5.2 Real structured eigenvalue backward errors with respect to |||·||| w,2 norm
5.2.2 Real T-palindromic polynomials
Now since λ?max is a simple eigenvalue of G +t?0H0 + · · ·+ t?mHm, if αj 6= 0 for any j ∈ {2, . . . , n(m+ 1)},the last inequality must be strict which is impossible. Thusue=α1ev where |α1|= 1. This implies
u=α1 kuk
kvkv =⇒ uj =α1 kuk
kvkvj forj = 0, . . . , m. (5.2.10) Using (5.2.10) in (5.2.4), we get
∆jM v0
w0
+λv1
w1
+· · ·+λmvm
wm
= vj
wj
for j = 0, . . . , m. (5.2.11) However by (5.2.2) we also have
∆jM v0
w0
+λv1
w1
+· · ·+λmvm
wm
= vj
wj
for j = 0, . . . , m. (5.2.12) By the uniqueness of each ∆j, j = 0, . . . , m, we have
∆j = ∆j, j = 0, . . . , m.
Hence ∆∈(HermR(n))m+1 and equality holds in (5.2.3).
Remark 5.2.3. The assumption that λ?maxis a simple eigenvalue of G+t?0H0+· · ·+t?mHm even when m = 1 in Theorem 5.2.2 is indeed necessary because the optimal Hermitian perturbations in Example 2.5.2 which attain the value of ηw,2Herm(P, λ) are not real. Note that in this case λ?max is not a simple eigenvalue of G+t?0H0+· · ·+t?mHm.
Remark 5.2.4. We note that Theorem 5.2.2 also holds for the case that some of the entries in weight vector w= (w0, . . . , wm) are zero. This is due to Theorem 2.4.1.
The main result of this section gives the structured backward error ηpalw,2T,R(P, λ) when λ ∈ R\ {0,1,−1} and P(z) is a real T-palindromic polynomial of any degree. It may be noted that despite the equality (5.2.13), this result is not a corollary of Theorem 3.1.6.
However the proof is based on similar arguments that make use of the real version of a combination of Theorem 1.2.18 and Theorem 1.2.21 withC replaced byR as given below.
We recall that a Hermitian matrix is indefinite if it has at least one positive and at least negative eigenvalues.
Theorem 5.2.5. Let G, H0, . . . , Hp ∈ Rn×n be Hermitian matrices. Assume that any nonzero linear combination α0H0 +· · ·+αpHp, (α0, . . . , αp) ∈ Rp+1 \ {0} is indefinite.
Then the following statements hold:
(1) The function L:Rp+1 →R, (t0, . . . , tp)7→λmax(G+t0H0+· · ·+tpHp) is convex and has a global minimum
λ?max= min
t0,...,tp∈RL(t0, . . . , tp).
(2) If p= 0 or the minimum λ?max of L is attained at (ˆt0, . . . ,ˆtp)∈Rp+1 and is a simple eigenvalue ofH? :=G+ˆt0H0+· · ·+ˆtpHp, then there exists an eigenvectoru∈Rn\{0} of H? associated with λ?max satisfying
uTHju= 0 for j = 0, . . . , p. (5.2.14) (3) Under the assumptions of (2) we have
sup
uTGu uTu
u∈Rn\ {0}, uTHju= 0, j = 0, . . . , p
=λ?max. (5.2.15) In particular, the supremum of the left hand side of (5.2.15) is a maximum and attained for the eigenvector u from (5.2.14).
Proof. The proofs of (1) and (3) follow from the proofs of the corresponding parts of Theorem 1.2.18.
Ifp= 0,then the proof of part (2) follows by arguing as in the proof of Theorem 1.2.21 due to the fact that the set
{xTH0xx∈Rn, xTx= 1}
is convex. Ifp >0 andλ?max is a simple eigenvalue ofH? then, part (2) of Theorem 1.2.18 implies that there exists a non zero (possibly complex) eigenvector u corresponding to
λ?max of H? that satisfies (5.2.14). However as H? is real and λ?max is real and simple, the eigenvector u can be chosen to be real. This proves part (2) and completes the proof of the theorem.
It is important to note that the assumption of simplicity ofλ?maxmade in the hypothesis of Theorem 5.2.5 is necessary even when p= 1.This is evident from the following example which is a slight modification of Example 1.2.24.
Example 5.2.6. LetG= diag(α, α, β) where α > β≥0. Also let
H0 =
1 0 0
0 −1 0
0 0 0
and H1 =
0 1 0 1 0 0 0 0 0
.
