Download What Does the Spectral Theorem Say?

What Does the Spectral Theorem Say?
Author(s): P. R. Halmos
Source: The American Mathematical Monthly, Vol. 70, No. 3 (Mar., 1963), pp. 241-247
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2313117 .
Accessed: 29/07/2011 13:16
Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp. JSTOR's Terms and Conditions of Use provides, in part, that unless
you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you
may use content in the JSTOR archive only for your personal, non-commercial use.
Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at .
http://www.jstor.org/action/showPublisher?publisherCode=maa. .
Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed
page of such transmission.
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact [email protected].
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to
The American Mathematical Monthly.
http://www.jstor.org
WHAT DOES THE SPECTRAL THEOREM SAY?
of Michigan
P. R. HALMOS, University
Most students of mathematicslearn quite early and most mathematicians
remembertill quite late that every Hermitianmatrix (and, in particular,every
real symmetricmatrix)may be put into diagonal form.A moreprecisestatement
of the resultis that every Hermitianmatrixis unitarilyequivalent to a diagonal
one. The spectraltheoremis widelyand correctlyregardedas the generalization
of this assertionto operatorson Hilbert space. It is unfortunatethereforethat
even the bare statementof the spectraltheoremis widelyregardedas somewhat
mysteriousand deep, and probably inaccessible to the nonspecialist.The purpose of this paper is to try to dispel some of the mystery.
Probably the main reason the generaloperator theoremfrightensmost people is that it does not obviously include the special matrixtheorem.To see the
situationhas
relationbetween the two, the descriptionof the finite-dimensional
to be distorted almost beyond recognition.The result is not intuitive in any
language; neitherStieltjes integralswith unorthodoxmultiplicativeproperties,
nor bounded operatorrepresentationsof functionalgebras,are in the daily toolkit of everyworkingmathematician.In contrast,the formulationof the spectral
theoremgiven below uses only the relativelyelementaryconcepts of measure
theory.This formulationhas been part of the oral traditionof Hilbert space for
quite some time (foran explicittreatmentsee [6]), but it has not been called the
spectral theorem; it usually occurs in the much deeper "multiplicitytheory."
Since the statementuses simple concepts only,this aspect of the presentformulation is an advantage, not a drawback; its effectis to make the spiritof one of
the harder parts of the subject accessible to the student of the easier parts.
Anotherreason the spectral theoremis thoughtto be hard is that its proof
is hard. An assessment of difficultyis, of course, a subjective matter,but, in
any case, thereis no magic new techniquein the pages that follow.It is the statement of the spectral theoremthat is the main concernof the exposition,not the
proof.The proofis essentiallythe same as it always was; most of the standard
methods used to establish the spectral theoremcan be adapted to the present
formulation.
Let 4 be a complex-valuedbounded measurable functionon a measure space
X with measure ,t. (All measure-theoreticstatements,equations, and relations,
e.g., "4 is bounded," are to be interpretedin the "almost everywhere"sense.)
An operatorA is definedon the Hilbert space 22(p) by
(Af )(x)-=*t(x)f (x),
x E- X;
the operatorA is called the multiplicationinduced by 4. The study of the relation between A and 4 is an instructiveexercise. It turnsout, forinstance, that
241
242
WHAT DOES THE SPECTRAL THEOREM SAY?
[March
the adjoint A* of A is the multiplicationinduced by the complex conjugate +
of 4. If ifr
also is a bounded measurable functionon X, with induced multiplication B, then the multiplicationinduced by the product function044is the product operatorAB. It followsthat a multiplicationis always normal; it is Hermitian if and only if the functionthat induces it is real. (For the elementaryconcepts of operatortheory,such as Hermitianoperators,normaloperators,projections, and spectra, see [3]. For presentpurposes a concept is called elementary
ifit is discussed in [3] beforethe spectral theorem,i.e., beforep. 56.)
As a special case let X be a finiteset (with n points, say), and let,t be the
"counting measure" in X (so that,t({x}) =1 for each x in X). In this case
?2(,g) is n-dimensionalcomplex Euclidean space; it is customaryand convenient
to indicate the values of a functionin ?2(M) by indices instead of parenthetical
arguments.With this notation the action on f of the multiplicationA induced
by 4 can be describedby
A(f1, ... ,fn) = (Oifi, ...* ,fn)
To say this with matrices,note that the characteristicfunctionsof the singletons in X forman orthonormalbasis in ?2(M); the assertionis that the matrix
of A with respect to that basis is diag (41, * , On)4
The general notation is now established and the special role of the finitedimensional situation within it is clear; everythingis ready for the principal
statement.
