On Beruling-Gelfand's Spectral Radius Theorem

Home , Spectral radius, Spectral theorem

Mathematics & Statistics Auburn University, Alabama, USA

Introduction Basics Jan 24, 2007 Proof Extensions On Beruling-Gelfand’s spectral radius theorem Home Page Title Page

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Speaker: Tin-Yau Tam J I

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2007 Spring Graduate Seminar Go Back

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[email protected] Quit Outline Introduction Basics Proof Extensions Introduction Home Page Basics Title Page JJ II Proof J I

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Quit 1. Introduction

Introduction Beruling-Gelfand’s spectral radius theorem Basics Proof Extensions lim kXmk1/m = r(X), m→∞ Home Page where Title Page

X is a square complex matrix, JJ II r(X) is the spectral radius of X, and J I Page 3 of 31 kXk is the spectral norm of X. Go Back We will prove the result in an elementary way and Full Screen Close discuss its extensions. Quit 2. Basics

n C = the space of complex n-tuples. n×n = the space of n × n complex matrices. C Introduction Basics A number λ ∈ C is said to be an eigenvalue of X ∈ Proof n Extensions Cn×n if there is a nonzero vector v ∈ C such that Xv = λv. Home Page Title Page Spectral radius: r(X) is the maximum eigenvalue JJ II J I modulus. Page 4 of 31

∗ 1/2 Go Back 2-norm: kvk2 = (v v) . Full Screen Spectral norm: Close

Quit kXk := max kXvk2 kvk2=1 Fact: spectral norm is submultiplicative

1. kXvk2 ≤ kXk kvk2, for all X ∈ Cn×n and v ∈ n C . Introduction Basics kABk ≤ kAk kBk A, B ∈ Proof 2. , for all Cn×n. Extensions Proof. (1) Home Page

kXvk2 Title Page = kXwk2 ≤ kXk kvk2 JJ II v J I where w := is a unit vector. Page 5 of 31 kvk2 (2) Go Back Full Screen

kABk = max kABvk2 ≤ max kAk kBvk2 = kAk kBk. Close kvk2=1 kvk2=1 Quit Question: But how to compute the spectral norm? The deﬁnition

kXk = max kXvk2 kvk2=1 Introduction Basics Proof is no good. Extensions ∗ A matrix H ∈ Cn×n is said to be Hermitian if H = H. Home Page Title Page

Fact: Eigenvalues of a Hermitian H are real. JJ II

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Proof. Suppose Hv = λv for some unit vector v ∈ Page 6 of 31 n ∗ C . Then v Hv = λ and by taking ∗ on both sides Go Back Full Screen ∗ ∗ ∗ λ = v Hv = v H v = λ.¯ Close

Quit Theorem 2.1. (Rayleigh-Ritz) Let H ∈ Cn×n be a Hermitian matrix. Then max v∗Hv is the kvk2=1 Introduction Basics largest eigenvalue of H. Proof Extensions Sketch of proof: Spectral theorem for Hermitian matrix H says that

Home Page there is an orthonormal basis {v1, . . . , vn} such that Hvi = λivi, i = 1, . . . , n. Title Page

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By Rayleigh-Ritz’s theorem, J I

∗ ∗ 1/2 ∗ ∗ 1/2 Page 7 of 31 kXk = max kXvk2 = max (v X Xv) = ( max v X Xv) kvk2=1 kvk2=1 kvk2=1 Go Back is the square root of the largest eigenvalue of X∗X. Full Screen Close

Quit Singular values of X are the square roots of eigenval- ∗ ues of the positive semi-deﬁnite matrix X X. Introduction Basics So kXk is the largest singular value of X. Proof Extensions Arrange the singular values of X in descending order

Home Page s (X) ≥ s (X) ≥ · · · ≥ sn(X) 1 2 Title Page and the eigenvalue moduli in descending order JJ II J I

Page 8 of 31 |λ (A)| ≥ |λ (A)| ≥ · · · ≥ |λn(A)|. 1 2 Go Back

Full Screen So r(X) = |λ1|. Close

Quit m 1/m Fact: r(X) ≤ kX k ≤ kXk for all m ∈ N.

