Weak Solutions for a Class of Quadratic
Operator-Differential Equations
Luis Antonio Gómez Ardila
UNIVERSIDAD DE LOS ANDES FACULTAD DE CIENCIAS
DEPARTAMENTO DE MATEMÁTICAS 2016
Weak Solutions for a Class of Quadratic
Operator-Differential Equations
Trabajo de Grado presentado como requisito parcial para optar al título de Magíster en Matemáticas
Luis Antonio Gómez Ardila
Directora: Monika Winklmeier
UNIVERSIDAD DE LOS ANDES FACULTAD DE CIENCIAS
DEPARTAMENTO DE MATEMÁTICAS 2016
Abstract
"Mathematics is the most beautiful and most powerful creation of the human spirit" (Stefan Banach)
This monograph presents results of existence and uniqueness of generalized solutions for two classes of quadratic operator-differential equations with constant coefficients:
Au(t) +Bu0(t)−Du00(t) = 0 (1)
and
Au(t) +iBu0(t) +Du00(t) = 0, (2) where A, B and D are self-adjoint operators which satisfy certain conditions under which the equation (1) is called elliptic-hyperbolic and the equation (2) is calledhyperbolic. The main result is the existence and uniqueness of weak solutions, on the positive real axis or on the negative real axis, for the class of operator differential equations called elliptic-hyperbolic. These solutions, on the positive real axis, decay exponentially to zero in the infinity. A similar result is obtained on the negative real axis.
We also give results of existence of solutions of (1) and (2) in the case in which A, B and D
are self-adjoint bounded operator. These results, specifically for the elliptic-hyperbolic case, are related with a factorization of the quadratic pencil A+λB +λ2C (where C = −D) and of its
associated pencil L(λ) =λ2A+λB+C. The factorization of this latter pencil is related to the
existence of roots for the operator equation
AZ2+BZ+C = 0. (3)
In order to prove that (3) has a solution Z, we define the so-called linearizer L of the pencil. Then we use the theory of Krein spaces to establish existence of certain L-invariant subspaces which finally lead to the existence of solutions Z of (3).
Additionally, some results on completeness and basis property of the system of eigenvectors as-sociated to the pencil A+λB−λ2Dare presented.
This work is based on the papers of Shkalikov [11] and Langer [4], and on the books of Bognár [1], Azizov [2], and Markus [6].
Dedication
To my mother, Alix Ardila.
Acknowledgements
My gratitude forever to...
Prof. Dr. Monika Winklmeier, my teacher and my adviser in this grade work, for giving me the opportunity to work with her, but specially for introducing me to that beautiful part of the mathematics called Operator Theory.
Prof. PhD. Jean Carlos Cortissoz, my graduate adviser, for his guidance during these two years.
The Department of Mathematics of Universidad de los Andes, for the financial support received through of graduate assistance.
My friends and teachers inUniversidad Industrial de Santander, the ProfessorsRicardo Monturiol, Edilberto Reyes,Rafael Isaacs,Rafael Castro andSonia Sabogal for their constant encouragement and help to achieve this goal.
Contents
Contents vii
1 Introduction 1
1.1 Preliminaries . . . 2
2 Inner Product Spaces 5 2.1 Vector Spaces . . . 5
2.2 Inner Products on Vector Spaces . . . 6
2.3 Orthogonality . . . 10
2.4 Isotropic Vectors . . . 12
2.5 Maximal Non-Degenerate Subspaces . . . 14
2.6 Maximal Semi-Definite Subspaces . . . 15
2.7 Projections of Vectors on Subspaces . . . 17
2.8 Ortho-Complemented Subspaces . . . 19
2.9 Fundamental Decompositions . . . 20
2.10 Orthonormal Systems . . . 22
3 Linear Operators on Inner Product Spaces 24 3.1 Linear Operators . . . 24
3.2 Isometric Operators . . . 26
3.3 Symmetric Operators . . . 28
3.4 Orthogonal and Fundamental Projectors . . . 30
3.5 Fundamental Symmetries . . . 32
3.6 Angular Operators . . . 37
4 Krein Spaces 41 4.1 Krein Spaces . . . 41
4.2 Subspaces in Krein Spaces . . . 45
4.3 The Gram Operator of a Closed Subspace . . . 50
4.4 Linear Operators in Krein Spaces . . . 53
5 Riesz Basis 56 5.1 Bessel sequences . . . 56
5.2 Basis . . . 58
6 Quadratic Operator Pencils 65
6.1 Quadratic Pencils of Bounded Self-adjont Operators . . . 65
6.2 Factorization of Quadratic Pencils . . . 68
6.3 Strongly Damped Pencils . . . 69
6.4 Operators Associated to a Pencil . . . 80
6.5 Operator-Differential Equations Associated to a Quadratic Pencil . . . 93
6.6 A special case . . . 96
7 Hyperbolic and Elliptic-hyperbolic Pencils 98 7.1 Hyperbolic Pencils . . . 98
7.2 Operator-Differential Equation Associated to a Hyperbolic Pencil . . . 104
7.3 Elliptic-hyperbolic Pencils and their Associated Operator-Differential Equation . . 106
8 Pencils of Unbounded Self-adjoint Operators and their Associated Differential Equations 113 8.1 Hyperbolic and Elliptic-hyperbolic Pencils of Unbounded Self-adjoint Operators . 113 8.2 Equation Associated with a Hyperbolic Pencil . . . 116
8.3 Equation Associated with a Elliptic-hyperbolic Pencil . . . 119
8.4 Completeness and Basis Property . . . 123
A Some Results on Functional Analysis and Operator Theory 124 A.1 Functional Analysis and Operator Theory . . . 124
A.2 Normal Points of a Bounded Operator . . . 125
A.3 Some Results about Definitizable Operators . . . 126
A.4 Distributions and Operators . . . 128
Chapter 1
Introduction
"No one shall expel us from Paradise that Cantor has created for us." (David Hilbert)
This monograph is based on the papers of Shkalikov [11] and Langer [4], and on the books of Bognár [1], Azizov [2], and Markus [6]. The main result is the existence and uniqueness of weak solutions, on the positive real axis or on the negative real axis, for a class of operator differential equations calledelliptic-hyperbolic. These solutions, on the positive real axis, decay exponentially to zero in the infinity. A similar result is obtained on the negative real axis.
Operator-differential equations of the form (1) and (2) can be considered as a generalization of ordinary lineal differential equations to the infinite-dimensional case [21]. They arise for example in the study of strongly damped mechanical systems [20], electrical systems and signal processing [23], in the study of partial differential equations (Wave and Heat Equations, for example) and in the study of Abstract Cauchy Problems [22].
The spectral analysis of the associated pencil L(λ) =λ2A+λB+C (whereD=−C) generalizes
the study of thequadratic eigenvalue problem which is very important in the previously mentioned applications to physics [23]. For the case where A, B and C are bounded self-adjoint operators acting in a separable Hilbert space H, such that L(λ) is strongly damped, the spectral analysis and linearization of the pencil L(λ) can be realized with geometric methods using Krein space theory [4]. Analogous to the finite-dimensional case, the problem of linearization of the pencil
L(λ) is related to the problem of factorization of the operator equation (3) and this latter is related to the problem of existence of invariant maximal semi-definite subspaces for an associated linearizing operator L [4] [6].
1.1
Preliminaries
1. By a vector space we shall always mean a complex vector space. Asubspace of a vector space
X is a set U ⊆X such that it is itself a vector space with the operations defined in X.
Anorm on a vector space X is a functionk·k:X −→Rsuch that, for everyx, y∈X andα∈C:
i) kxk= 0 impliesx= 0, ii) kαxk=|α|kxk,
iii) kx+yk ≤ kxk+kyk.
The pair (X,k·k) is called a normed space.
A subspaceU of a normed space(X,k·k)is calledclosed (open) if and only if it is a closed (open) subset in the topology induced on X by the norm k·k. Note that the subspace U ⊆ X is closed if, and only if, for every sequence(un)n∈N⊆U and u0 ∈X, un→u0 in the norm impliesu0 ∈U. A normed space X is called a Banach space if, and only if, every Cauchy sequence in X is convergent. A subspace U of a Banach space (X,k·k)is closed if and only if (U,k·k) is in itself a Banach space.
A pre-Hilbert space is a pair (X,h·,·i) where X is a vector space and h·,·i : X ×X −→ C is a funtion such that for every x, y, z ∈X and α, β ∈C:
i) hαx+βy, zi=αhx, zi+βhy, zi, ii) hy, xi=hx, yi,
iii) hx, xi>0 if x6= 0.
The function h·,·i is called a positive definite inner product (see section 2.2 for the general defi-nition of an inner product).