Then any non zero real linear combination of H0 and H1 is indefinite and for t0, t1 ∈ R, the matrix
H(t0, t1) =G+t0H0+t1H1 =
α+t0 t1 0 t1 α−t0 0
0 0 β
has eigenvalues α±p
t20+t21 and β.Clearly L:R2 →R defined by L(t0, t1) =λmax(H(t0, t1)) = α+
q t20+t21
has its minimum λ?max at (t0, t1) = (0,0), i.e., when H(0,0) =G. But the maximal eigen- valueαofGis a double eigenvalue with corresponding eigenvectorse1 ande2 which are the first two basis vectors of R3. Therefore, the matrix whose columns form an orthonormal basis of the eigenspace ofGcorresponding toαisU = [ e1 e2 ]∈R3×2.There exists a real non zero vectorxin the eigenspace of Gcorresponding toα satisfyingx∗H0x=x∗H1x= 0 if and only if the real joint numerical range of the matrices
U1 :=UTH0U =
"
1 0
0 −1
#
and U2 :=UTH1U =
"
0 1 1 0
#
defined by
{(xTU1x, xTU2x)x∈R2,kxk2 = 1}
contains 0.Clearly this is not true in this case as this set represents the unit circle.
Note that this example does not contradict Theorem 1.2.18 as the eigenvector of G with respect toα that satisfiesu∗H0u=u∗H1u= 0 is the complex vector u= [1 i]T.
The following theorem gives a formula for the structured eigenvalue backward error ηpalw,2T,R(P, λ) when P(z) is a real T-palindromic polynomial and λ ∈R\ {0,±1}.
Theorem 5.2.7. Let P(z) = Pm
j=0zjAj be a real T-palindromic polynomial. Choose λ ∈ R\ {0,±1}. Suppose that det(P(λ)) 6= 0 so that M = (P(λ))−1 exists and set k =bm−21c. Then
λ?max:= min
t0,...,tk∈Rλmax(G+t0S0+· · ·+tkSk)
is attained for some (ˆt0, . . . ,ˆtk)∈Rk+1 where G is defined by (3.1.15) andSj, j = 0, . . . , k are defined by (3.1.17), respectively. Whenever m ≤ 2 or λ?max is a simple eigenvalue of G+ ˆt0S0+· · ·+ ˆtkSk, then
ηw,2palT,R(P, λ) = 1 pλ?max =
t0,...,tmink∈Rλmax(G+t0S0+· · ·+tkSk) −1/2
.
Proof. Consider the Hermitian matrices Sej := Cej +CejT, j = 0, . . . , k, where Cej are as defined in (3.1.7). Observe that k = 0 whenever m = 1 or m = 2. Therefore, the proof follows from Theorem 5.2.5 if it is established that each nontrivial linear combination of S0, . . . , Sk given by (3.1.17), or equivalently, of Se0, . . . ,Sek is indefinite. Suppose there exists h
α0, . . . , αk
iT
∈ Rk+1 such that S := Pk
j=0αjSej is semidefinite. Then recalling that Λm = [1, λ, . . . , λm]∈C1×(m+1), we have
S =
Xk j=0
αj
ΛTm eTj+1−eTm+1−j
⊗MT +
ej+1−em+1−j Λm
⊗M
= (ΛTmαT)⊗MT + (αΛm)⊗M
where the vector α is given by α := [α0, . . . , αk, −αk, . . . ,−α0]T when m is odd and by α:= [α0, . . . , αk,0,−αk, . . . ,−α0]T whenm is even.
Setting
Q:=
1 −λ 0 . . . 0 0 1 −λ . .. ...
... . .. ... ... 0 ... . .. ... −λ 0 . . . 0 1
∈C(m+1)×(m+1) (5.2.16)
and a= [a0, . . . , am]T :=QTα, since ΛmQ=eT1,we have
(Q⊗In)TS(Q⊗In) = (QTΛTmαTQ)⊗MT + (QTαΛmQm)⊗M
= (e1aT)⊗MT + (aeT1)⊗M
=
a0(M +MT) a1MT · · · amMT
a1M 0 · · · 0
... ... ...
amM 0 · · · 0
.
IfSis semidefinite, then a1 =· · ·=am = 0 which implies thatQTα=a=a0e1.Therefore, a0 =α0 and
−λαj−1+αj = 0, j = 1, . . . , k. (5.2.17) Also,
(λ+ 1)αk = 0 when m is odd and (5.2.18)
λαk = 0 when m is even. (5.2.19)
The identities (5.2.17) imply that
αk=λkα0. (5.2.20)
When m is odd, we have αk = 0 from (5.2.18) since λ 6= −1. Similarly when m is even, (5.2.19) gives αk = 0 as λ6= 0. In either case, (5.2.20) implies that a0 =α0 = 0 asλ 6= 0 and this completes the proof.
Remark 5.2.8. An optimal real T-palindromic perturbation to P(z) corresponding to ηpalw,2T,R(P, λ) in Theorem 5.2.7 can be constructed by following the procedure mentioned in Remark 3.1.9.
An analogue of Theorem 5.2.7 that obtains ηpalw,2b T,R(P, λ) for a real T-palindromic poly- nomial P(z) with a restricted perturbation set palT,R(I) defined by
palT,R(I) :=
(∆0, . . . ,∆m)∈palT,R
∆j = 0 for j /∈I
may also be derived.
