1

]= ρ2

LS

where the vector d = UT c and ρ2LS = ‖(c − AxLS)‖2

The secular equation can also be written as

σ2n+1 = cT c − cTA(ATA− σ2

n+1I )−1AT c

This is obtained by writing(ATA AccTA cT c

) (x−1

)= (σn+1)

2

(x−1

)and eliminating x

For data least squares (DLS) when only the matrix is perturbed,the secular equation is

cT c − cTA(ATA− σ2I )−1AT c = 0

This can also be written as

cT (AAT − σ2I )−1c = 0

−5 0 5 10 15 20 25−10

−8

−6

−4

−2

0

2

4

6

8

10TLS secular function as a function of σ2

Example of TLS secular function as a function of σ2

Approximation of the TLS secular equation

We approximate the quadratic form in the TLS secular equation byusing one of the Golub-Kahan bidiagonalization algorithms with cas a starting vectorIt reduces A to lower bidiagonal form and generates a matrix

Ck =

γ1

δ1. . .. . .

. . .

. . . γk

δk

a k + 1 by k matrix such that CT

k Ck = Jk the tridiagonal matrixgenerated by the Lanczos algorithm for the matrix ATA

At iteration k we approximate the TLS secular equation by

cT c − ‖c‖2(e1)TCk(CTk Ck − σ2I )−1CT

k e1 = σ2

This corresponds to the Gauss quadrature rule

We use the SVD of Ck = UkSkV Tk . Let σ

(k)i be the singular values

of Ck and ξ(k) = UTk e1

(ξ(k)k+1)

2

σ2−

k∑i=1

(ξ(k)i )2

(σ(k)i )2 − σ2

=1

‖c‖2

We need to compute the smallest zero. Secular equation solversuse rational interpolationWhen an approximate solution σ2

tls has been computed, we solve

xtls = (ATA− σ2tls I )

−1AT c

The Gauss–Radau rule

We implement the Gauss–Radau rule by using the otherGolub-Kahan bidiagonalization algorithm with AT c as a startingvectorIt reduces A to upper bidiagonal form. If

Bk =

γ1 δ1

. . .. . .

γk−1 δk−1

γk

the matrix Bk is the Cholesky factor of the Lanczos matrix Jk

To obtain the Gauss–Radau rule we must modify Bk to have aprescribed eigenvalue zLet ω be the solution of

(BTk Bk − zI )ω = (γk−1δk−1)

2ek

Let

ωk = (z + ωk)− (γk−1δk−1)2

γ2k−1

= (z + ωk)− δ2k−1

The modified matrix giving the Gauss–Radau rule is

Bk =

γ1 δ1

. . .. . .

γk−1 δk−1

γk

where γk =

√ωk

Using Bk we solve the secular equation

‖c‖2 − ‖AT c‖2(e1)T (BTk Bk − σ2I )−1e1 = σ2

with the SVD of Bk

Numerical experiments

As = UsΣsVTs , Us = I − 2

usuTs

‖us‖2, Vs = I − 2

vsvTs

‖vs‖2

where us and vs are random vectorsΣs is an m × n diagonal matrix with elements [1, · · · ,

√n]

Let xs be a vector whose ith component is 1/i and cs = Asxs

A = As + ξ randn(m, n)

The right hand side is

c = cs + ξ randn(m, 1)

A small example

Example TLS1, m = 100, n = 50, BNS1 ε = 10−6

ξ L it. s it. sol. exact sol.

0.3 10−2 30 57 0.01703479103104873 0.01703478979190218

0.3 10−1 26 49 0.169448388286749 0.1694483528865543

0.3 28 73 1.464892131470029 1.464891451263777

30 33 64 88.21012648624229 88.21012652906667

A larger example

I We are not able to store A which is a dense matrix in Matlab

I We use the vectors us and vs to do matrix multiplies with As

or ATs

I We perturb the singular values in the same way as the righthand side

Example TLS3, m = 10000, n = 5000, noise=0.3, BNS1

ε L it. s it. min it. max it. av. it. solution

10−6 250 273 1 2 1.09 1.418582932414374

10−10 328 660 1 3 2.01 1.418576233569240

It works fine but it is too expensive

The Gauss–Radau rule

Example TLS1, m = 100, n = 50, Gauss–Radau, noise=0.3,ε = 10−6

Met. L it.√

z s it. min it. max it. solution

Newt 28 σmin 130 2 14 1.464891376927382

Newt 28 σmax 79 2 4 1.464892626809155

Rat 28 σmin 98 2 5 1.464891376927382

Rat 28 σmax 74 2 3 1.464892626809155

Example TLS3, m = 10000, n = 5000, Gauss–Radau, noise=0.3,ε = 10−6

Met. L it.√

z s it. min it. max it. solution

Newt 250 σmin 2572 3 31 1.418576232676234

Newt 250 σmax 1926 3 26 1.418583305908228

Rat 250 σmin 837 2 5 1.418576232676233

Rat 250 σmax 653 2 4 1.418583305908227

Optimization of the algorithm

To reduce the cost

I We monitor the convergence of the smallest singular value ofA

I For this we solve a secular equation at every Lanczos iteration

I We use a third order rational approximation and tridiagonalsolves

I The Gauss and Gauss–Radau estimates are only computed atthe end

Example TLS3, m = 10000, n = 5000, noise=0.3, ε = 10−6

Met. L it. trid√

z s it. solution

- 250 551

Gauss - 2 1.418582932414440

G–R σmin(Bk) 2 1.418582932414443

G–R σmax(Bk) 3 1.418583305908306

Example TLS4, m = 100000, n = 50000, noise=0.3, ε = 10−6

Met. L it. trid√

z s it. solution

- 755 1775

Gauss - 1 0.8721122166701496

G–R σmin(Bk) 2 0.8721122166735605

G–R σmax(Bk) 3 0.8721124331415380

For example TLS3 with m = 10000, n = 5000 and ε = 10−6

I The computing time when solving for Gauss and Gauss–Radauat each iteration was 117 seconds

I With the last algorithm it is 12 seconds

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