Every pre-Hilbert space (X,h·,·i) is a normed space with the norm induced by h·,·i; this norm is defined by the formula kxk=phx, xi, x∈ X. In this case, if (X,k·k) is a Banach space then (X,h·,·i)is called a Hilbert space.
2. By a linear operator from a vector space X into a vector space Y we mean a function
T :D(T)⊆X −→Y, where D(T) ⊆X is a subspace of X (called the domain of T), such that
T(αu+βv) =αT u+βT v for every u, v ∈ D(T) and α, β ∈C. In the case Y =X we say that
Let X be a vector space. The linear operator P : X −→ X is called a projector if and only if
P2 =P.
If T :D(T) ⊆X −→ Y is a linear operator then the subspaces ker(T) ⊆X and R(T) ⊆Y are defined by:
ker(T) ={x∈D(T)|T x= 0},
R(T) ={T x|x∈D(T)}.
Let(X,k·kX)and (Y,k·kY) be normed spaces and letT be a linear operator fromX intoY. The
operatorT is calledbounded if and only if there existsM ≥0such thatkT xkY ≤MkxkX for every
x∈D(T). The bounded operator T is calledcompact if, and only if, for every bounded sequence (xn)n∈N ∈ D(T), the sequence (T xn)n∈N has a convergent subsequence in Y. An operator T is
called closed if, and only if, for every sequence (xn)n∈N ⊆ D(T) such that (xn)n∈N is convergent
in X and (T xn)n∈N is convergent in Y, then: lim
n→∞xn ∈D(T) and nlim→∞T xn =T nlim→∞xn
.
If X and Y are normed spaces then we set:
• B(X, Y) = {T :X −→Y |T is a bounded linear operator}.
• B∞(X, Y) = {T :X −→Y |T is a compact linear operator}.
• B(X) = B(X, X).
• B∞(X) = B∞(X, Y).
3. LetT be a densely defined linear operator acting in a Hilbert space H. Define theadjoint T∗
of T by:
• D(T∗) = {y∈ H |x7→ hT x, yiis a bounded operator onD(T)}.
• If y∈D(T∗), then T∗y is the unique u such that hT x, yi=hx, ui for every x∈D(T).
Clearly, since T is densely defined, T∗ is well-defined. Moreover T∗ is a densely defined closed operator.
If T ∈ B(H), then we have
H= ker(T)⊕R(T∗) = ker(T∗)⊕R(T).
4. The densely defined linear operator T acting in a Hilbert space H is called:
• symmetric if and only if T ⊆T∗.
• essentially self-adjoint if and only ifT =T∗.
• self-adjoint if and only if T =T∗.
Chapter 2
Inner Product Spaces
2.1
Vector Spaces
Definition 2.1.1. LetX be a vector space and U1, . . . , Un subspaces of X. Then:
(i) U1 +· · ·+Un := {x1 +· · ·+xn | xj ∈ Uj, j = 1, . . . , n} is the vector sum of subspaces
U1, . . . , Un.
(ii) U1, . . . , Un are linearly independent subspaces if, and only if,x1+· · ·+xn= 0 with xj ∈Uj
(j = 1, . . . , n) implies x1 =· · ·=xn= 0.
Remarks:
• Clearly, the vector sum of subspaces of X is again a subspace ofX.
• In the definition 2.1.1, (ii) is equivalent to:
(U1 +· · ·+Uj−1)∩Uj ={0} for every j = 2, . . . , n.
• The vector sum of linearly independent subspaces U1, . . . , Un ⊆ X is called a direct vector
sum and is denoted by U1u· · ·uUn.
• By Zorn’s lemma, if U ⊆ X is a subspace then there exists a subspace V ⊆ X such that
with respect to X. Moreover, if W ⊆X is a subspace withU ∩W ={0}then there exists
V ⊇W subspace of X such that U uV =X.
Definition 2.1.2. Let X be a vector space, U ⊆ X a subspace and X/U the quotient space of
X with respect toU. Then the dimension ofX/U is called the codimension of U with respect to
X and is denoted by codim(U); this is,
codim(U) := dim(X/U).
Proposition 2.1.3. LetX be a vector space andU ⊆X a subspace. Then every complementary subspace for U with respect to X is isomorphic to X/U.
Proof. Let V ⊆ X be a complementary subspace for U with respect to X. Let T : V −→ X/U
be such that, for every v ∈ V,T v := [v] (here [v] denote the class of equivalency ofv); clearly T
is well-defined and it is a linear transformation. T is injective since ker(T) ={v ∈V : [v] = 0}=
{v ∈ V | v ∈ U} = U ∩V = {0}. T is surjective because if w ∈ X/U then there exists x ∈ X
and, therefore, u∈U and v ∈V such that w= [x] = [u+v] = [v] =T v. So T is an isomorphism between V and X/U.
Corollary 2.1.4. The dimension of every complementary subspace for a subspace U is equal to codim(U).
2.2
Inner Products on Vector Spaces
Definition 2.2.1. An inner product on a vector spaceX is a function
[·,·] :X×X −→C
such that:
(i) For every x, y inX, [y, x] = [x, y].
Remark. The pair(X,[·,·]) whereX is a vector space and[·,·]is an inner product on X will be called an inner product space. If (X,[·,·]) is an inner product space then (X,−[·,·]) is called the anti-space of (X,[·,·]).
Definition 2.2.2. Let (X,[·,·]) and (X0,[·,·]0) be inner product spaces. A linear operator T :
X → X0 is called an isometry if, and only if, [T x, T y]0 = [x, y] for every x, y ∈ X. If further T
is a linear isomorphism then X and X0 are said be isometrically isomorphic. In this case, T is called an isometrical isomorphism between X and X0.
Proposition 2.2.3 (Polarization Formula). Let X be an inner product space. Then, for every
x, y inX:
4[x, y] = [x+y, x+y]−[x−y, x−y] +i[x+iy, x+iy]−i[x−iy, x−iy].
The proof of the polarization formula is a straightforward calculation of the right hand side. Remark. If X is an inner product space then, for every x∈X,[x, x]∈R since [x, x] = [x, x].
Definition 2.2.4. LetX be an inner product space and x∈X. Then xis called:
(i) positive if [x, x]>0.
(ii) negative if [x, x]<0.
(iii) neutral if [x, x] = 0.
Remark. Clearly, 0 is neutral. Note that if X is an inner product space and x is a positive (negative, neutral) vector then, for every α ∈C with α 6= 0, the vector αx is positive (negative, neutral) too.
Definition 2.2.5. Let(X,[·,·]) be an inner product space. [·,·] is called indefinite if and only if there exist x,y in X such that[x, x]>0and [y, y]<0.
Example 1. Let(H,h·,·i)be a Hilbert space and letU,V be nontrivial subspaces ofHsuch that
H = U⊕V˙ (where ⊕˙ denotes the orthogonal direct sum of subspaces). Then, for every x ∈ H,
x = u1 +v1 and y = u2 +v2, then define [x, y] := hu1, u2i − hv1, v2i. Clearly, [·,·] is an inner
product on H. Note that, for every u∈U \ {0}, [u, u]>0and, for every v ∈V \ {0}, [v, v]<0. So, [·,·] is an indefinite inner product on H.
Proposition 2.2.6. Let X be an indefinite inner product space. Then there exists a non-zero neutral vector in X.
Proof. Let x,y ∈X such that [x, x]>0 and [y, y]<0. Note that, for every λ ∈R:
[x+λy, x+λy] = [y, y]λ2+ 2 Re([x, y])λ+ [x, x].
SinceRe([x, y])2−[x, x][y, y]>0, the discriminant of the quadratic equation[y, y]λ2+2 Re([x, y])λ+ [x, x] = 0 is positive and, therefore, there exists λ0 ∈R such that [x+λ0y, x+λ0y] = 0, hence
z0 :=x+λ0y is a neutral vector. Suppose, towards a contradiction, that z0 = 0; then x =−λ0y
and, therefore, [x, x] =λ2
0[y, y] .
Definition 2.2.7. Let(X,[·,·])be an inner product space. If[·,·]is not indefinite then it is called semi-definite and (X,[·,·]) is called a semi-definite inner product space.
Remark. If [·,·] is a semi-definite inner product on a vector space X then, for every x ∈ X, [x, x] ≥ 0 (and in this case [·,·] is called positive semi-definite) or, for every x ∈ X, [x, x] ≤ 0 (and in this case [·,·]is called negative semi-definite). If, for every x∈ X, [x, x] = 0 then [·,·] is called neutral (note that a neutral inner product is both a positive and a negative semi-definite inner product).
Proposition 2.2.8. Let X be a semi-definite inner product space and let x ∈ X be a neutral vector. Then, for every y∈X,[x, y] = 0.
