6
which shows thatPjis hermitian.
Consider an orthonormal basisBM = {|ei}mi=1 for M, and extend it to a basisB= {|ei}Ni=1forV. Now construct a (hermitian) projection opera- torP=M
i=1|eiei|. This is the operator that projects an arbitrary vector inV onto the subspaceM. It is straightforward to show that 1−Pis the projection operator that projects ontoM⊥(see Problem6.1).
An arbitrary vector|a ∈Vcan be written as
|a =(P+1−P)|a =P|a
inM
+(1 −P)|a
inM⊥
.
Furthermore, the only vector that can be in bothMand M⊥ is the zero vector, because it is the only vector orthogonal to itself. We thus have Proposition 6.1.3 IfVis an inner product space,thenV=M⊕M⊥ for any subspaceM.Furthermore,the projection operators corresponding to MandM⊥are hermitian.
This section explores the possibility of obtaining subspaces by means of the action of a linear operator on vectors of an N-dimensional vector space V. Let |a be any vector in V, and A a linear operator onV. The vectors
|a,A|a,A2|a, . . . ,AN|a
are linearly dependent (there areN+1 of them). LetM≡Span{Ak|a}Nk=0. It follows that,m≡dimM≤dimV, andMhas the property that for any vector |x ∈M the vector A|x also belongs to M (show this!). In other words, no vector inM“leaves” the subspace when acted on byA.
Definition 6.1.4 A subspaceMis aninvariant subspace of the operator invariant subspace;
reduction of an operator AifAtransforms vectors ofMinto vectors ofM. This is written succinctly asA(M)⊂M. We say thatMreducesAif bothMandM⊥are invariant subspaces ofA.
Starting with a basis ofM, we can extend it to a basisB= {|ai}Ni=1ofV whose firstmvectors spanM. The matrix representation ofAin such a ba- sis is given by the relationA|ai =N
j=1αj i|aj,i=1,2, . . . , N. Ifi≤m, thenαj i=0 forj > m, becauseA|aibelongs toMwheni≤mand there- fore can be written as a linear combination of only {|a1,|a2, . . . ,|am}. Thus, the matrix representation ofAinBwill have the form
A=
A11 A12 021 A22
,
whereA11 is anm×mmatrix,A12anm×(N−m)matrix,021 the(N− matrix representation of
an operator in a subspace
m)×mzero matrix, andA22 an(N−m)×(N−m)matrix. We say that A11representsthe operatorAin them-dimensional subspaceM.
It may also happen that the subspace spanned by the remaining basis vectors inB, namely|am+1,|am+2, . . . ,|aN, is also an invariant subspace
ofA. ThenA12will be zero, andAwill take ablock diagonalform:1
block diagonal matrix defined
A=
A11 0 0 A22
.
If a matrix representing an operator can be brought into this form by a suitable choice of basis, it is calledreducible; otherwise, it is called ir-
reducible and irreducible matrices
reducible. A reducible matrixAis denoted in two different ways:2 A=
A1 0 0 A2
⇔ A=A1⊕A2. (6.3)
For example, whenMreducesAand one chooses a basis the firstmvectors of which are inMand the remaining ones inM⊥, thenAis reducible.
We have seen on a number of occasions the significance of the hermitian conjugate of an operator (e.g., in relation to hermitian and unitary operators).
The importance of this operator will be borne out further when we study the spectral theorem later in this chapter. Let us now investigate some properties of the adjoint of an operator in the context of invariant subspaces.
Lemma 6.1.5 A subspaceMof an inner product spaceVis invariant under
condition for invariance the linear operatorAif and only ifM⊥is invariant underA†.
Proof The proof is left as a problem.
An immediate consequence of the above lemma and the two identities (A†)†=Aand(M⊥)⊥=Mis contained in the following theorem.
Theorem 6.1.6 A subspace of V reduces A if and only if it is invariant under bothAandA†.
Lemma 6.1.7 Let Mbe a subspace ofV andP the hermitian projection operator ontoM.ThenMis invariant under the linear operatorAif and only ifAP=PAP.
Proof SupposeMis invariant. Then forany|xinV, we have P|x ∈M ⇒ AP|x ∈M ⇒ PAP|x =AP|x. Since the last equality holds for arbitrary|x, we haveAP=PAP.
Conversely, supposeAP=PAP. For any|y ∈M, we have P|y = |y ⇒ AP
=PAP
|y =A|y =P AP|y
∈M.
Therefore,Mis invariant underA.
1From now on, we shall denote all zero matrices by the same symbol regardless of their dimensionality.
2It is common to use a single subscript for submatrices of a block diagonal matrix, just as it is common to use a single subscript for entries of a diagonal matrix.
