z∈σ(A+4A)andηS(z, A) =k4Ak=η(z, A). Consequently, we have σ²S(A)∩iR=σ²(A)∩iR.
Proof: Consider the caseM A∈HermwhenM =MH.Then forλ∈R, M(A−λI)∈S.Since M(A−λI) is hermitian, we have the spectral decompositionM(A−λI) =Udiag(µ1,· · · , µn)UH, whereU is unitary andµj’s appear in descending order of their magnitudes. Note that|µn|= σmin(M(A−λI)) =σmin(A−λI) =η(λ, A). Now defining4A:=−µnM−1U(:, n)U(:, n)H, we haveλ∈σ(A+4A), 4A∈Sand k4Ak=η(λ, A) =ηS(λ, A).Hence the result follows.
The proof is similar for the case when M A∈skew-HermandMH =−M.
Finally, consider the case whenM A∈skew-HermandMH =M.Then forλ∈iR,the set of purely imaginary numbers, we haveM(A−λI)∈S,that is,M(A−λI) is skew hermitian.
Hence the result follows from spectral decomposition of M(A−λI). The proof is similar for the case when M A∈Herm andMH=−M.¥
Similar results hold for some other important structured matrices such as Toeplitz and Hankel matrices which are not described in the setting of Jordan or Lie algebras.
2.4 Structured backward error for approximate invariant
isometry X∈Rn×k such thatRange(X) =X. Then ηSF(X, A) =√
2k(I−XXT)AXkF, η2S(X, A) =k(I−XXT)AXk2.
Define E := −XXTA−AXXT + 2XXTAXXT. Then E ∈ S,kEk2,F = ηSF(X, A) and (A+E)X ⊆ X.
Proof: By Theorem 2.1 it is easy to verify that for given A ∈ sym and X ∈ Cn×k there always existsE ∈symsuch that
(A+E)X =XD⇒EX =XD−AX (2.6)
for some D=DT of orderk.
Now construct a real isometry Q1 such that Q = [X, Q1] is orthogonal and QT1X = 0.
Define E=Q
"
E11 E12T E12 E22
#
QT.Hence from (2.6) we obtain
"
E11 E12T E12 E22
# "
XT QT1
# X=
"
XT QT1
#
(XD−AX)
which gives E11=D−XTAX, E12=−QT1AX.Therefore we have E=Q
"
D−XTAX (−QT1AX)T
−QT1AX E22
#
QT. (2.7)
This giveskEk2F =kD−XTAXk2F+2kQT1AXk2F+kE22k2F.To minimizekEkFwe fixE22= 0.
Consequently we have kEk2F = kD −XTAXk2F + 2kQT1AXk2F, where D is a complex symmetric matrix of orderk.Setting D:=XTAXwe obtain
ηF(X,S) =√
2kQT1AXkF =√ 2
q
kXTAk2F − kXTAXk2F. given byE=−XXTA−AXXT + 2XXTAXXT.
Next, consider the spectral norm. By (2.7) and (2.5) we have
kEk2≥
°°
°°
°
"
0 (−QT1AX)T
−QT1AX 0
#°°
°°
°2
=kQT1AXk2 (2.8)
for anyD=DT of order k.SettingD=XTAX,by (2.7) we have E=Q
"
0 (−QH1 AX)T
−QH1AX E22
# QH.
Now by DKW Theorem 1.2.5 we construct infinitely many symmetric matricesE22∈C(n−k)×(n−k)
such thatkEk2=kQT1AXk2=k(I−XXT)AXk2.Note that the lower bound is attained by this particular choice ofD.Then by (2.8) we have
η2S(X, A) =k(I−XXT)AXk2.
Moreover
E22=µ(I−KKH)1/2Z(I−KKT)1/2
where K :=−µ−1QT1AX, µ=k(I−XXT)AXk2 and Z =ZT is an arbitrary contraction.