Proof. Suppose x6= 0 (since the casex= 0 is clear) and suppose, without loss of generality, that the space is positive semi-definite (otherwise take −[·,·]). Let y∈X be any vector.
If[y, y] = 0then, for everyλ ∈R,0≤[x+λy, x+λy] = 2λRe([x, y])and0≤[x+iλy, x+iλy] = 2λIm([x, y]); so Re([x, y]) = 0 and Im([x, y]) = 0. Now, if [y, y] > 0 then, for every λ ∈ R,
q(λ) := [x +λy, x +λy] = λ2[y, y] + 2λRe([x, y]) ≥ 0 and p(λ) := [x +iλy, x + iλy] =
λ = 0 and so Re([x, y])2 and Im([x, y])2, the discriminants, are zero; this is, Re([x, y]) = 0 and
Im([x, y]) = 0. Therefore, [x, y] = 0.
Corollary 2.2.9. Let X be an inner product space and let x, y ∈ X such that [x, x] = 0 and [x, y]6= 0. Then the subspace Span({x, y}) is indefinite.
Proof. Let U := Span({x, y}). If U is not indefinite then U is a semi-definite subspace and, by proposition 2.2.8, [x, y] = 0 .
Corollary 2.2.10. IfX is a neutral inner product space then, for everyx, y∈X, [x, y] = 0.
Proposition 2.2.11 (Schwarz Inequality). LetX be a semi-definite inner product space. Then, for every x, y∈X:
|[x, y]|2 ≤[x, x][y, y].
Proof. Suppose, without loss of generality, that the inner product is positive semi-definite. If x
or y is a neutral vector then, by proposition 2.2.8, the inequality holds.
Suppose that x and y are not neutral vectors. Let u = [[y,xx,x]]x and v = y− [[x,xy,x]]x, then [u, v] =
h
[y,x]
[x,x]x, y− [y,x] [x,x]x
i
= 0, y = u+v and, therefore, [y, y] = [u, u] + [v, v] ≥ [u, u] = |[[x,yx,x]|]2. So,
|[x, y]|2 ≤[x, x][y, y].
Definition 2.2.12. An inner product on a vector space X is called definite if and only if, for every x∈X, [x, x] = 0 implies x= 0.
Remarks:
(i) IfU is a definite subspace of an inner product space (X,[·,·]), thenU can be endowed with a norm, k·kU, defined by the formula kxkU =
p
|[x, x]|, x ∈ U. This norm is called the
norm induced by the inner product onU.
(ii) By proposition 2.2.6, a definite inner product is not indefinite; this is, a definite inner product is always semi-definite. So, if the inner product is positive definite then, for every
(iii) A positive definite inner product space is what is known classically as a pre-Hilbert space (see page 2).
(iv) Clearly, if [·,·] is a positive definite inner product on a vector space X then −[·,·] is a negative definite inner product on X.
Proposition 2.2.13. LetXbe an inner product space. IfXcontains at least one positive vector, then every element of X is the sum of two positive vectors.
Proof. Let x0 ∈X be a positive vector. Then, for every α ∈R\ {0}, αx0 is positive. If x ∈X,
then, clearly, there isα∈R\ {0}such that [x+αx0, x+αx0] = [x, x] + 2 Re([x, x0])α+ [x0, x0]α2
is positive. So there exists α ∈ R\ {0} such that −αx0 and x +αx0 are positive, and x =
(x+αx0) + (−αx0).
Remark. The proposition 2.2.13 holds if positive is replaced by negative.
2.3
Orthogonality
Definition 2.3.1. LetX be an inner product space; x, y∈X and A, B ⊆X. Then:
(i) x and y are orthogonal (x⊥y) if and only if[x, y] = 0.
(ii) A and B are orthogonal (A⊥B) if and only if, for every x∈A and y∈B, x⊥y.
Definition 2.3.2. LetX be an inner product space andA⊆X. The orthogonal complement of
A, denoted A⊥, is defined by
A⊥ :={x∈X :x⊥A}={x∈X :x⊥y for ally∈A}.
The following proposition is clear so we omit its proof.
Proposition 2.3.3. LetX an inner product space. Then:
(i) For every A⊆X, A⊥ is a suspace of X.
(iii) For every U, V subspaces of X, (U +V)⊥=U⊥∩V⊥.
(iv) For every A⊆X, A⊆(A⊥)⊥ =:A⊥⊥.
(v) For every A⊆X, A⊥=A⊥⊥⊥.
Example 2. Let X be a two-dimensional vector space with basis {e1, e2}. On X define an inner
product by the relations [e1, e1] = 1, [e2, e2] =−1 and [e1, e2] = 0. Let U := Span(e1+e2). Then
U is a subspace ofX such that U⊥=U (because,[αe1+βe2, e1 +e2] = 0 ⇔α=β).
Definition 2.3.4. LetX be an inner product space and let U, V ⊆X be subspaces. U and V
are dual companions (or U and V form a dual pair) if and only ifU ∩V⊥ =U⊥∩V ={0}.
Remarks:
• IfXis a semi-definite inner product space andx∈Xis a neutral vector then, by proposition
2.2.8, x∈A⊥ for every A⊆X.
• If x, y ∈X are orthogonal then [x+y, x+y] = [x, x] + [y, y].
• If U is a subspace of X then it is possible that U ∩U⊥6={0}(see example 2).
• By example 2, if U ⊆ X is a subspace and V ⊆ X is its orthogonal complement with
respect to X then U and V are not necessarily linearly independent (nevertheless, ifX is a definite inner product space and U ⊆X is a subspace then its orthogonal complement is a complementary subspace for U).
• The vector sum (resp. direct vector sum) of pairwise orthogonal subspacesU1, . . . , Un⊆X
is called anorthogonal sum (resp. orthogonal direct sum) and it is denoted byU1⊕ · · · ⊕Un
(resp. U1⊕˙ . . .⊕U˙ n).
• Let (Xj,[·,·]j) be inner product spaces, j = 1, . . . , n. Let X := X1 × · · · ×Xn and for
x= (x1,· · · , xn), y= (y1,· · · , yn)inX define [x, y] :=Pnj=1[xj, yj]j. Clearly(X,[·,·])is an
inner product space. Now, for everyk= 1, . . . , n, letX(k) :={(x1, . . . , xn)∈X :xj = 0, j 6=
k}. So,X(k) is a subspaces of X which is isometrically isomorphic to Xk (k = 1, . . . , n) and
2.4
Isotropic Vectors
Definition 2.4.1. Let X be an inner product space and U ⊆ X a subspace. The isotropic part of U is the subspace U0 :=U ∩U⊥
.
Remarks:
• If x∈U0 then x is called an isotropic vector of U.
• Ifxis an isotropic vector for some subspace thenxis a neutral vector, but not every neutral
vector is isotropic (in the example 2, e1+e2 is neutral but is not isotropic for X).
• If U0 6={0} then U is called degenerate.
• X is degenerate if, and only if, its isotropic part, X0 =X⊥, is not equal to{0}.
• In the example 2, X is not degenerate (since X⊥={0}) butU is a degenerate subspace of
X.
• If U is a definite subspace of an inner product space X, then U is non-degenerate.
• In X/X0 an inner product is defined by [˜x,y˜]∼ := [x, y], where x˜, y˜ ∈ X/X0, x ∈ x˜ and y∈ y˜. Note that[·,·]∼ is well-defined since if x, u∈x˜ and y, v∈y˜then x−u, y−v ∈X0
and, therefore, [x, y]−[u, v] = [x−u, y] + [u, y −v] = 0. That [·,·]∼ is an inner product is
clear since [·,·]is so.
Theorem 2.4.2. Let(X,[·,·])be an inner product space and letU be a non-degenerate subspace of X such that U =U++U− with U+ a positive semi-definite subspace and U− a negative
semi-definite subspace. Then U =U+uU−, U+ is maximal positive semi-definite in (U,[·,·])and U−
is maximal negative semi-definite in (U,[·,·]).
Proof. Suppose, towards a contradiction, that there exists w ∈U+∩U− such that w 6= 0. Then w is neutral and, therefore, w∈U+⊥ and w∈U−⊥ (see proposition 2.2.8); thus w ∈U0 . Hence,
Suppose, towards a contradiction, that U+ is not maximal positive semi-definite in U, and let
V ⊆U be a positive semi-definite subspace such that V )U+. Since U =U+uU−, there exists
v ∈ V ∩U− (therefore, v is neutral) such that v 6= 0. Thus, v ⊥ V, v ⊥ U− and, therefore,
v ⊥U . Hence, U+ is maximal positive semi-definite and, analogously, U− is maximal negative
semi-definite.