SPECTRAL
THEOREM.
multiplication..
Every Hermitian operatoris unitarilyequivalentto a
In complete detail the theoremsays that if A is a Hermitian operator on a
Hilbert space 3C,then thereexists a (real-valued) bounded measurable function
4 on some measure space X with measure u, and there exists an isometry U
from?2(M) onto JC,such that
(U-1AUf)(x) = 4(x)f(x),
x E X,
foreachf in ?2(M). What followsis an outlineof a proofof the spectral theorem,
a briefdiscussionof its relationto the versioninvolvingspectral measures,and
an illustrationof its application.
Three tools are needed forthe proofof the spectral theorem.
(1) The equalityof normand spectralradius. If the spectrumof A is A(A),
then the spectralradius r(A) is definedby
r(A) = sup {I XI : X E A(A)}A
It is always true that r(A)_?IIAI([3, Theorem 2, p. 52]); the useful fact here
thenr(A) =IIAII
is thatifA is Hermitian,
([3, Theorem2, p. 55]).
theorem
(2) The Riesz representation
for compactsetsin theline. If L is a positive linear functionaldefinedforall real-valued continuousfunctionson a com-
19631
WHAT DOES THE SPECTRAL THEOREM SAY?
243
pact subset X of the real line, then thereexists a unique finitemeasure u on the
Borel sets of X such that
L(f)
=
f
fd1.
forall f in the domain of L. (To say that L is linear means of course that
L(af + fig)= aL(f) + AL(g),
wheneverf and g are in the domain of L and axand 3 are real scalars; to say that
L is positive means that L(f) 0 wheneverfis in the domain of L and f> 0.) For
a proof,see [4, Theorem D, p. 247].
(3) The Weierstrassapproximationtheorem
for compactsetsin theline. Each
real-valued continuousfunctionon a compact subset of the real line is the uniformlimit of polynomials.For a pleasant elementarydiscussion and proofsee
[I, p. 102].
Consider now a Hermitian operator A on a Hilbert space SC. A vector t in
XC is a cyclicvectorforA if the set of all vectors of the formq(A)t, where q runs
over polynomialswith complexcoefficients,
is dense in SC. Cyclic vectors may
not exist, but an easy transfiniteargumentshows that XC is always the direct
sum of a familyof subspaces, each of which reduces A, such that the restriction
of A to each of them does have a cyclic vector. Once the spectral theoremis
knownforeach such restriction,it followseasily forA itself;the measure spaces
that serve forthe directsummandsof a have a natural directsum, whichserves
forSCitself.Conclusion: there is no loss of generalityin assuming that A has a
cyclic vector,say t.
For each real polynomialp write
L(p) = (p(A)%, t).
Clearly L is a linear functional;since
= r(p(A))
I L(p) I ? IIP(A)II
-IIt1I2
_11t112
= sup {JX: XE A(p(A))} IIII2
= sup p(X)I : X E A(A)} .I|1I2,
the functionalL is bounded for polynomials. (The last step uses the spectral
mapping theorem; cf. [3, Theorem 3, p. 551.) It follows (by the Weierstrass
theorem)that L.has a bounded extensionto all real-valued continuousfunctions
on A(A). To prove that L is positive,observe firstthat ifp is a real polynomial,
then
((p(A))2, {)=IP(A){I!2
? 0
If f is an arbitrarypositive continuousfunctionon A(A), then approximate V/f
uniformlyby real polynomials;the inequality just proved impliesthat L(f) >0
244
WHAT DOES THE SPECTRAL THEOREM SAY?
[March
(since f is then uniformlyapproximated by squares of real polynomials). The
Riesz theoremnow yields the existenceof a finitemeasureA.tsuch that
(p(A)t, t)
=
fpdlk
forevery real polynomialp.
For each (possibly complex) polynomialq write
Uq = q(A) .
Since A is Hermitian,(q(A))*(=
'
qf2(A));
it followsthat
f
{(A)) is a polynomialin A, and so is (q(A)) *q(A)
q
qI2dl = (q(A)q(A)%, )
=
((q(A))*q(A)%, ) = jq(A)t|12==|Uq|j2.