Proof. Since kXk := max kXvk and if Xv = kvk2=1 2 Introduction λv for some unit vector v, then Basics Proof Extensions |λ| = kXvk2 ≤ kXk.

It is not hard to see that r(Xm) = r(X)m. So for all Home Page Title Page m ∈ , N JJ II r(X)m = r(Xm) ≤ kXmk. J I Page 9 of 31

Go Back k · k By submultiplicativeness of , Full Screen

Close kXmk ≤ kXkm. Quit 3. Proof Introduction Basics Theorem 3.1. (Beruling, 1938, Gelfand, 1941) Let Proof Extensions A ∈ Cn×n. m 1/m Home Page lim kX k = r(X). (1) Title Page m→∞ JJ II where r(X) is the spectral radius of X and kXk is J I the spectral norm of A. Page 10 of 31 Go Back

We now provide an elementary proof which is differ- Full Screen ent from those in the literature. Close Quit Some ingredients: (1) Schur triangularization theorem asserts that if ∗ X ∈ Cn×n, then there is a unitary matrix U (U = Introduction −1 ∗ Basics U ) such that T := U XU is upper triangular and Proof Extensions the diagonal entries of T are the eigenvalues of X and can be preordered. Proof: Induction on n. Home Page Title Page (2) r(T ) = r(X) Proof: Similar matrices have the JJ II same eigenvalues. J I (3) T m = U∗XmX so that kT mk = kU∗XmUk = Page 11 of 31 Go Back m n kX k (exercise! Hint: kUvk2 = kvk for all v ∈ C ). Full Screen Pn Pn Close (4) kXk ≤ |xij| (exercise!). i=1 j=1 Quit We may assume that X = T is upper triangular with ascending diagonal moduli Introduction Basics |t | ≤ · · · ≤ |t |. Proof 11 nn Extensions When T is nilpotent, that is, r(T ) = 0, we have a Home Page m strictly upper triangular matrix T . Thus T = 0 for Title Page m ≥ n and hence (1) is obviously true. So we assume JJ II J I that T is not nilpotent so that r(X) = |tnn|= 6 0. Page 12 of 31 Since both sides of (1) are homogenous, by appropri- Go Back Full Screen ate scaling, we may assume that |t | ≥ 1. nn Close

Quit m (m) Write T = [tij ] ∈ Cn×n which is also upper triangular. For 1 ≤ i ≤ j ≤ n, m (m) X Y Introduction tij = tp`−1p`. (2) Basics Proof i=p0≤p1≤···≤pm=j `=1 Extensions (m) m Clearly tii = tii , i = 1, . . . , n. Let us estimate Home Page

(m) Title Page |tij | for ﬁxed i < j. JJ II

The number of (m+1)-tuples (p0, p1, ··· , pm), where J I i = p0 ≤ p1 ≤ · · · ≤ pm = j are integers, is equal Page 13 of 31 Go Back to j−i+m−1. For each of such (p , ··· , p ), there m−1 0 m Full Screen are at most j − i numbers `’s in {1, ··· , m} such that Close

Quit p`−1 =6 p`. Let c := max |tpq| ≥ 1 1≤p,q≤n denote the maximal entry modulus of T . By (2) and Introduction Basics the fact that |t11| ≤ · · · ≤ |tnn|, when m ≥ n, Proof m Extensions (m) X Y |t | ≤ |tp p | ij `−1 ` Home Page

i=p0≤p1≤···≤pm=j `=1 Title Page X j−i m−j+i ≤ c tjj JJ II J I i=p0≤p1≤···≤pm=j Page 14 of 31 j − i + m − 1 j−i m−j+i Go Back = c |tjj| (3) m − 1 Full Screen

Close n + m − 2 n−1 m ≤ c |t | . Quit m − 1 nn By (3), for m ≥ n, m m m |tnn| = r(T ) ≤ kT k n n X X (m) Introduction ≤ |t | Basics ij Proof j=1 i=1 Extensions 2 n−1 n + m − 2 m ≤ n c |tnn| Home Page

m − 1 Title Page 2 n−1 n−1 m ≤ n c (n + m − 2) |tnn| JJ II J I Taking m-th roots on all sides and taking limits for Page 15 of 31 m → ∞ lead to Go Back Full Screen

m 1/m Close lim kT k = |tnn| = r(T ). m→∞ Quit 4. Extensions

Theorem 4.1. Let A, B, X ∈ Cn×n such that A, B Introduction are nonsingular. Then Basics Proof lim kAXmBk1/m = r(X) (4) Extensions m→∞