Proposition 2.4.3. Let X be an inner product space and U, U1, . . . , Un subspaces of X such
that U =U1⊕ · · ·˙ ⊕U˙ n. ThenU0 =U10⊕ · · ·˙ ⊕U˙ n0.
Proof. Let i, j ∈ {1, . . . , n} with i 6= j. If x ∈ Ui0 and y ∈ Uj0 then x ∈ Ui and y ∈ Uj. So
[x, y] = 0 since Ui and Uj are orthogonal. Hence Ui0 and Uj0 are orthogonal.
If x1 +· · ·+xn = 0 with xj ∈ Uj0 then x1 = · · · = xn = 0 since xj ∈ Uj and U1, . . . , Un are
linearly independent. Hence U0
1, . . . , Un0 are linearly independent.
So far, U0
1, . . . , Un0 are mutually orthogonal and linearly independent subspaces of X.
Let x = x1 +· · ·+xn ∈ U10⊕˙ . . .⊕U˙ n0. Then x ∈ U1⊕˙ . . .⊕U˙ n = U and, for every uj ∈ Uj
(j = 1, . . . , n),[u1+· · ·+un, x] = [u1, x] +· · ·+ [un, x] = [u1, x1] +· · ·+ [un, xn] = 0 sincexj ∈Uj⊥
(j = 1, . . . , n). Then x∈U⊥ and, therefore, x∈U0. So, U10⊕˙ . . .⊕U˙ 0
n ⊆U0.
Let x∈ U0 =U ∩U⊥
, then there exists xj ∈ Uj (j = 1, . . . , n) such that x =x1+· · ·+xn. Fix
j ∈ {1, . . . , n}; then for every u∈Uj ⊆U we have [x, u] = 0. But, if u∈Uj then [x, u] = [xj, u];
so xj ∈Uj⊥ and, therefore,xj ∈Uj0. Then, x∈U10⊕˙ . . .⊕U˙ n0. So, U0 ⊆U10⊕˙ . . .⊕U˙ n0.
Therefore, U0 =U0
1⊕˙ . . .⊕U˙ n0.
Corollary 2.4.4. The orthogonal direct sum of non-degenerate subspaces is non-degenerate.
Proposition 2.4.5. LetX be a semi-definite inner product space. Then,
X0 ={x∈X|[x, x] = 0}.
Proof. Clearly, X0 ⊆ {x ∈ X | [x, x] = 0} and, by proposition 2.2.8, {x ∈ X | [x, x] = 0} ⊆
X⊥ =X0. So,X0 ={x∈X : [x, x] = 0}.
Corollary 2.4.6. Let X be a neutral inner product space (this is, [x, x] = 0 for every x ∈ X). Then, X0 =X.
Proof. This follows from proposition 2.4.5 because every neutral space is a semi-definite space.
Proposition 2.4.7. Let X be an inner product space and U ⊆ X a subspace. If U is neutral then U⊥⊥ is neutral.
Proof. If U is neutral then, by proposition 2.2.8, U ⊆ U⊥. Therefore, U⊥⊥ ⊆ (U⊥⊥)⊥, which shows that U⊥⊥ is neutral.
Proposition 2.4.8. Let U be a degenerate subspace of an inner product space X. ThenU⊥⊥ is degenerate.
Proof. As U ⊆U⊥⊥ and U⊥ =U⊥⊥⊥, it follows that {0} 6=U ∩U⊥ ⊆U⊥⊥∩U⊥⊥⊥. So U⊥⊥ is degenerate.
2.5
Maximal Non-Degenerate Subspaces
Proposition 2.5.1. LetX be an inner product space; X0 its isotropic part andV a complemen-tary subspace for X0. Then V is a non-degenerate subspace and X =X0⊕V˙ .
Proof. SinceX0 =X⊥then, for everyu∈X0andv ∈V,[u, v] = 0. So,X0 andV are orthogonal
subspaces and, as V is a complementary subspace for X0, we have X =X0⊕V˙ .
Suppose, towards a contradiction, thatV is degenerate, and letx6= 0be such thatx∈V0. Then,
for every u ∈ X0 and v ∈ V, [u+v, x] = [u, x] + [v, x] = 0 + 0 = 0. So x ∈ X0, but this is a
contradiction since X0 and V are linearly independent.
Proposition 2.5.2. LetX be an inner product space; X0 its isotropic part andV a complemen-tary subspace for X0. Then V is isometrically isomorphic to X/X0.
Proof. By proposition 2.1.3,V is isomorphic to X/X0 and an isomorphism is given byT :V −→
X/X0, T x := ˜x. Note that, for every x, y ∈ V, [T x, T y]
∼ = [˜x,y˜]∼ = [x, y] (see remark before
definition 2.4.1). So T is an isometrical isomorphism.
Theorem 2.5.4. LetX be an inner product space and V ⊆X. V is a complementary subspace of X0 if and only if V is a maximal non-degenerate subspace of X.
Proof. ⇒ If V is a complementary subspace for X0 then, by proposition 2.5.1, V is a
non-degenerate subspace of X and X =X0⊕V˙ .
Suppose, towards a contradiction, that there exists W ⊆ X non-degenerate subspace such that
V (W. As X =X0uV and V
(W then W ∩X0 6={0}. Let w∈W ∩X0 with w6= 0; then, for every x∈W, [x, w] = 0. So w∈W0 and, therefore, W0 6={0} .
⇐ As V is a non-degenerate subspace of X it follows that V ∩X0 = {0}, so V and X0 are linearly independent. If X0
uV 6=X then, by the remark before definition 2.1.1, there exists a subspace W ⊆X with V (W such thatX =X0uW. So W
)V and, by proposition 2.5.1, W is non-degenerate .
Corollary 2.5.5. IfX is an inner product space thenX contains maximal non-degenerate sub-spaces and every non-degenerate subspace admits a maximal non-degenerate extension.
2.6
Maximal Semi-Definite Subspaces
Proposition 2.6.1. Let X be an inner product space and U, V ⊆ X be two maximal positive semi-definite subspaces (or two maximal negative semi-definite subspaces). Then (U +V)0 =
{x∈U ∩V |[x, x] = 0}.
Proof. Suppose, without loss of generality, that U andV are both maximal positive semi-definite subspaces.
Let x ∈ U ∩V be such that [x, x] = 0; clearly x ∈ U +V. As U and V are semi-definite subspaces then, by proposition 2.2.8, x ∈ U0∩V0 and, therefore, for every u ∈ U and v ∈ V,
[u+v, x] = [u, x] + [v, x] = 0 + 0 = 0. So x∈(U +V)0.
Let x ∈ (U +V)0, then [x, x] = 0 and x ∈ U⊥∩V⊥. Suppose, towards a contradiction, that
x6∈U; as, for everyu∈U andα ∈C,[u+αx, u+αx] = [u, u]≥0thenU+ Span({x}))U and
U+ Span({x})is a positive semi-definite subspace (since U is a maximal positive semi-definite subspace). Sox∈U and, analogously,x∈V. Therefore,(U+V)0 ={x∈U∩V |[x, x] = 0}.
Proposition 2.6.2. LetX be an inner product space and let U, V ⊆X be subspaces such that
X =UuV. IfU is positive definite andV is negative semi-definite, then U is a maximal positive definite subspace and V is a maximal negative semi-definite subspace.
Analogously, if U is positive semi-definite and V is negative definite, thenU is a maximal positive semi-definite subspace and V is a maximal negative definite subspace.
Proof. Suppose that U is positive definite and V is negative semi-definite, and suppose, towards a contradiction, that U is not a maximal positive definite subspace. Let W ⊆ X be a positive definite subspace such thatU (W; asX =UuV then W∩V 6={0} and, therefore, there exists
w∈W such that[w, w]>0and[w, w]≤0 . SoU is maximal positive definite and, analogously,
V is maximal negative semi-definite.
The case where U is positive semi-definite and V is negative definite is similar.
Proposition 2.6.3. Let X be an inner product space and U ⊆ X. If U is a maximal positive definite subspace, then U⊥ is negative semi-definite.
Analogously, if U is a maximal negative definite subspace then U⊥ is positive semi-definite.
Proof. Let U be a maximal positive definite subspace. Suppose, towards a contradiction, that there exists v ∈ U⊥ such that [v, v] > 0. Then U + Span({v}) is a positive definite subspace (since, for every u ∈ U and α ∈ C, [u+αv, u+αv] = [u, u] +|α|2[v, v]) and, as v 6∈ U (since
U0 ={0}), U + Span({v}))U .
The case where U is maximal negative definite is similar.
Corollary 2.6.4. Let X be an inner product space and U ⊆ X. If U is a positive definite and maximal positive semi-definite (resp., negative definite and maximal negative semi-definite) subspace, then U⊥ is negative definite (resp., positive definite).