This means that the linear transformationU froma dense subset of 22(yt) into
SC is an isometry,and hence that it has a unique isometricextensionthat maps
22(y) into SC.The assumptionthat t is a cyclic vector implies that the range of
U is in fact dense in, and hence equal to, the entirespace E.
It remains only to prove that the transformof A by U is a multiplication.
Write +(X) =X for all X in A(A). Given a complex polynomial q, write g(X)
=Xq(X) =45(X)q(X); then
U-'AUq
= U-'Aq(A)t
= UIq(A)t-
U-=
U = Q.
In otherwords U-'A U agrees, on polynomials,with the multiplicationinduced
by q, and that is enough to conclude that U'-1AU is equal to that multiplication. This completes the outline of the proofof the spectral theoremfor Hermitian operators.
The formulationof the spectral theoremgiven above yields fairlyeasily all
the informationcontained in the morecommonversions.Thus ifA is the multiplication on S2(,) induced by the real function4 on X, and if F is a (complex)
Borel measurable functionthat is bounded on A(A), then F(A) can be defined
as the multiplicationinduced by the composite functionF o 4. The mapping
F- ,F(A) is the homomorphismthat is frequentlyknown by the impressive
name of "the functionalcalculus." If, in particular,F= FM is the characteristic
functionof a Borel set M in the real line, and if E(M) is the multiplicationinduced by FM o 0, then E is the spectral measure of A. The verificationthat E
is indeed a spectral measure is easy. To prove that it belongs to A (i.e., that
A =fXdE(X)), proceed as follows.Fix f and g in 22(yg) and write
v(M) = (E(M)f, g)
foreach Borel set M; it is to be proved that
(Af, g) =
Xdv(X).
1963]
WHAT DOES THE SPECTRAL THEOREM SAY?
245
Since (E(M)f, g) =f(FM o 0)fgdyand v(M) =fFMdv, it followsthat
f
(FM o )fgdy=
Fmd
forall Borel sets M. This implies that
(E o q$)fgdy
=
Fd
wheneverF is a simple function,and hence,by approximation,wheneverF is a
bounded Borel measurable function.This conclusion (for F(X)
is just what
OX)
was wanted.
The multiplicationversionof the spectraltheoremimpliesthe spectralmeasure version,but the latteris canonical (E is uniquely determinedby A) whereas
the formeris not. Consider, forinstance, the identityoperator on a separable
infinite-dimensional
Hilbert space in the role of A. It is unitarilyequivalent to
multiplication by the constant function 1 on, say, the unit interval (with
Lebesgue measure); it is also unitarilyequivalent to multiplicationby the constant function1 on the set of positive integers(with the countingmeasure).
There is a spectral theoremfornormal operatorsalso; its statementcan be
obtained fromthe one given above by substituting"normal" for "Hermitian."
It is a well-knowntechnical nuisance that the proofof the spectral theoremfor
normal operators involves some difficulties
that do not arise in the Hermitian
case. The sourceof the troubleis that it is not enoughjust to replace polynomials
in a real variable by polynomialsin a complexvariable; theWeierstrasstheorem
demands the considerationof polynomialsin two real variables. There is a consequent difficulty
in extendingthe spectralmappingtheoremto the kindof functions (polynomialsin the real and imaginaryparts of a complex variable) that
arise in the imitationof the proofabove. Even the equality of normand spectral
radius, while true fornormaloperators,requiresa proofquite a bit deeper than
in the real case. One way around all this is not to imitate the proofbut to use
the result. In [3, p. 72], forinstance,the spectral theoremfornormaloperators
(spectral measure version) is derived from the Hermitian theorem (spectral
measure version); the only additional tool needed is an essentiallyclassical extension theoremformeasures in the plane.
In any case, all this talk about proofis somewhat beside the point in this
paper. The reason a proofis outlined above is not so much to induce belief in
the resultas to clarifyit. The emphasis here is not on howbut on what,not on
proofbut on statement,
not on How does thespectraltheoremcomeabout?but on
What does thespectraltheoremsay?