Home Page Remark: All norms on Cn×n are equivalent, i.e., if n Title Page k · kα and k · kβ are norms C , there are constants JJ II c1, c2 > 0 such that J I Page 16 of 31 c1kXkα ≤ kXkβ ≤ c2kXkα, Go Back Full Screen for all X ∈ Cn×n. So the theorem is true for all Close Quit norm, not just the spectral norm. Proof. It sufﬁces to prove (4) for B = In and for all X ∈ n×n since Introduction C Basics m −1 m Proof AX B = AB(B XB) Extensions −1 and the spectrum of X and B XB are identical. Home Page

Since k · k is submultiplicative, for all m ∈ N, Title Page 1 JJ II kXmk1/m ≤ kAXmk1/m ≤ kAk1/mkXmk1/m. kA−1k1/m J I −1 1/m 1/m Page 17 of 31 Since kA k and kAk converge to 1, we have Go Back the desired result. Full Screen Close

Quit Rewrite Beurling-Gelfand :

m 1/m m 1/m lim kX k = r(X) ⇔ lim [s1(X )] = |λ1(X)| m→∞ 2 m→∞

Yamamoto (1967): Introduction Basics m 1/m Proof lim [s (X )] = |λ (X)|, i = 1, . . . , n. Extensions m→∞ i i

• a natural generalization of Beurling-Gelfand (ﬁnite Home Page dim. case). Title Page • Loesener (1976) rediscovered Yamamoto JJ II J I • Mathias (1990 another proof) Page 18 of 31 • Johnson and Nylen (1990 generalized singular val- Go Back Full Screen ues) Close • Nylen and Rodman (1990 Banach algebra) Quit Numerical experiments: Computing the discrepancy between

[s(Xm)]1/m and |λ(X)| Introduction Basics of randomly generated X ∈ GLn(C). Here s(X) := diag (s1(X), . . . , sn(X)) Proof Extensions and λ(X) := diag (λ1(X), . . . , λn(X)).

The graph of Home Page Title Page m 1/m 100 k[s(X )] − |λ(X)|k2 JJ II

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Close 20 0 20 40 60 80 100 Quit If we consider m 1/m |s1(X ) − |λ1(X)|| for the above example, convergence occurs. But divergence occurs for m 1/m |s2(X ) − |λ2(X)||

Introduction Basics Proof Extensions

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0 20 40 60 80 100 Quit Theorem 4.2. Let A, B, X ∈ Cn×n such that A, B are nonsingular. Then Introduction m 1/m Basics lim [s (AX B)] = |λ (X)|, k = 1, . . . , n. Proof m→∞ k k Extensions Proof. One may use the inequality (needs a proof) Home Page

Title Page si(AB) ≤ s1(A)si(B), JJ II A, B ∈ Cn×n and continuity argument to have J I Page 21 of 31 s (A)s (Xm) ≤ s (AXm) ≤ s (A)s (Xm), i = 1, . . . , n. n i i 1 i Go Back

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By Yamamoto we get the result. Close

Quit Banach algebra A Banach algebra A is a complex algebra and a Ba- nach space with respect to a norm that satisﬁes the Introduction Basics submultiplicative property: Proof Extensions kxyk ≤ kxk kyk, x, y ∈ A

Home Page and A contains a unit element e such that Title Page

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kek = 1 J I

Page 22 of 31 and Go Back xe = ex = x Full Screen Close for all x ∈ A. Quit Example: Cn×n is a Banach algebra with In as the identity element with respect to the spectral norm. Introduction Basics Proof Extensions Example: H = a Banach space. Then B(H) = the algebra of all bounded linear operators on H is Home Page Title Page a Banach algebra with respect to the usual operator JJ II norm. The identity operator I is the unit element. J I Page 23 of 31 B(H) I Every closed subalgebra of that contains is Go Back also a Banach algebra. Full Screen

Quit Spectrum σ(x) of x ∈ A is the set of all λ ∈ C such Introduction that λe − x is not invertible. Basics Proof Spectral radius of x: Extensions

r(x) = sup{|λ| : λ ∈ σ(x)} Home Page

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Remark: The spectrum and the spectral radius of an JJ II x ∈ A are deﬁned in terms of the algebraic structure J I Page 24 of 31 of A, regardless of any metric (or topological) con- Go Back sideration. Full Screen

Quit Theorem 4.3. Let A be a Banach algebra and x ∈ A.