Proof. Suppose thatU is a positive definite and maximal positive semi-definite subspace; thenU∩ U⊥ ={0}, U is maximal positive definite (by the maximality as positive semi-definite subspace) and, by proposition 2.6.3, U⊥ is negative semi-definite. Let v ∈ U⊥ such that [v, v] = 0 and suppose, towards a contradiction, that v 6= 0; then U + Span({v}) is positive semi-definite and
The case where U is negative definite and maximal negative semi-definite is analogous.
2.7
Projections of Vectors on Subspaces
Definition 2.7.1. Let X be an inner product space; U ⊆X be a subspace and x∈X. If there exists u∈U and v ∈U⊥ such thatx=u+v then uis called a projection of xon U.
Remarks:
• Note that, given a subspace U of an inner product space X and given x∈X then, x has a
projection on U if and only if x ∈ U +U⊥. So a element x ∈ X does not need to have a projection on U since, in general, it is possible thatX 6=U +U⊥ (see example 2).
• Ifu∈U is a projection of the vectorxon the subspace U andv ∈U⊥ is such thatx=u+v
then, for everyw∈U0, u+w is a projection of xonU (sinceu+w∈U and v−w∈U⊥). • By the previous remark, the projection of a vector on a subspace, if it exists, is not unique
in general; this is the case when the subspace is degenerate. So, if there existsx∈X having exactly one projection on a subspace U ⊆ X then U is non-degenerate. Moreover, if the subspace U is non-degenerate then every vector xhas at most one projection on U.
• Clearly, if u1, u2 ∈U are projections of the vector x on the subspace U then u1−u2 ∈U0
(since there existv1, v2 ∈U⊥such thatx=u1+v1 =u2+v2 and sou1−u2 =v2−v1 ∈U⊥;
therefore u1−u2 ∈U ∩U⊥ =U0).
Proposition 2.7.2. Let X be an inner product space; U ⊆ X a neutral subspace and x ∈ X. Then x admits a projection on U if and only if x ⊥ U. In this case, every element of U is a projection of xon U.
Proof. As U is neutral, by corollary 2.2.10, U ⊆ U⊥. So, by the preceding remark, x has a projection on U if and only in x∈U⊥.
Theorem 2.7.3. LetX be an inner product space;U ⊆Xa positive definite subspace andx∈X. Then x admits a projection on U if and only if the function ϕ:U −→R, ϕ(u) := [x−u, x−u] attains its minimum at a unique u0 ∈U. This u0 is the unique projection of x onU.
Proof. As U is a definite subspace, U is non-degenerate.
⇒ Suppose that x admits a projection on U; as U is non-degenerate, the projection of x onU
is unique. Let u0 and v0 be the unique vectors in U and U⊥, respectively, such that x=u0+v0.
Note thatϕ(u0) = [v0, v0]and, for everyu∈U withu6=u0, asU is positive definite andu−u0 6= 0,
ϕ(u) = [(u0+v0)−u,(u0+v0)−u] = [v0+ (u0−u), v0+ (u0−u)] = [v0, v0] + [u0−u, u0−u]>
[v0, v0] =ϕ(u0). So u0 is the unique point of minimum for ϕ.
⇐ Suppose thatϕ has a unique point of minimum u0 ∈ U. Suppose, towards a contradiction,
that x−u0 6∈ U⊥, and let u ∈ U such that [x−u0, u] 6= 0 (therefore u 6= 0). Then, for every
λ ∈ C, λ 6= 0, [x −(u0 +λu), x − (u0 + λu)] > [x −u0, x − u0]; this is, for every λ 6= 0,
|λ|2[u, u]−λ[u, x−u
0]−λ¯[x−u0, u] > 0. So, taking λ = α[x−u0, u], we have that for every
α ∈ R with α 6= 0, α2[u, u]−α >0 . Therefore x−u0 ∈ U⊥ and so u0 is a projection of x on
U. Clearly this projection is unique since U is non-degenerate.
Corollary 2.7.4. LetX be an inner product space; U ⊆X be a negative definite subspace and
x ∈ X. Then x admits a projection on U if and only if the function ϕ : U −→ R, ϕ(u) := [x−u, x−u] attains its maximum at a uniqueu0 ∈U. This u0 is the unique projection of xon
U.
Theorem 2.7.5. Let X be an inner product space and U, U1, . . . , Un ⊆ X be subspaces such
that U = U1⊕ · · ·˙ ⊕U˙ n. Let x ∈ X and u ∈ U; then u is a projection of x on U if and only if
there exists ui projection of x onUi (i= 1, . . . , n) such that u=u1+· · ·+un.
Proof. Note that, for every i= 1, . . . , n, U⊥ ⊆Ui⊥ since Ui ⊆U.
⇒ Suppose that u is a projection of x on U and let v ∈ U⊥ such that x = u+v. As U =
U1⊕ · · ·˙ ⊕U˙ n, there exist unique vectors u1, . . . , un such that u = u1+· · ·+un and, for i 6= j,
[ui, uj] = 0. So x=u1+· · ·+un+v and u1, . . . , un,v are orthogonal to each other. Therefore,
for every i= 1, . . . , n,ui is a projection of x onUi.
⇐ Suppose that there exists ui ∈ Ui projection of x on Ui (i = 1, . . . , n) such that u =
u1+· · ·+un. Note that for every i= 1, . . . , n, if wi ∈Ui then [wi, x−u] = [wi, x−ui] = 0 since
x−ui ∈Ui⊥; so x−u⊥Ui, i = 1, . . . , n, and therefore x−u⊥ U. Henceu is a projection of x
2.8
Ortho-Complemented Subspaces
Definition 2.8.1. Let X be an inner product space andU ⊆X be a subspace. If U +U⊥ =X
then U is called an ortho-complemented subspace of X.
Remarks:
• Note that a subspace is ortho-complemented if and only if it admits an orthogonal
comple-mentary subspace (this is a complecomple-mentary subspace contained in U⊥).
• If U ⊆ X is a non-degenerate subspace then U is ortho-complemented if and only if X =
U⊕U˙ ⊥.
• IfU is a neutral subspace then, asU ⊆U⊥,U is ortho-complemented if and only ifU⊥=X
if and only if U ⊆X0.
• If U, U1, . . . , Un ⊆ X are subspaces such that U = U1⊕ · · ·˙ ⊕U˙ n, then U is
ortho-complemented if and only if each Ui (i= 1, . . . , n) is ortho-complemented.
• Clearly, if U is an ortho-complemented subspace of an inner product space X then every
x∈X has at least one projection on U.
• If U is ortho-complemented then, as U ⊆U⊥⊥,U⊥ is ortho-complemented too.
Proposition 2.8.2. Let X be an inner product space and U ⊆ X be a subspace. If U is ortho-complemented then U0 ⊆X0.
Proof. Suppose thatU is ortho-complemented, that isX =U+U⊥, and letu∈U0 =U∩U⊥. For
every w∈U and z ∈U⊥ we have [w+z, u] = [w, u] + [z, u] = 0 + 0 = 0; so U0 ⊆X⊥=X0. Corollary 2.8.3. IfX is a non-degenerate inner product space then every ortho-complemented subspace is non-degenerate.
Proposition 2.8.4. Let X be a non-degenerate inner product space and U ⊆ X an ortho-complemented subspace. Then U⊥⊥=U.
Proof. By corollary 2.8.3, as X is non-degenerate,U and U⊥ are non-degenerate subspaces such that X =U⊕U˙ ⊥.
ClearlyU ⊆U⊥⊥. Suppose, towards a contradiction, thatU (U⊥⊥; thenU⊥∩U⊥⊥ 6={0} .
Proposition 2.8.5. Every definite subspace of finite dimension is ortho-complemented.
Proof. Let U be a definite subspace of an inner product space X such that n = dim(U) < ∞. Suppose, without loss of generality, that U is positive definite. Since U is finite-dimensional, it has a orthonormal basis (u1, . . . , un). Note that if x∈X then u= [x, u1]u1 +· · ·+ [x, un]un is a
projection of xof U since [x−u, u] = 0. Hence, U is ortho-complemented.
2.9
Fundamental Decompositions
Definition 2.9.1. An inner product spaceX isdecomposable if and only if there exists a positive definite subspaceX+ ⊆Xand a negative definite subspaceX− ⊆Xsuch thatX =X0⊕X˙ +⊕X˙ −.
A decomposition of this form is called a fundamental decomposition of X.
Remark. If X is a non-degenerate and decomposable inner product space, then every funda-mental decomposition ofX is of the formX =X+⊕X˙ − whereX+ is a positive definite subspace
and X− is a negative definite subspace.
