MATRIX THEORY CHAPTER L FALL 2017 1.EIGENVALUE,EIGENVECTOR AND CHARACTERISTIC POLYNOMIAL Definition 1.For AE M (C), Eigenvalue and eigenvector associated with: Ax=x,λ∈C,0≠x∈C" ·Spectrum a(A）and radius p(A） a(A)=:is an cigenvaluesof A} p(A)=max{N:A∈a(A)} .Characteristic polynomial: pa(t)=det(tI-A) Theorem 1.Suppose Ar=Ar.For any polynomial q(t)=agt+ast+.+at+ao,denote q(4=a4+ak-14-1+…+a1A+a0 Then q(A)z=q(A). Theorem 2.AM(C),the charucteristic polynomial pa()=det(I-A)=(t-)t-2)…(t-入n) =-E(4)t-1+E(4)tn-1-…+(-1)En(4 =-S(d1,入n)m-1+S2(d1,,入n)-1-…+(-1)nSn(A1,入） E(A)=Sk(1...) In particular, E1(4)=tr(A)=1+…+Xm=S(1,,Am), En(A）=det(A）=1…入m=Sn(,n Theorem 3.Let AE M and pA(t)is the characteristic polynomial of A.Then pA(A)=0.(Here means a zero n×n matrir.) Example 1.A pa()=dt(AI-A)=X2-8A+15=(A-3)(A-5) .Observation:A1 =3,2=5; 8=A1+A2=tr(A),15=A12=det(A) 0]o =a=[la=oa
MATRIX THEORY - CHAPTER 1 FALL 2017 1. Eigenvalue, eigenvector and characteristic polynomial Definition 1. For A ∈ Mn(C), • Eigenvalue λ and eigenvector x associated with λ: Ax = λx, λ ∈ C, 0 6= x ∈ C n . • Spectrum σ(A) and radius ρ(A) σ(A) = {λ : λ is an eigenvaluesof A} ρ(A) = max{|λ| : λ ∈ σ(A)} • Characteristic polynomial: pA(t) = det(tI − A) Theorem 1. Suppose Ax = λx. For any polynomial q(t) = akt k + ak−1t k−1 + · · · + a1t + a0, denote q(A) = akA k + ak−1A k−1 + · · · + a1A + a0. Then q(A)x = q(λ)x. Theorem 2. A ∈ Mn(C), the characteristic polynomial pA(t) = det(tI − A) = (t − λ1)(t − λ2)· · ·(t − λn) = t n − E1(A)t n−1 + E2(A)t n−1 − · · · + (−1)nEn(A) = t n − S1(λ1, ..., λn)t n−1 + S2(λ1, ..., λn)t n−1 − · · · + (−1)nSn(λ1, ..., λn) where Ek(A) is the summation of k × k principle minors and Sk(λ1, ..., λn) is the k-th symmetric functions. Hence for all 1 ≤ k ≤ n Ek(A) = Sk(λ1, ..., λn) In particular, E1(A) = tr(A) = λ1 + · · · + λn = S1(λ1, ..., λn), En(A) = det(A) = λ1 · · · λn = Sn(λ1, ..., λn). Theorem 3. Let A ∈ Mn and pA(t) is the characteristic polynomial of A. Then pA(A) = 0. (Here 0 means a zero n × n matrix.) Example 1. A = 7 −2 4 1 • Characteristic polynomial: pA(λ) = det(λI − A) = λ 2 − 8λ + 15 = (λ − 3)(λ − 5) • Observation: λ1 = 3, λ2 = 5; 8 = λ1 + λ2 = tr(A), 15 = λ1λ2 = det(A) • Eigenvalues and eigenvectors: λ1 = 3, α1 = 1 2 , Aα1 = λ1α1 λ2 = 5, α2 = 1 1 , Aα2 = λ2α2 1
2 FALL 2017 .Notice that A is a scaling on eigenvectors.For any vectorR2,write+bo,then Av=aA101+bA202 For erample, e=[8]=3[2l+2[1=a+2ae 4e=3ha+2a:=4a+i0ag=[28] Direct computation -[][8]-[] ·Define s=a,al=[21] Then 4s-[2][:]-[&l[88]-sm AS A[a1,a2]=[A1a1.A2a2][a1,a21A=SA Here S is invertible: s-[1] Hence S-AS-05-A. We say A is similar to a diagonal matrirA. Example 2.A= .Characteristic polymomial: PA()=det(M-A)=X3-9A2+24A-16=(A-1(A-4)2. .Oberservation. 9=1+4+4=tr(4)=S(1,4,4) 16=1×4×4=det(4)=S5(1,4,4) =1×4+4×4+4×1=S2(1.4,4) .Eigenvalues and eigenvectors: 「1 1=1,01=1 :Ao1 =A01: 1 2=4,a2 0 -1 0 g=4,a3= Ao3 =Ag03 -1 ·Define S=[a1,02,03]=
