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149IX. The spectrum of a lm** point spectrum **An eigenpair for A ∈ L(X) is any (x, z) ∈ (X\0) × F satisfying Ax = zx. Thescalar z in such an eigenpair is called an eigenvalue for A and the vector x is called aneigenvector for A belonging to the eigenvalue z. The collection of all eigenvectors of Abelonging to z is ker(A − z)\0. Thus z is an eigenvalue of A iff (A − z) fails to be 1-1.The collection of all eigenvalues of A is called the point spectrum of A and is denotedbyσP(A).Eigenstructure of a matrixIf X is finite-dimensional, then A is a matrix and one is naturally led to look into theeigenstructure of A when one looks for a basis V , i.e., an invertible lm V : Fn→ X,AX → XV ↑ ↑ VFn→ FnˆAfor which the corresponding matrix representationˆA = V−1AV for A is particularlysimple. Ideally, one wantsˆA to be a diag onal matrix. If there i s such V , then A iscalled diagona(liza )ble. Assuming thatˆA is diagonal,ˆA = diag⌈z1, . . . , zn⌋ say, thennecessarily Avj= zjvj, all j, i.e., the basis V must consist of (nontrivial) eigenvectors ofA.Whether or notˆA is diagonal,ˆA is said to be similar to A.** who cares about eigenstru c ture? **If you look into t he question as t o why one might want a particularly simple matrixrepresentation for A in the first place, you will find that it is useful for understanding thepowers of A, of importance in the analysis of fixed-point iteration for solving linear (andnonlinear) systems, the solution of a system of first-order ODEs, and in t he numericalsolution of evolution equations.For example, a square matrix A is powerbounded, i.e., {Ak: k ∈ N} i s a boundedset iff ∀{z ∈ σP(A)} |z| ≤ 1 with equality o nly if z is not defecti ve, i.e., only if ran(A−z)∩ker(A − z) = { 0}. Further, A is convergent, i.e., li mk→∞Akexists iff ∀{z ∈ σP(A)} | z| ≤1 with equality only if z is not defective and z = 1. Finally, A is convergent to 0, i.e.,limk→∞Ak= 0 iff ∀{z ∈ σP(A)} |z| < 1 (as was mentioned already in Chapter 2 in thediscussion of fixed point it erat ion).polynomials in a lmc2002 Carl de Boor150 IX. The spectrum of a lm** polynomials in a lm **More generally, one is i nterested in understanding the behavior of l inear combinationsPj≤ka(j)Ajof such powers, i.e., of polynomials p(A) in A (with p :=Pj()ja(j)), and,ultimately, of functions f(A) in A, to the extent that f can be approximated arbitrarilyclosely by polynomials p, hence f(A) can be understood as the limit of p(A) as p → f.E.g., y(t) = exp(tA)y0is the unique (vector-valued) solution at t of the first-order ODEDy = Ay w ith side condition y(0) = y0.Having a complete understanding of the eigenstructure of A vastly simplifies all deal-ings with p(A). Indeed, if A = VˆAV−1, then, for any p ∈ Π,p(A) = V p(ˆA)V−1,while, for a diagonal mat rixˆA = diag⌈. . . , zj, . . .⌋,p(ˆA) = diag⌈. . . , p(zj), . . .⌋.Thus, for a diagonalizable A,σP(p(A)) = p(σP(A)).This is a particular example of the Spectral Mapping Theorem.Work with polynomials in the lm A i s materially helped by the seemingly trivial factthat any two polynomials in the same linear map commute:(1) ∀{p, q ∈ Π; A ∈ L(X)} p(A)q(A) = q(A)p(A).H.P.(1) Prove (1).As an illustration, here is a proof of the basic fact that every A ∈ L(X) with 0 <dim X < ∞ and F = C has eigenvalues. Indeed, there is x ∈ X\0 and, for any suchx, [x, Ax, A2, . . . , Adim Xx] must fail to b e 1-1, hence there is a 6= 0 so that p(A)x :=Pja(j)Ajx = 0, showing that p(A) fails to be 1-1, even though p 6= 0. Let d := max{j :a(j) 6= 0}. Then, wlog, a(d) = 1, i.e., p is monic. Further, d > 0 since x 6= 0. SinceF = C, we can therefore write p as the productQj(· − zj) of d > 0 linear factors. But,since p(A) =Qj(A − zj) fails to be 1-1, at least one of the factors A − zjmust fail to be1-1.** A-invariant direct sum decompositions **As a start toward a simplest ma trix representation, assume that P is a spe c tralprojector for the lm A, i.e., a lprojector that commutes with A,P A = AP.Then the corresponding direct sum decompositionX = X1˙+X2, X1:= ran P, X2:= ker Pis A-invariant in the sense that its summands are A-invariant,AXi⊂ Xi, all i.A-invariant direct sum decompositionsc2002 Carl de BoorEigenstructure of a matrix 151Conversely, for any such A-invariant direct sum decomposition X = X1˙+X2, the corre-sponding lprojector, g iven by ran P = X1, ker P = X2, is spectral for A si nce ran(AP ) =AX1⊆ X1= ran P , hence P AP = AP , while also ran(A(1 − P )) = AX2⊆ X2= ker P ,hence P A(1 − P ) = 0, therefore, altogether, P A = P AP + P A(1 − P ) = AP .If now X is finite-dimensional, then so are the Xi, a nd, with Viany basis for Xi, V :=[V1, V2] is a basis for X with the happy property thatˆA = V−1AV = [V1, V2]−1[AV1, AV2]is block-diagonal, in particular,ˆA =ˆA100ˆA2,ˆAi:= V−1iAXiVi,since the columns of AViare in Xi, hence their coordinates wrto V have nonzero entriesonly corresponding to the columns of Viin V .If you conclude from this that a search for ‘simple’ matrix representations for A ∈L(X) is equivalent to a search for A-invariant direct sum decompositions for A with manysummands or, equivalently, a search for a many-termed sequence (Pi) in L(X) wi th PiPj=δijand APi= PiA, all i, j, then you would be quite right.** primary decomposition **Here, for the record, is a first step in that direction that goes back to Frob enius. Tobe sure, this first step does not, in general, do the complete job. For that, just skip to theheading ‘A finest A-invariant direct sum decomposition’.Assuming dim X to be finite-dimensional, so is L(X), hence [Ar: r = 0, . . . , dim L(X)]cannot be 1-1, therefore there are polynomials p 6= 0 that annihilate A in the sense thatp(A) = 0. Let p b e any such monic annihilating polynomial and assume, for simplicity,that F = C. Thenp =:Yipi,with pi= (· − zi)mi, and zi6= zjfor i 6= j. It follows that the polynomialsℓi:= p/pi, all i,do not have a common zero, hence1 =Xiℓihifor certain polynomials hi. Indeed, l et hibe the unique polynomial of degree < miforwhich ℓihiagrees mi-fold with 1 at zi. Then 1−Piℓihiis a poly nomi al of degree <Pimiand vanishes mi-fold at zi, all i, hence must be zero.SetPi:= ℓi(A)hi(A), Xi:= ker pi(A), all i.Then1 =XiPi,primary decompositionc2002 Carl de Boor152 IX. The spectrum of a lmand, for j 6= i, Pjvanishes on Xi(since then ℓjhas pias a factor), hence 1 = Pion


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