Lemma 2.9.2. Let X be an inner product space such that X = N⊕X˙ +⊕X˙ − where N is a
neutral subspace, X+ is a positive definite subspace and X− is a negative definite subspace.
Then N =X0 and thereforeX is decomposable.
Proof. Since N is neutral then N0 = N (by corollary 2.2.10) and, as X+ and X− are definite subspaces, (X+)0 = (X−)0 = {0}. Then, by proposition 2.4.2, X0 = N0⊕˙(X+)0⊕˙(X−)0 = N.
So X =X0⊕X˙ +⊕X˙ − and therefore X is decomposable.
Theorem 2.9.3. Let X be an inner product space and U ⊆ X be a positive definite (negative definite) subspace. Then X has a fundamental decomposition with X+ = U (X− = U) if and
i) U is maximal positive definite (maximal negative definite);
ii) U is ortho-complemented.
Proof. Suppose, without loss of generality, that U is positive definite.
⇒ Suppose thatX has a fundamental decomposition X =X0⊕X˙ +⊕X˙ −
with X+=U; clearly,
U is ortho-complemented. Suppose, towards a contradicition, that U is not maximal positive definite and let W ⊆X be a positive definite subspace such that W )U; then W∩(X0⊕X˙ −)6= {0}. Let w 6= 0 be such that w ∈W ∩(X0⊕X˙ −), then [w, w]> 0 (since W is positive definite)
and [w, w]≤0 (sinceX0⊕X˙ − is negative semi-definite) .
⇐ Suppose that U is a maximal positive definite and ortho-complemented subspace; then U⊥
is negative (by proposition 2.6.3) and there exists V ⊆ U⊥ subspace such that X = U⊕V˙ . Let N ⊆ V a maximal neutral subspace of V, then there exists X− ⊆ V subspace such that
V = N⊕X˙ − (by proposition 2.2.8, N ⊥ V since V is semi-definite); clearly X− is negative
definite.
So, settingX+ =U, we have X =N⊕X˙ +⊕X˙ − whereX+ is positive definite andX− is negative
definite. By lemma 2.9.2, N = X0. Therefore X has a fundamental decomposition with X+ =
U.
Remarks:
• By theorem 2.9.3, an inner product space X is decomposable if and only if it contains
ortho-complemented maximal definite subspaces.
• IfX is decomposable with fundamental decompositionX =X0⊕X˙ +⊕X˙ −then, by theorem
2.9.3, X+ is maximal positive definite and X−
is maximal negative definite.
• Let X be a non-degenerate and decomposable inner product space and let X = X+⊕X˙ +
be a fundamental decomposition of X. Let k·kX+ and k·kX− be the norms induced by the
inner product on X+ and X−, respectively (see remarks before definition 2.2.12). Then X is endowed with a norm given by the formula kxk = kx+k
X+ +kx−kX−, where x ∈ X,
Corollary 2.9.4. Every inner product space of finite dimension is decomposable.
Proof. LetX be an inner product space of finite dimension. LetU be a maximal positive definite subspace of X (such a subspace exists sinceX is finite-dimensional); thenU is finite-dimensional and, by proposition 2.8.5, U is orthocomplemented. Hence, by the theorem 2.9.3, X is decom-posable.
Corollary 2.9.5. Every finite dimensional non-degenerate subspace of an inner product space is ortho-complemented.
Proof. Let X be an inner product space and let U be a non-degenerate and finite-dimensional subspace of X. Then, by corollary 2.9.4,U is decomposable; that is,U =U+⊕U˙ −. SinceU+ is a
definite subspace ofXwith finite dimension then, by proposition 2.8.5,U+is ortho-complemented in X. So, X =U+⊕˙(U+)⊥
.
Note that U− ⊆ (U+)⊥ and, again by proposition 2.8.5, U− is ortho-complemented in (U+)⊥;
that is, there existsV ⊆(U+)⊥such that(U+)⊥ =U−⊕V˙ . Therefore,X =U+⊕U˙ −⊕V˙ =U⊕V˙ ;
that is, U is ortho-complemented.
2.10
Orthonormal Systems
Definition 2.10.1. Let(X,[·,·]) be a inner product space and let(eγ)γ∈Γ be a indexed family of
vectors of X. The family(eγ)γ∈Γ is called an orthonormal system if and only if |[eγ, eγ0]|=δγ,γ0.
Remark. Clearly, every orthonormal system in an inner product space is a linearly independent
subset.
Lemma 2.10.2. Let X be a non-degenerate inner product space and let x ∈ X be a non-zero neutral vector. Then there exists an orthonormal system (u, v)⊆X such thatx∈Span({u, v}).
Proof. Since X is non-degenerate, there exists y ∈ X such that [x, y] 6= 0 and, by the corollary 2.2.9, U = Span({x, y}) is a indefinite subspace. Clearly, dim(U) = 2.
Let u ∈ U be such that [u, u] = 1. Since Span({u}) is a positive definite subspace of U, it is ortho-complemented (by the proposition 2.8.5) and, moreover, it is a maximal positive definite
subspace ofU. So, by the proposition 2.6.3,Span({u})⊥is a negative semi-definite 1-dimensional subspace of U and, sinceU is indefinite, it is negative definite. Let v ∈Span({u})⊥ be such that [v, v] = −1. Then (u, v) ⊆ U ⊆X is an orthonormal system such that U = Span({u, v}) (since
{u, v} is a linearly independent subset) and, therefore, x∈Span({u, v}).
Theorem 2.10.3. LetX be a non-degenerate inner product space and(xn)n∈N ⊆X. Then there exists a countable orthonormal system (ej)j such that, for every n∈N, xn∈Span((ej)j).
Proof. Suppose, without loss of generality, that xn6= 0 for every n∈N.
• If x1 is not neutral, let e1 = |[x1, x1]|−1/2x1. Otherwise, by the lemma 2.10.2, there is an
orthonormal system {e1, e2)} such thatx1 ∈Span({e1, e2}).
• Suppose that for some n ∈ N there is a finite orthonormal system {e1, . . . , ek} such that
its span Uk contains the vectors x1, . . . , xn. If xj ∈ Uk for every j ∈ N, then the
as-sertion is clear. Otherwise, let jn be the first index with xjn 6∈ Uk. Let yn = xjn −
Pk
j=1[ej, ej][xjn, ej]ej; then yn6= 0 and yn ⊥Uk.
Since Uk = Span({e1}) ˙⊕ · · ·⊕˙ Span({ek}) then, by corollary 2.4.3, Uk is non-degenerate.
By corollary 2.9.5, Uk is ortho-complemented. Therefore, Uk⊥ is non-degenerate. If yn is
not neutral then let ek+1 = |[yn, yn]|−1/2yn. Otherwise, by the lemma 2.10.2, there is a
orthonormal system {ek+1, ek+2} ⊆ Uk⊥ such that yn ∈ Span({ek+1, ek+2}). In any case,
a finite extension for the orthonormal system {e1, . . . , ek} is obtained such that its span
Chapter 3
Linear Operators on Inner Product Spaces
3.1
Linear Operators
Definition 3.1.1. LetX be a vector space and let T be a linear operator acting in X. Then:
i) T isinvertible if and only if T is injective [that is,ker(T) = {0}].
ii) T iscompletely invertible if and only if D(T) = R(T) =X and T is invertible.
Definition 3.1.2. LetT be a linear operator acting in a vector space X and let λ∈C. Then:
i) The subspaceXλ :=
S
n∈Nker(T −λI)
nis called theprincipal subspace (orroot subspace) of
T associated to λ, and its non-zero vectors are called the principal vectors (or root vectors) of T associated to λ. The dimension of Xλ is called the algebraic multiplicity of λ.
ii) The complex numberλis called aneigenvalue of T if and only if the linear operator T−λI
is not invertible. In this case the subspaceker(T−λI)is called theeigenspace associated to
λ, and its non-zero elements are called the eigenvectors ofT associated toλ. The dimension of ker(T −λI) is called the geometric multiplicity of λ.
Remarks:
• Clearly ker(T −λI)⊆Xλ and, therefore, the geometric multiplicity is less than or equal to
• Clearly ifker(T −λI)6={0} thenXλ 6={0}; moreover, ifXλ 6={0}then ker(T −λI)6={0}
[let x 6= 0 be such that x ∈ Xλ and let p ∈ N the minimum positive integer such that
(T −λI)px= 0; ifp= 1 then x∈ker(T −λI)and if p >1 theny:= (T −λI)p−1x6= 0 and
y∈ker(T −λI)].
• An eigenvalue λ of T is called semi-simple if, and only if,Xλ = ker(T −λI).
