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What Does the Spectral Theorem Say? Author(S): P 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? P. R. HALMOS, Universityof Michigan Most students of mathematics learn quite early and most mathematicians remembertill quite late that every Hermitian matrix (and, in particular,every real symmetricmatrix) may be put into diagonal form.A moreprecise statement of the resultis that every Hermitianmatrix is unitarilyequivalent to a diagonal one. The spectral theoremis widely and correctlyregarded as the generalization of this assertion to operatorson Hilbert space. It is unfortunatetherefore that even the bare statementof the spectral theoremis widelyregarded as somewhat mysteriousand deep, and probably inaccessible to the nonspecialist.The pur- pose of this paper is to try to dispel some of the mystery. Probably the main reason the general operator theoremfrightens most peo- ple is that it does not obviously include the special matrixtheorem. To see the relation between the two, the descriptionof the finite-dimensionalsituation has to be distorted almost beyond recognition.The result is not intuitive in any language; neitherStieltjes integralswith unorthodoxmultiplicative properties, nor bounded operatorrepresentations of functionalgebras, are in the daily tool- kit of everyworking mathematician. In contrast,the formulationof the spectral theorem given below uses only the relatively elementaryconcepts of measure theory.This formulationhas been part of the oral traditionof Hilbert space for quite some time (foran explicittreatment see [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 presentformu- lation 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 state- ment 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 operator A is definedon the Hilbert space 22(p) by (Af )(x)-=*t(x)f (x), x E- X; the operator A is called the multiplicationinduced by 4. The study of the rela- tion between A and 4 is an instructiveexercise. It turns out, 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 ifralso is a bounded measurable functionon X, with induced multiplica- tion B, then the multiplicationinduced by the product function044 is the prod- uct operatorAB. It followsthat a multiplicationis always normal; it is Hermi- tian if and only if the functionthat induces it is real. (For the elementarycon- cepts of operatortheory, such as Hermitianoperators, normal operators, projec- tions, and spectra, see [3]. For presentpurposes a concept is called elementary if it 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 described by A(f1, ... ,fn) = (Oifi, ...* ,fn) To say this with matrices,note that the characteristicfunctions of the single- tons in X forman orthonormalbasis in ?2(M); the assertion is 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 finite- dimensional situation within it is clear; everythingis ready for the principal statement. SPECTRAL THEOREM. Every Hermitian operatoris unitarilyequivalent to a multiplication.. In complete detail the theoremsays that if A is a Hermitian operator on a Hilbert space 3C,then there exists 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 outline of a proofof the spectral theorem, a briefdiscussion of its relation to the version involvingspectral 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 is thatif A is Hermitian,then r(A) =IIAII([3, Theorem2, p. 55]). (2) The Riesz representationtheorem for compactsets in theline. If L is a posi- tive linear functionaldefined for all real-valued continuousfunctions on 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) = ffd1. for all 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 Weierstrassapproximation theorem for compactsets in theline. Each real-valued continuous functionon a compact subset of the real line is the uni- formlimit of polynomials. For a pleasant elementarydiscussion and proof see [I, p. 102]. Consider now a Hermitian operator A on a Hilbert space SC. A vector t in XC is a cyclicvector for A if the set of all vectors of the formq(A)t, where q runs over polynomials with complexcoefficients, is dense in SC. Cyclic vectors may not exist, but an easy transfiniteargument shows 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 known foreach such restriction,it followseasily forA itself;the measure spaces that serve forthe directsummands of a have a natural directsum, which serves forSC itself.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 I L(p) I ? IIP(A)II-IIt1I2 = r(p(A)) _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 continuous functionon A(A), then approximate V/f uniformlyby real polynomials; the inequality just proved implies that 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 finitemeasure A.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))*(= {(A)) is a polynomialin A, and so is (q(A)) *q(A) qf2(A));' it followsthat f qqI 2dl = (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 isometricextension that maps 22(y) into SC.
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