Example 5.2.9. L(z) =A+zAT is a real T-palindromic pencil of size 3 with eigenvalues
−1 and −0.5954±0.8034i all on the unit circle. For λ = −1.6656 and w = (1,1), the
eigenvalue backward error is 0.6563 with respect to real T-palindromic perturbations (ob- tained in Theorem 5.2.7) and 0.5614 with respect to complex T-palindromic perturbations (obtained in Theorem 3.1.10). With respect to arbitrary perturbations, the eigenvalue backward error is 0.3177.Figure 5.2.1 illustrates the effect of real T-palindromic, complex T-palindromic and arbitrary perturbations on the eigenvalues of L(z) so that they move to form an eigenvalue at λ for the respective perturbed pencils.
−2 −1 0 1 2
−2
−1 0 1 2
−2 −1 0 1 2
−2
−1 0 1 2
Figure 5.2.1: Eigenvalue perturbation curves for the real T-palindromic pencil L(z) of Ex- ample 2.5.3 with respect to real and complex T-palindromic perturbations (left) and arbitrary perturbations (right).
The plot on the left of Figure 5.2.1 shows the effect of perturbationsL(z)+t∆L(z) on the eigenvalues ofL(z) (in thick curves) ast moves from 0 to 1,∆L(z) being the minimal real T-palindromic perturbation toL(z) corresponding toηpalw,2T,R(L, λ) that induces eigenvalues at (λ,1/λ). In this case eigenvalue curves starting from the eigenvalues −0.5954±0.8034i (each marked by a star surrounded by a circle) on the unit circle coalesce on the unit circle and split out with one of the branches moving over ∞ to form the eigenvalue λ (marked by a star surrounded by a diamond) and the other moving out to form the eigenvalue 1/λ (inside the unit circle) as t moves from 0 to 1.
The movement of the perturbed eigenvalues under the given real T-palindromic pertur- bations may partly be attributed to two facts. The first is that−1 is always an eigenvalue of a T-palindromic polynomial of odd degree and odd size (since P(−1) is then a skew symmetric matrix of odd size and thus singular) of which the given pencil L(z) is a par- ticular case. The second fact is that eigenvalues of real T-palindromic polynomials occur in quadruples (µ,µ,¯ 1/µ,1/¯µ).This symmetry breaks down only on the unit circle and on the real line where it reduces to the pairing (µ,1/µ). Since the only eigenvalues of L(z) other than −1 are also on the unit circle, in order to maintain the eigenvalue symmetry,
they have to pass through the intersection of the unit circle and the real line to form the eigenvalues atλ and 1/λ.
The plot on the left of Figure 5.2.1 also shows the effect of perturbationsL(z) +t∆L(z)d on the eigenvalues of L(z) (in thin curves) as t moves from 0 to 1, d∆L(z) being the minimal complex T-palindromic perturbation to L(z) corresponding to ηpalw,2T(L, λ) that induces eigenvalues at (λ,1/λ). Also in this case −1 cannot be moved toλ with respect to T-palindromic perturbations. Instead, an eigenvalue curve starting from−0.5954 + 0.8034i moves to λ while another starting from −0.5954−0.8034i moves to 1/λ (inside the unit circle) as t moves from 0 to 1.
Finally, the plot on the right of Figure 5.2.1 shows the effect of perturbations L(z) + t∆L(z) on the eigenvalues ofg L(z) as t moves from 0 to 1, ∆L(z) being an optimalg perturbation corresponding to ηw,2(L, λ) that induces an eigenvalue at λ and is not T- palindromic. In this case, the nearest eigenvalue −1 of L(z) moves to form the eigenvalue λ of L(z) +∆L(z).g
Table 5.2.1 compares the backward errors ηw,2(L, λ), ηpalw,2T,R(L, λ) and ηw,2palT(L, λ) for the T-palindromic pencil L(z) in Example 5.2.9 as λ converges to −1 along the real line.
Observe that while both ηw,2(L, λ), and ηw,2palT(L, λ) decrease, ηw,2palT,R(L, λ) increases as λ approaches−1.This leads to large differences between ηw,2palT,R(L, λ) and the other backward errors at values of λ close to−1.
Table 5.2.1: Values ofηw,2(L, λ), ηpalw,2T(L, λ)andηw,2palT,R(L, λ)for the T-palindromic pencilL(z) in Example 5.2.9 asλ→ −1.
λ ηw,2(P, λ) ηpalw,2T(P, λ) ηw,2palT,R(P, λ)
−1.6656 0.1692 0.3177 0.5614
−1.5500 0.1501 0.3076 0.5623
−1.4500 0.1308 0.2992 0.5631
−1.3500 0.1086 0.2912 0.5638
−1.2500 0.0827 0.2842 0.5644
−1.1500 0.0528 0.2788 0.5649