In particular, taking Z = 0 and simplifying the expression E we obtain the same as that of for Frobenius norm.¥
Note thatE ∈symis unique such thatkEkF =ηSF(X, A) and (A+E)X ⊂ X.Also notice that for the spectral norm we have
E=−XXTA−AXXT + 2XXTAXXT +µQ1(I−KKH)1/2Z(I−KKT)1/2QT1, where QT1X = 0, K := −µ−1QT1AX, µ = k(I−XXT)AXk2 and Z = ZT is an arbitrary contraction, such thatkEk2=η2S(X, A) and (A+E)X ⊂ X.
Corollary 2.4.2. LetS=sym.Let A∈SandX be a subspace of Cn.Then any E∈S such that (A+E)X ⊆ X is given by
E=XDXT −XXTA−AXXT +XXTAXXT + (I−XXT)Z(I−XXT) where DT =D∈Ck×k, ZT =Z ∈C(n−k)×(n−k).
Proof: The proof is followed by simplifying (2.7). ¥
Next we considerX is a subspace ofCn and A ∈ skew-sym. To determineηS(X, A), we assume that there is an isometry X∈Rn×k such that Range(X) =X.
Theorem 2.4.3. Let S= skew-sym and A∈ S. Let X be a subspace ofCn such that there exist a real isometry X ∈Cn×k andRange(X) =X.Then
ηSF(X, A) =√
2k(I−XXT)AXkF, η2S(X, A) =k(I−XXT)AXk2.
DefineE:=−XXTA−AXXT+ 2XXTAXXT.ThenkEkF,2=ηS2,F(X, A)and(A+E)X ⊆ X.
Proof: The proof is similar to Theorem 2.4.1. ¥
Note thatE∈skew-symis a unique matrix such thatkEkF =ηSF(X, A) and (A+E)X ⊆ X. Also notice that
E=XXTA−AXXT+ 2XXTAXXT+µQ1(I−KKH)1/2Z(I−KKT)1/2QT1, where QT1X = 0, K :=−µ−1QT1AX, µ=kAXk2 andZ =−ZT is an arbitrary contraction, is such that E∈skew-sym,kEk2=η2S(X, A) and (A+E)X ⊆ X.
Corollary 2.4.4. Let S=skew-sym.Let A∈S andX be a subspace ofCn.Then any E∈S such that (A+E)X ⊆ X is given by
E=XDXT +XXTA+XXTAXXT −AXXT + (I−XXT)Z(I−XXT) where DT =−D∈Ck×k, ZT =−Z ∈C(n−k)×(n−k).
Now we considerS=Herm.
Theorem 2.4.5. Let S=Herm and A∈S. Let X be a subspace ofA and X ∈Cn×k be an isometry such that Range(X) =X. Then
ηSF(X, A) =√
2k(I−XXH)AXkF, η2S(X, A) =k(I−XXH)AXk2.
Define E = −XXHA−AXXH + 2XXHAXXH. Then (A+E)X ⊆ X and kEk2,F = ηS2,F(X, A)for both the Frobenius and the spectral norms..
Proof: Let X :=
h
x1 . . . xk
i
∈ Cn×k be an isometry such that Range(X) = X. By Theorem 2.1 it follows that there exist E∈Ssuch that
(A+E)X =XD⇒EX =XD−AX (2.9)
for some D = DH of order k. Now construct a unitary matrix Q = [X, Q1] such that QH1 X= 0.DefineE=Q
"
E11 E12H E12 E22
#
QH. Hence by (2.9) we obtain
"
E11 E12H E12 E22
# "
XH QT1
# X=
"
XH QH1
#
(XD−AX)
which gives E11=D−XHAX, E12=−QH1AX.Therefore we have E=Q
"
D−XHAX (−QH1AX)H
−QH1AX E22
#
QH. (2.10)
This giveskEk2F =kD−XHAXk2F+2kQH1AXk2F+kE22k2F.To minimizekEkF we fixE22= 0.