1. The spectrum σ(x) is compact and nonempty. Introduction Basics Proof 2. Extensions m 1/m m 1/m lim kx k = inf kx k = r(x) Home Page m→∞ m≥1 Title Page

Remark: r(x) ≤ kxk is contained in the formula. JJ II m 1/m J I Remark: limm→∞ kx k depends obviously on Page 25 of 31 metric properties of A. This is a very remarkable fea- Go Back ture of the spectral radius formula. Full Screen Close

Quit Summary: lim kxmk1/m = inf kxmk1/m = r(x) m→∞ m≥1 Introduction Basics −→ Banach algebra Proof Cn×n Extensions matrix −→ Banach algebra element spectral norm −→ norm on a Banach algebra Home Page Title Page

JJ II Remark: However Yamamoto’s type result doesn’t J I make sense in Banach algebra. Even we have * oper- Page 26 of 31 Go Back ∗ ∗ ation (then a C algebra) so that we have σ(x x), the Full Screen spectrum is not ﬁnite nor discrete in general. Close Quit Review Yamamoto’s result:

lim [s(Xm)]1/m = |λ(X)|, m→∞ Introduction Basics s(X) := diag (s1(X), . . . , sn(X)) and λ(X) := diag (λ1(X), . . . , λn(X)). Proof Extensions

Singular value decomposition (SVD): Home Page Given X ∈ Cn×n there are unitary matrices U, V such that Title Page

X = USV JJ II

J I where S = diag (s1, . . . , sn). Page 27 of 31

Go Back GLn( ), called the general linear group, is the group of nonsingular matrices C Full Screen in . It is a Lie group. Cn×n Close

Quit Lie groups

• g = real semisimple Lie algebra with connected noncompact Lie group G. Introduction • g = k + p a ﬁxed (algebra) Cartan decomposition of g Basics • K ⊂ G k Proof the connected subgroup with Lie algebra . Extensions • a ⊂ p a maximal abelian subspace. • Fix a closed Weyl chamber a in a and set + Home Page

Title Page A+ := exp a+,A := exp a JJ II

An example J I

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Each A ∈ Cn×n has Hermitian decomposition: Go Back A − A∗ A + A∗ Full Screen A = + . 2 2 Close

Quit (Group) Cartan decomposition

G = KA+K • k , k ∈ K not unique in g = k a (g)k , the element a (g) ∈ A is unique. 1 2 1 + 2 + + Introduction Basics Proof CMJD for real semisimple G: Extensions • h ∈ G is hyperbolic if h = exp(X) where X ∈ g is real semisimple, that is, ad X ∈ End (g) is diagonalizable over R. Home Page • u ∈ G is unipotent if u = exp(N) where N ∈ g is nilpotent, that is, Title Page ad N ∈ End (g) is nilpotent. JJ II • e ∈ G is elliptic if Ad(e) ∈ Aut (g) is diagonalizable over C with eigenval- J I ues of modulus 1. Page 29 of 31 Each g ∈ G can be uniquely written as Go Back

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g = ehu, Close

Quit where e, h, u commute. Extension of Yamamoto:

Introduction Write g = e(g)h(g)u(g) Basics Proof Fact: h(g) is conjugate to b(g) ∈ A+. Extensions

Home Page Theorem 4.4. (Huang and Tam, 2006) Given g ∈ G, let b(g) ∈ A+ be the Title Page unique element in A+ conjugate to the hyperbolic part h(g) of g. Then JJ II m 1/m lim [a+(g )] = b(g). J I m→∞ Page 30 of 31

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Journal of Japan Math. Soc. 58 (2006) 1197-1202. Full Screen

Quit Introduction T HAN K YOU FOR Basics Proof YOURATTENTION Extensions

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