To see how the multiplication point of view can be used, consider the
Fuglede commutativitytheorem [2]. A possible statementis this: if A is normal and ifB is an operatorthat commuteswithA, then B commuteswith F(A)
for each Borel measurable functionbounded on A(A). (An alternative state-
246
[March
WHAT DOES THE SPECTRAL THEOREM SAY?
ment,only apparentlyweaker, is that ifB commuteswith A, then B commutes
with A*; for a recent elegant proofsee [5].) The spectral theoremshows that
there is no loss of generalityin assuming that A is the multiplicationinduced
by X, say, on a measure space X with measure,u.If FM is, foreach Borel set M
in the complex plane, the characteristicfunctionof M, and ifE(M) is the multiplication induced by FM o 0, then it is sufficient
to prove that B commuteswith
each E(M). (Approximate the general F by simple functions,as before.) If
8(M) is the range of the projectionE(M), then the desired result is that 8(M)
reduces B, but it is, in fact, sufficient
to prove that 8(M) is invariant under B.
Reason: apply the invariance conclusion,once obtained, to the complementof
M, and inferthat both 8(M) and (8(M))' are invariant under B.
Observe now that &(M) is the set of all those functionsin 2Q(,u)
that vanish
outside 4-1(M), and considerfirstthe case of the closed unit disc,
M=
{X: I XI < 11;
then
?-1(M)=
Assertion: 8(M)
consists of all
{X: I+(X)|
f in
22(yA)
?1}.
for which the sequence
{I Afl!,
*
} is bounded.Indeed,iff vanishesoutside+-1(M), then
IIA2fll
II||Afll,
I|A f112=
f
I
|?nfI I2dA =J fM
nl 2 . If!l2dA
<
IfIf2dA.
If, on the other hand, there is a set S of positive measure on whichfF0 and
I01>1, then
I!Anfll2
=f
IpnfI2dy
?f
I&0InIfI2dA
X*c,.
The assertion is proved, and the invariance of 8(M) under B follows: if
c for all n.
?c for all n, then JJAnBfll
_?|IBIIJ
JAnfII5
=|JBAnfII<JJBJ.JJAnfJ
If Mis any closeddisc,M= {X: IX-XoIr ,then
0-1(M) =x:
I (X)-o
<I r}{x:
) (x)
< i}
Since B commutes with multiplicationby q, it commutes with multiplication
by (O-Xo)/r also, and it follows fromthe preceding paragraph that 8(M) is
invariant under B.
The rest of the proofis easy measuretheory;fromthis point of view spectral
measures behave even betterthan numericalmeasures. Since 8(M) is invariant
under B whenever M is a disc, the same is true whenever M is the union of
countably many discs. This implies that 8(M) is invariant under B whenever
M is open, and hence (regularity)forarbitraryBorel sets M.
1963]
AN EXTENSION
247
OF THE FERMAT THEOREM
References
1. R. P. Boas,A primer
ofrealfunctions,
Math.Assoc.ofAmerica,CarusMonographno. 13,
1960.
2. B. Fuglede,A commutativity
theorem
fornormaloperators,
Proc.Nat. Acad. Sci. U.S.A.,
36 (1950) 35-40.
3. P. R. Halmos,Introduction
to Hilbertspace,Chelsea,New York,1951.
4.
, Measuretheory,
Van Nostrand,New York,1950.
5. M. Rosenblum,On a theoremof Fugledeand Putnam,J. LondonMath. Soc., 33 (1958)
376-377.
6. I. E. Segal,Decomposition
ofoperatoralgebras,II, Memoirs,Amer.Math.Soc., 1951.
AN EXTENSION OF THE FERMAT THEOREM
L. CARLITZ, Duke University
1. If p is a prime,e > 1 and (a, p)=1 then
(1)
aw _ 1 (modpe)
provided pe-l(p(2)
1) w. It followsfrom(1) that
(_1)( r-
an+sw =
an(aw - 1)r _ 0 (mod
pre)
forall r2?0. This congruenceis useful forexample in derivingKummer's congruence forthe Euler and Bernoulli numbers [1, Ch. 14].
It may be of interestto examine the sum
>
r
Arank=
(3)
(_ 1)r
(
a (n+8w)k
s
8=0
where k is a fixedinteger ? 1. We remarkthat (3) is suggestedby some recent
work [2] on colored graphs. We shall prove
THEOREM 1. Let pe-l(p
-1) I w and letX> e be thelargestintegersuch that
(4)
1 (mod px).
Then
(5)
r
E (- 1)r
( )
a(n+sw)k
where
(6)
rk=
the greatestinteger _ (r+k - 1)/k.
[(r+k-1)/k],
_ 0 (modpXrk),