Consequently we havekEk2F =kD−XHAXk2F+ 2kQH1 AXk2F,whereD is any Hermitian matrix of orderk. SettingD:=XHAXwe obtain
ηFS(X, A) =√
2kQH1AXkF =√ 2
q
kXHAk2F− kXHAXk2F.
Simplifying the expressions ofE we obtainE=−XXHA−AXXH+ 2XXHAXXH. Next, consider the spectral norm. By (2.7) and (2.5) we have the following lower bound
kEk2≥
°°
°°
°
"
0 (−QH1AX)H
−QH1AX 0
#°°
°°
°2
=kQ1QH1AXk2=k(I−XXH)AXk2 (2.11)
for anyD=DH of orderk. SetD=XHAX.Then by (2.10) we have
E=Q
"
0 (−QH1 AX)H
−QH1AX E22
# QH.
Using dilation Theorem 1.2.6 for Hermitian matrices we can construct infinitely many Hermi- tian matrices E22 ∈C(n−k)×(n−k) such thatkEk2=kQH1 AXk2 =k(I−XXH)AXk2. Note that the lower bound is attained by this particular choice of D. Consequently, by (2.11) we
obtain
ηS2(X, A) =k(I−XXH)AXk2. Moreover
E22=µ(I−KKH)1/2Z(I−KKH)1/2
where K :=−µ−1QH1AX, µ=k(I−XXH)AXk2 and Z =ZH is an arbitrary contraction.
In particular, taking Z = 0 and simplifying the expression E we obtain the same as that of for Frobenius norm.¥
It follows from the proof thatEis unique such thatkE F =ηFS(X, A) and (A+E)X ⊆ X. For the spectral norm,E is given by
E=−XXHA−AXXH+ 2XXHAXXH+µQ1(I−KKH)1/2Z(I−KKH)1/2QH1 , where QH1 X = 0, K :=−µ−1QH1 AX, µ= k(I−XXH)AXk2 and Z =ZH is an arbitrary contraction, such thatkEk2=η2S(X, A) and (A+E)X ⊆ X.
Corollary 2.4.6. Let S=HermandA∈S. Let X be a subspace ofCn andX ∈Cn×k is an isometry such that Range(X) =X. Then anyE∈S such that (A+E)X ⊆ X is given by
E=XDXH−XXHA−AXXH+XXHAXXH+ (I−XXH)Z(I−XXH) where DH =D∈Ck×k, ZH=Z∈C(n−k)×(n−k).
Proof: The proof is followed by simplifying (2.10). ¥ Now we considerS=skew-Herm.
Theorem 2.4.7. Let S=skew-Herm. X be a subspace of Cn andX ∈Cn×k is an isometry such that Range(X) =X.Then
ηSF(X, A) =√
2k(I−XXH)AXkF, η2S(X, A) =k(I−XXH)AXk2.
DefineE=XXHA−AXXH.ThenE∈Ssuch that(A+E)X ⊆ X andkEk2,F =η2,FS (X, A) for both the Frobenius and the spectral norms.
Proof: The proof is followed by Theorem 2.4.5¥
Let S = skew-Herm. Define E = XXHA−AXXH. Then E ∈ S is unique such that (A+E)X ⊆ X andkEkF =ηFS(X, A). Next define
E=XXHA−AXXH+µQ1(I−KKH)1/2Z(I−KKH)1/2QH1 ,
where QH1X = 0, K :=−µ−1QH1AX, µ=k(I−XXH)AXk2 andZ =−ZH is an arbitrary contraction. then notice that (A+E)X ⊆ X andkEk2=ηS2(X, A).
Corollary 2.4.8. LetS=skew-Herm.LetA∈SandX be a subspace ofCn.Then anyE∈S such that (A+E)X ⊆ X is given by
E=XDXH+XXHA−AXXH−XXHAXXH+ (I−XXH)Z(I−XXH)
where DH =−D∈Ck×k, ZH =−Z ∈C(n−k)×(n−k).
We mention that the results obtained above can easily be extended to the case when S⊂ {J,L}.