Linear Algebra, Theory And Applications, 2012a

396 NUMERICAL METHODS FOR FINDING EIGENVALUES
wherethisistheQR factorization of SL k . Then
where F k → 0. Let I + F k = Q ′ k R′ k . Then
Q (k) R (k) = (Q k R k + SE k ) D k U
(
= Q k I + Q
∗
k SE k R −1 )
k Rk D k U
= Q k (I + F k ) R k D k U
Q (k) R (k) = Q k Q ′ kR ′ kR k D k U
By Lemma 15.2.3
Q ′ k → I and R k ′ → I. (15.18)
Now let Λ k be a diagonal unitary matrix which has the property that
Λ ∗ kD k U
is an upper triangular matrix which has all the diagonal entries positive. Then
Q (k) R (k) = Q k Q ′ kΛ k (Λ ∗ kR ′ kR k Λ k )Λ ∗ kD k U
That matrix in the middle has all positive diagonal entries because it is itself an upper
triangular matrix, being the product of such, and is similar to the matrix R ′ k R k which is
upper triangular with positive diagonal entries. By Lemma 15.2.3 again, this time using the
uniqueness assertion,
Q (k) = Q k Q ′ kΛ k ,R (k) =(Λ ∗ kR ′ kR k Λ k )Λ ∗ kD k U
Note the term Q k Q ′ k Λ k must be real because the algorithm gives all Q (k) as real matrices.
By (15.18) it follows that for k large enough
Q (k) ≈ Q k Λ k
where ≈ means the two matrices are close. Recall
and so for large k,
A k = Q (k)T AQ (k)
A k ≈ (Q k Λ k ) ∗ A (Q k Λ k )=Λ ∗ kQ ∗ kAQ k Λ k
As noted above, the form of Λ ∗ k Q∗ k AQ kΛ k in terms of which entries are large and small is
not affected by the presence of Λ k and Λ ∗ k
. Thus, in considering what form this is in, it
suffices to consider Q ∗ k AQ k.
This could get pretty complicated but I will consider the case where
if |λ i | = |λ i+1 | , then |λ i+2 | < |λ i+1 | . (15.19)
This is typical of the situation where the eigenvalues are all distinct and the matrix A is real
so the eigenvalues occur as conjugate pairs. Then in this case, L k above is lower triangular
with some nonzero terms on the diagonal right below the main diagonal but zeros everywhere
else. Thus maybe
(L k ) s+1,s ≠0
Recall (15.17) which implies
Q k = SL k R −1
k
(15.20)