Theorem 6.1.8 LetMbe a subspace ofV,Pthe hermitian projection op- erator ofVontoM,andAa linear operator onV.ThenMreducesAif and only ifAandPcommute.
Proof SupposeMreducesA. Then by Theorem6.1.6,Mis invariant under bothAandA†. Lemma6.1.7then implies
AP=PAP and A†P=PA†P. (6.4)
Taking the adjoint of the second equation yields(A†P)†=(PA†P)†, orPA= PAP. This equation together with the first equation of (6.4) yieldsPA=AP.
Conversely, suppose that PA=AP. Then P2A=PAP, whence PA= PAP. Taking adjoints gives A†P=PA†P, because P is hermitian. By Lemma6.1.7,Mis invariant under A†. Similarly, from PA=AP, we get PAP=AP2, whencePAP=AP. Once again by Lemma6.1.7,Mis invari-
ant underA. By Theorem6.1.6,MreducesA.
6.2 Eigenvalues and Eigenvectors
The main goal of the remaining part of this chapter is to prove that certain kinds of operators, for example a hermitian operator, is diagonalizable, that is, that we can always find an (orthonormal) basis in which it is represented by a diagonal matrix.
Let us begin by considering eigenvalues and eigenvectors, which are gen- eralizations of familiar concepts in two and three dimensions. Consider the operation of rotation about the z-axis by an angle θ denoted by Rz(θ ).
Such a rotation takes any vector (x, y) in the xy-plane to a new vector (xcosθ−ysinθ, xsinθ+ycosθ ). Thus, unless(x, y)=(0,0)orθ is an integer multiple of 2π, the vector will change. Is there a nonzero vector that is so special (eigen, in German) that it does not change when acted on by Rz(θ )? As long as we confine ourselves to two dimensions, the answer is no.
But if we lift ourselves up from the two-dimensionalxy-plane, we encounter many such vectors, all of which lie along thez-axis.
The foregoing example can be generalized to any rotation (normally specified by Euler angles). In fact, the methods developed in this section can be used to show that a general rotation, given by Euler angles, always has an unchanged vector lying along the axis around which the rotation takes place. This concept is further generalized in the following definition.
Definition 6.2.1 Let A∈End(V) be a linear transformation, and |a a nonzerovector such that
eigenvector and
eigenvalue A|a =λ|a, (6.5)
withλ∈C. We then say that|ais aneigenvectorofAwitheigenvalueλ.
Proposition 6.2.2 Add the zero vector to the set of all eigenvectors ofA belonging to the same eigenvalueλ,and denote the span of the resulting set byMλ.ThenMλ is a subspace ofV,and every(nonzero)vector inMλis an eigenvector ofAwith eigenvalueλ.
Proof The proof follows immediately from the above definition and the def-
inition of a subspace.
Definition 6.2.3 The subspaceMλ is referred to as the eigenspaceof A eigenspace corresponding to the eigenvalueλ. Its dimension is called the geometric
multiplicityofλ. An eigenvalue is calledsimpleif its geometric multiplic-
ity is 1. The set of eigenvalues ofAis called thespectrumofA. spectrum By their very construction, eigenspaces corresponding to different eigen-
values have no vectors in common except the zero vector. This can be demonstrated by noting that if|v ∈Mλ∩Mμforλ=μ, then
0=(A−λ1)|v =A|v −λ|v =μ|v −λ|v =(μ−λ)
=0
|v ⇒ |v =0.
An immediate consequence of this fact is Mλ+Mμ=Mλ⊕Mμ ifλ=μ.
More generally,
Proposition 6.2.4 If{λi}ri=1are distinct eigenvalues of an operatorAand Mi is the eigenspace corresponding toλi,them
M1+ · · · +Mr =M1⊕ · · · ⊕Mr ≡ r
i=1
Mi. (6.6)
In particular,by Proposition2.1.15,the eigenvectors ofAcorresponding to distinct eigenvalues are linearly independent.
Let us rewrite Eq. (6.5) as(A−λ1)|a =0. This equation says that|a is an eigenvector ofAif and only if |a belongs to the kernel ofA−λ1.
If the latter is invertible, then its kernel will consist of only the zero vector, which is not acceptable as a solution of Eq. (6.5). Thus, if we are to obtain nontrivial solutions,A−λ1must have no inverse. This is true if and only if
det(A−λ1)=0. (6.7)
The determinant in Eq. (6.7) is a polynomial inλ, called thecharacteris-
tic polynomialofA. The roots of this polynomial are calledcharacteristic characteristic polynomial and characteristic roots of an operator
rootsand are simply the eigenvalues ofA. Now, any polynomial of degree greater than or equal to 1 has at least one (complex) root. This yields the following theorem.