• Clearly, the principal subspaces of a linear operator are linearly independent.
• The set of all eigenvalues of the operator T is denoted by σp(T).
Definition 3.1.3. Let T be a linear operator acting in a vector space X and let λ ∈ C be an eigenvalue of T. A Jordan chain of T associated to λ with length p ∈ N is a finite sequence of non-zero vectors in Xλ, (x1, . . . , xp), such that xk+1 = (T −λI)xk (k = 1, . . . , p −1) and
(T −λI)xp = 0.
Proposition 3.1.4. Let(x1, . . . , xp)be a Jordan chain of T associated to the eigenvalue λ∈C.
Then x1, . . . , xp are linearly independent vectors.
Proof. Letα1, . . . , αp ∈Csuch thatα1x1+· · ·+αpxp = 0; applying(T−λI)p−1 we haveα1xp = 0
and, as xp 6= 0, α1 = 0. Letk ≤p and suppose thatα1 =· · · =αk−1 = 0. Applying (T −λI)p−k
we obtain αkxp = 0 and thus αk= 0. Therefore x1, . . . , xp are linearly independent vectors.
Definition 3.1.5. Let T be a linear operator acting in a vector space X and let U ⊆ X be a subspace. We say that the subspace U is invariant under T (or is T-invariant) if and only if
T(U ∩D(T))⊆U.
Definition 3.1.6. Let T be a linear operator acting in X and U1, . . . , Un ⊆ X subspaces such
that X =U1u· · ·uUn. The direct decomposition X =U1u· · ·uUn reduces the operator T if
and only if:
1) The subspacesU1, . . . , Un are T-invariant;
2) D(T) = (D(T)∩U1)u· · ·u(D(T)∩Un).
Remarks:
• Note that if Y, U1, . . . , Un ⊆X are subspaces with X =U1u· · ·uUn then Y ⊇(Y ∩U1)u
· · ·u(Y ∩Un); but is not necessary thatY = (Y ∩U1)u· · ·u(Y ∩Un). For example take
X =C2,U1 ={(u,0) :u∈C}, U2 ={(0, v) :v ∈C}) andY ={(u, u) :u∈C}.
• LetU1, . . . , Unbe subspaces ofX such thatX =U1u· · ·uUnand letT be a linear operator
acting in X with D(T) = (D(T)∩U1)u· · ·u(D(T)∩Un). Then, given u ∈ D(T) there
exist uniqueui, vi ∈Ui (i= 1, . . . , n) such that u=u1+· · ·+un and T(u) = v1+· · ·+vn.
Let Tji :D(T)∩Ui −→Uj be the linear operator such that u∈ Ui 7−→Tji(u) ∈Uj where
Tji(u) is the j-component in the direct decomposition of T(u) in U1 u · · ·uUn. So, if
u=u1+· · ·+un∈(D(T)∩U1)u· · ·u(D(T)∩U1) = D(T) then:
T(u) = Pn
i=1T(ui) =
Pn
i=1
Pn
j=1Tji(ui) =
Pn
j=1
Pn
i=1Tji(ui).
3.2
Isometric Operators
Recall 3.2.1. A linear operatorT :D(T)⊆X→X is anisometryif and only if[T x, T y] = [x, y] for every x, y ∈D(T).
Remarks:
• Clearly, an isometric operator acting in an inner product space maps positive (negative,
neutral) vectors in positive (negative, neutral) vectors.
• Note that an isometric linear operator need not be invertible. For example, in a non-zero
neutral inner product space the mapping 0is isometric but it is not invertible.
• If T is an invertible isometric linear operator, then it is clear that T−1 is isometric.
Proposition 3.2.2. Let T be an isometric linear operator acting in X and let u, v be isotropic and non-isotropic vectors of D(T), respectively. Then T u, T v are isotropic and non-isotropic vectors of R(T), respectively.
Proof. For every x ∈ D(T), as T is isometric and u is an isotropic vector of D(T), [T x, T u] = [x, u] = 0; so T u is an isotropic vector of R(T).
As v ∈D(T)is a non-isotropic vector of D(T), letx0 ∈D(T)such that [x0, v]6= 0; then, sinceT
is isometric, [T x0, T v] = [x0, v]6= 0. So T v is a non-isotropic vector ofR(T).
Corollary 3.2.3. If T is an isometric linear operator acting in an inner product space X and D(T) is non-degenerate, thenT is invertible.
Proof. Let x ∈ D(T) with x 6= 0; as x is non-isotropic then, by proposition 3.2.2, T x is non-isotropic and therefore T x6= 0.
Theorem 3.2.4. LetT be a isometric operator acting in a inner product spaceXand letλ, µ∈C
be eigenvalues of T such that λµ6= 1. Then Xλ ⊥Xµ.
Proof. Given u∈Xλ and v ∈Xµ there are p, q ∈Nsuch that (T −λ)pu= 0and (T −µ)qv = 0,
then we have show that [u, v] = 0. By induction on n:=p+q:
• If n = 2 then p = q = 1; so if u ∈ Xλ and v ∈ Xµ are such that (T −λ)u = 0 and
(T −µ)v = 0, then [u, v] = [T u, T v] = [λu, µv] =λµ[u, v]. Thus, as λµ6= 1, [u, v] = 0.
• If n ≥ 2, suppose that the affirmation is true for all p+q ≤n. Now assume that u ∈Xλ,
v ∈ Xµ are such that (T −λ)pu = 0 and (T −µ)qv = 0 with p+q = n+ 1. So with
u0 = (T −λ)u and v0 = (T − λ)v we have [u, v0] = [u0, v] = [u0, v0] = 0 and, therefore, [u, v] = [T u, T v] = [u0+λu, v0+µv] =λµ[u, v]. Thus [u, v] = 0 since λµ6= 1, .
Corollary 3.2.5. If T is an isometric operator acting in an inner product space X and λ ∈ C
with |λ| 6= 1 is an eigenvalue of T, then Xλ is neutral.
Proposition 3.2.6. LetT be an isometric operator acting in an inner product space X and let
U ⊆X be a subspace such thatU ⊆T(U∩D(T)). Then U⊥ isT-invariant.
Proof. Let v ∈ U⊥∩D(T). If u ∈ U then there exists w ∈ U ∩D(T) such that u = T w; so [u, T v] = [T w, T v] = [w, v] = 0. Thus, T v ∈U⊥.
Proposition 3.2.7. Let T be an isometric operator acting in an inner product space X and let λ ∈ C be an eigenvalue of T. If ker(T −λI) is a definite subspace then |λ| = 1 and λ is semi-simple.
Proof. Suppose that ker(T −λ) is a definite subspace; then, by corollary 3.2.5, |λ| = 1. Now suppose, towards a contradiction, that λ is not semi-simple. Let x∈ Xλ \ker(T −λ) and p≥2
such that (T −λ)px= 0 and (T −λ)p−1x6= 0. Then,
0 = −[(T −λ)px, λT(T −λ)p−2x]
=−[T(T −λ)p−1x, λT(T −λ)p−2x] +λ[(T −λ)p−1x, λT(T −λ)p−2x] =−[(T −λ)p−1x, λ(T −λ)p−2x] + [(T −λ)p−1x, T(T −λ)p−2x] = [(T −λ)p−1x, (T −λ)p−1x]6= 0 ( ),
where the last inequality follows because(T−λ)p−1xbelongs to the definite spaceker(T−λ).
3.3
Symmetric Operators
Definition 3.3.1. Let T be a linear operator acting in an inner product space X. Then T is called symmetric if and only if [T x, y] = [x, T y]for every x, y ∈D(T).
Remarks:
• If T is a symmetric operator acting in a Hilbert space then σp(T)⊆R, but in an arbitrary
inner product space this can be false. For example, in a two-dimensional inner product space (X,[·,·]) with basis {e1, e2} such that [e1, e1] = [e2, e2] = 0 and [e1, e2] = 1, we
have that the linear operator T which maps e1 to ie1 and e2 to −ie2 is symmetric (since
[T(α1e1 +α2e2), β1e1 +β2e2] = [iα1e1 −iα2e2, β1e1 + β2e2] = iα1β2 − iα2β1 = [α1e1 +
α2e2, iβ1e1−iβ2e2] = [α1e1+α2e2, T(β1e1 +β2e2)]), but its eigenvalues are ±i∈C\R.
• IfT is a symmetric operator acting in a Hilbert spaceH and λ,µare eigenvalues of T with
λ6=µthenHλ ⊥ Hµ. The following proposition generalizes this for arbitrary inner product
Proposition 3.3.2. LetT be a symmetric operator acting in an inner product spaceX and let
λ, µ∈C be eigenvalues ofT such thatλ =6 µ. ThenXλ ⊥Xµ.
