4.4. DEFLATION

That is, we have to compute the spectral decomposition of a matrix that is a diagonal plus a rank-one update. Let

(4.7) D + ρvvT = QΛQT

be this spectral decomposition. Then, the spectral decomposition of the tridiagonal T is

Forming the product (Q1 Q2)Q will turn out to be the most expensive step of the algorithm. It costs n3 + O(n2) floating point operations

4.4Deflation

There are certain solutions of (4.7) that can be given immediately, by just looking carefully at the equation.

If there are zero entries in v then we have

(4.9) vi = 0 vT ei = 0 = (D + ρvvT )ei = diei.

Thus, if an entry of v vanishes we can read the eigenvalue from the diagonal of D at once and the corresponding eigenvector is a coordinate vector.

If identical entries occur in the diagonal of D, say di = dj , with i < j, then we can find a plane rotation G(i, j, φ) (see (3.4)) such that it introduces a zero into the j-th position

So, if there are multiple eigenvalues in D we can reduce all but one of them by introducing zeros in v and then proceed as previously in (4.9).

When working with floating point numbers we deflate if

(4.10) |vi| < CεkT k or |di − dj | < CεkT k, (kT k = kD + ρvvT k) where C is a small constant. Deflation changes the eigenvalue problem for D + ρvvT into

where D1 has no multiple diagonal entries and v1 has no zero entries. So, we have to compute the spectral decomposition of the matrix in (4.11) which is similar to a slight perturbation of the original matrix. G is the product of Givens rotations.

4.4.1Numerical examples

Then a little MATLAB experiment shows that

with

Here it is not possible to deflate.

Let us now look at an example with more symmetry,

In this example we have three double eigenvalues. Because the corresponding components of v (vi and vi+1) are equal we define

Therefore, (in this example) e4, e5, and e6 are eigenvectors of

D + GT v(GT v)T = D + GT vvT G

corresponding to the eigenvalues d4, d5, and d6, respectively. The eigenvectors of T corresponding to these three eigenvalues are the last three columns of

4.5The eigenvalue problem for D + ρvvT

We know that ρ 6= 0. Otherwise there is nothing to be done. Furthermore, after deflation, we know that all elements of v are nonzero and that the diagonal elements of D are all

distinct, in fact,

|di − dj | > CεkT k.

We order the diagonal elements of D such that

d1 < d2 < · · · < dn.

Notice that this procedure permutes the elements of v as well. Let (λ, x) be an eigenpair of

λ cannot be equal to one of the di. If λ = dk then the k-th element on the left of (4.13) vanishes. But then either vk = 0 or vT x = 0. The first cannot be true for our assumption about v. If on the other hand vT x = 0 then (D − dkI)x = 0. Thus x = ek and vT ek = vk = 0, which cannot be true. Therefore D − λI is nonsingular and

This equation shows that x is proportional to (λI − D)−1v. If we require kxk = 1 then

Since vT x 6= 0, λ is an eigenvalue of (4.12) if and only if

10

5

1 0

This equation is called secular equation. The secular equation has poles at the eigenvalues of D and zeros at the eigenvalues of D + ρvvT . Notice that

Thus, the derivative of f is positive if ρ > 0 wherever it has a finite value. If ρ < 0 the derivative of f is negative (almost) everywhere. A typical graph of f with ρ > 0 is depicted in Fig. 4.1. (If ρ is negative the image can be flipped left to right.) The secular equation implies the interlacing property of the eigenvalues of D and of D + ρvvT ,

So, we have to compute one eigenvalue in each of the intervals (di, di+1), 1 ≤ i < n, and a further eigenvalue in (dn, ∞) or (−∞, d1). The corresponding eigenvector is then given by (4.15). Evidently, these tasks are easy to parallelize.

Equations (4.17) and (4.15) can also been obtained from the relations

These are simply block LDLT factorizations of the first matrix. The first is the well-known one where the factorization is started with the (1, 1) block. The second is a ‘backward’ factorization that is started with the (2, 2) block. Because the determinants of the tridiagonal matrices are all unity, we have

Denoting the eigenvalues of D + ρvvT again by λ1 < λ2 < · · · < λn this implies