Proof. Suppose, without loss of generality, thatλ 6= 0. Letu∈Xλ,v ∈Xµandp, q ∈Nsuch that (T −λ)pu= 0 and (T −µ)qv = 0. We have to show that[u, v] = 0. By induction on n:=p+q:
• If n = 2 then p = q = 1; so if u ∈ Xλ and v ∈ Xµ are such that (T −λ)u = 0 and
(T −µ)v = 0, then [u, v] = λ1[λu, v] = λ1[T u, v] = 1λ[u, T v] = λ1[u, µv] = µλ[u, v]. Thus, as
µ
λ 6= 1,[u, v] = 0.
• If n ≥ 2, suppose that the affirmation is true for all p+q ≤ n. Now let u ∈ Xλ, v ∈ Xµ
and assume (T −λ)pu= 0and (T −µ)qv = 0for integers p, q such that p+q =n+ 1. So with u0 = (T −λ)u and v0 = (T −λ)v we have [u, v0] = [u0, v] = [u0, v0] = 0 and, therefore, [u, v] = 1
λ[T u−u
0, v] = 1
λ[T u, v] =
1
λ[u, T v] =
1
λ[u, v
0 +µv] = 1
λ[u, µv] = µ
λ[u, v]. Thus, as µ
λ 6= 1,[u, v] = 0.
Corollary 3.3.3. If λ is a non-real eigenvalue of a symmetric operator T acting in an inner product space X, then Xλ is neutral.
The following proposition is similar to the proposition 3.2.6.
Proposition 3.3.4. LetT be a symmetric operator acting in an inner product spaceX and let
U ⊆X be aT-invariant subspace such thatU ⊆D(T). Then U⊥ is T-invariant.
Proof. Let v ∈ U⊥ ∩D(T). For every u ∈ U, as U is T-invariant and U ⊆ D(T), [u, T v] = [T u, v] = 0. Thus, U⊥ is T-invariant.
Proposition 3.3.5. Let T be a symmetric operator acting in an inner product space X and let λ ∈ C be an eigenvalue of T. If ker(T −λI) is a definite subspace, then λ ∈ R and λ is semi-simple.
Proof. Suppose that ker(T −λ) is a definite subspace; then, by corollary 3.3.3,λ ∈R.
Now suppose, towards a contradiction, that λ is not semi-simple. Let x ∈ Xλ \ker(T −λ) and
[(T−λ)p−1x, (T−λ)(T−λ)p−2x] = [(T−λ)p−1x, T(T −λ)p−2x]−λ[(T−λ)p−1x, (T−λ)p−2x] =
[T(T −λ)p−1x, (T −λ)p−2x]−λ[(T −λ)p−1x, (T −λ)p−2x] = λ[(T −λ)p−1x, (T −λ)p−2x]− λ[(T −λ)p−1x, (T −λ)p−2x] = 0 .
Definition 3.3.6. Let(X,[·,·])be an inner product space and let G:X −→X be a symmetric linear operator. In X we define an inner product [·,·]G by [x, y]G := [Gx, y] for every x, y ∈ X.
This inner product is called the G-inner product onX.
Remark. Every concept related to the G-inner product is prefixed by G and every symbol has the subindex G. For example, the vectors x and y are G-orthogonal (write x ⊥G y) if and only
[x, y]G = 0.
Proposition 3.3.7. LetX be an inner product space; U, V ⊆X be subspaces andG:X −→X
be a symmetric operator. If U ⊥V and U isG-invariant, then U ⊥G V.
Proof. For every u∈U and v ∈V we have[u, v]G = [Gu, v] = 0 sinceG(U)⊆U and U ⊥V.
3.4
Orthogonal and Fundamental Projectors
Definition 3.4.1. Let X be an inner product space and let P : X −→X be a linear operator.
P is called an orthogonal projector in X if and only if P is symmetric and P2 =P.
Theorem 3.4.2. LetP be an orthogonal projector in an inner product spaceX. Then R(P)is ortho-complemented and, for every x∈X,P x is a projection of xon R(P).
Conversely, if X is a non-degenerate inner product space, U ⊆ X is an ortho-complemented subspace and PU is the mapping that carries each vector of X into its projection on U, then PU
is an orthogonal projector in X with R(PU) =U.
Proof. Letx∈X, thenx=P x+ (x−P x)and, for everyz ∈X, asP is an orthogonal projector, [P z, x−P x] = [z, P x−P2x] = [z, P x−P x] = 0. So x =P x+ (x−P x) with P x∈ R(P) and
x−P x∈ R(P)⊥. Thus R(P) is ortho-complemented and, for every x ∈ X, P x is a projection of x onR(P).
Conversely, suppose that X is a non-degenerate inner product space and U ⊆ X is an ortho-complemented subspace; then U is non-degenerate and X = U⊕U˙ ⊥. So the mapping P
U that
carries each vector of X into its projection on U is well-defined, clearly it is linear, R(PU) = U
andP2 =P. Now, ifx=u+v ∈U⊕U˙ ⊥andy=z+w∈U⊕U˙ ⊥then[P
Ux, y] = [u, z+w] = [u, z]
and [x, PUy] = [u+v, z] = [u, z]; thus PU is symmetric. Therefore PU is an orthogonal projector
in X.
Remark. Let X be a non-degenerate and decomposable inner product space and let X =
X+⊕X˙ −
be a fundamental decomposition of X. Then, by the theorem 3.4.2, there exist P+, P−
orthogonal projectors in X such that R(P+) = X+ and R(P−) = X−. Clearly, P++P− = I
and P+P− =P−P+= 0.
Definition 3.4.3. The orthogonal projectors, P+ and P−, in the preceding remark are called
the fundamental projectors associated with the fundamental decompositionX =X+⊕X˙ −.
Remark. LetX be a non-degenerate and decomposable inner product space and letP+ and P−
be the fundamental projectors associated with the fundamental decomposition X = X+⊕X˙ −. Let k·k, k·kX+ and k·kX− be the norms in the remarks after the theorem 2.9.3; then, for every
x ∈ X, kP+xk = kx+k = kx+k
X+ ≤ kxk and therefore P+ is a bounded linear operator in the
norm space (X,k·k). Analogously, P− is so.
Theorem 3.4.4. Let X be a non-degenerate and decomposable inner product space, let X =
X+⊕X˙ − be a fundamental decomposition of X, and let P+, P− be the fundamental projectors
associated with X = X+⊕X˙ −. If U, V ⊆ X are positive and negative semi-definite subspaces,
respectively, then P+
U and P−V are invertible. Further, dim(U) ≤ dim(X+) and dim(V) ≤
dim(X−).
Proof. LetU be a positive semi-definite subspace ofX. Let u∈U,u=u++u− where u+∈X+
and u− ∈ X−. If P+u= 0 then 0 = P+(u++u−) = u+, therefore u = u− and so u ∈ U ∩X−.
Hence u= 0; that isP+
U is invertible.
Now, clearly, dim(U) = dim(P+(U)) ≤ dim(X+) since P+(U) ⊆ X+. Analogously, dim(V) ≤
dim(X−).
Remark. Clearly, the theorem 3.4.4. holds if U and V are definite subspaces.
3.5
Fundamental Symmetries
Definition 3.5.1. Let X be a non-degenerate and decomposable inner product space, X =
X+⊕X˙ − a fundamental decomposition of X and P+, P− the fundamental projectors associated
with X =X+⊕X˙ −. The operatorJ :=P+−P− is called thefundamental symmetry associated
with X =X+⊕X˙ −. Remarks:
• If J is the fundamental symmetry associated with the fundamental decomposition X =
X+⊕X˙ −, then for everyx=x++x− we haveJ x =x+−x−, wherex+∈X+ andx− ∈X−. • Clearly, if X+ 6= {0} then 1 is an eigenvalue of J and its associated eigenspace is X+; similarly, if X− 6= {0} then −1 is an eigenvalue of J and its associated eigenspace is X−. Further, σ(J) = σp(J)⊆ {−1,1}.
Thus, if the fundamental symmetry associated with a fundamental decomposition is known then the fundamental decomposition can be determined, since thenx+ = 12(x+J x)andx−= 12(x−J x) for every x∈X.
Proposition 3.5.2. Every fundamental symmetryJ of a non-degenerate and decomposable inner product spaceX is completely invertible withJ−1 =J. Moreover,J is symmetric and isometric. Proof. Let X = X+⊕X˙ − be a fundamental decomposition of X and let J be its associated
fundamental symmetry. Clearly, D(J) = R(J) = X and J2 = I; therefore J is completely invertible with J−1 =J.
If x=x++x−
and y=y++y−
, with x+,y+∈X+ and x−
, y− ∈X−, then