2 FALL 2017 • Notice that A is a scaling on eigenvectors. For any vector v ∈ R 2 , write v = aα1 + bα2, then Av = aλ1α1 + bλ2α2 For example, v = 5 8 = 3 1 2 + 2 1 1 = 3α1 + 2α2, Av = 3Aα1 + 2Aα2 = 9α1 + 10α2 = 19 28 Direct computation: Av = 7 −2 4 1 5 8 = 19 28 • Define S = [α1, α2] = 1 1 2 1 Then AS = 7 −2 4 1 1 1 2 1 = 1 1 2 1 3 0 0 5 = SΛ AS = A[α1, α2] = [λ1α1, λ2α2] = [α1, α2]Λ = SΛ Here S is invertible: S −1 = −1 1 2 −1 . Hence S −1AS = 3 0 0 5 = Λ. We say A is similar to a diagonal matrix Λ. Example 2. A = 3 −1 −1 −1 3 −1 −1 −1 3 • Characteristic polynomial: pA(λ) = det(λI − A) = λ 3 − 9λ 2 + 24λ − 16 = (λ − 1)(λ − 4)2 . • Oberservation: 9 = 1 + 4 + 4 = tr(A) = S1(1, 4, 4) 16 = 1 × 4 × 4 = det(A) = S3(1, 4, 4) 24 = 3 −1 −1 3 + 3 −1 −1 3 + 3 −1 −1 3 = E2(A) = 1 × 4 + 4 × 4 + 4 × 1 = S2(1, 4, 4) • Eigenvalues and eigenvectors: λ1 = 1, α1 = 1 1 1 , Aα1 = λ1α1; λ2 = 4, α2 = 1 0 −1 , Aα2 = λ2α2; λ3 = 4, α3 = 0 1 −1 , Aα3 = λ3α3. • Define S = [α1, α2, α3] = 1 1 0 1 0 1 1 −1 −1
MATRIX THEORY-CHAPTER 1 AS A[a1,02,03]=[A101,A202,A303]=[01,02,031A =S a到5 inie and hene-As=A比Asmilr togol matris Q1.For AEMn,can we always find n different eigenvalues? A1.No.But we can find n eigenvalues if counting the algebraic multiplicity,i.e.if we write pA()=(t-)(t-2)2…(t-) where A...are distinct eigenvalues,then n1+n2+…+k=n ors? maximum number of independent eigenvectors associated with ,i.e. mi=dimNx,(A),NA,(A)={EC":Ar=} Here mi is called the geometric multiplicity of.Then for each 1≤m≤n: See the next example T2001 Example 3.A- 031 003 characteristic polynomial:pA(A)=(A-2)(A-3)2. ·eigenvectors 「1 h1=2,a1= 01 ,Aa1-2a1 [o 2=3,a2= A02=302 0 ·for1=2, algebraic multiplicity of1=geometric multiplicity of for A2=3, algebraic multiplicity of A2=2>1 geometric multiplicity of A2 2.SIMILARITY AND DIAGONALIZATION Definition 2.For AE M(C), If B=S-AS for some nonsingular matriz S.then we say B is similar to A. If S-AS=A for some nonsingular matrir S and diagonal matrir A,then we say A is diagonalisable. Theorem 4.Similarity is an equivalence relation,i.e.if denoting"B is similar to A"by BA,then ·refective:
MATRIX THEORY - CHAPTER 1 3 AS = 3 −1 −1 −1 3 −1 −1 −1 3 1 1 0 1 0 1 1 −1 −1 = 1 1 0 1 0 1 1 −1 −1 1 4 4 = SΛ AS = A[α1, α2, α3] = [λ1α1, λ2α2, λ3α3] = [α1, α2, α3]Λ = S Moreover, S is invertible and hence S −1AS = Λ, i.e. A is similar to a diagonal matrix Λ = diag(1, 4, 4) Q1. For A ∈ Mn, can we always find n different eigenvalues? A1. No. But we can find n eigenvalues if counting the algebraic multiplicity, i.e. if we write pA(t) = (t − λ1) n1 (t − λ2) n2 · · ·(t − λk) nk where λ1, ..., λk are distinct eigenvalues, then n1 + n2 + · · · + nk = n. Q2. For A ∈ Mn, can we always find n independent eigenvectors? A2. No. Suppose λ1, ..., λk are distinct eigenvalues and ni is the algebraic multiplicity of λi . Let mi be the maximum number of independent eigenvectors associated with λi , i.e. mi = dimNλi (A), Nλi (A) = {x ∈ C n : Ax = λix} Here mi is called the geometric multiplicity of λi . Then for each λi 1 ≤ mi ≤ ni See the next example Example 3. A = 2 0 0 0 3 1 0 0 3 • characteristic polynomial: pA(λ) = (λ − 2)(λ − 3)2 . • eigenvectors: λ1 = 2, α1 = 1 0 0 , Aα1 = 2α1 λ2 = 3, α2 = 0 1 0 , Aα2 = 3α2 • for λ1 = 2, algebraic multiplicity of λ1 = 1 = geometric multiplicity of λ1 for λ2 = 3, algebraic multiplicity of λ2 = 2 > 1 = geometric multiplicity of λ2 2. Similarity and diagonalization Definition 2. For A ∈ Mn(C), • If B = S −1AS for some nonsingular matrix S, then we say B is similar to A. • If S −1AS = Λ for some nonsingular matrix S and diagonal matrix Λ, then we say A is diagonalisable. Theorem 4. Similarity is an equivalence relation, i.e. if denoting ”B is similar to A” by B ∼ A, then • reflective: A ∼ A; • symmetric: B ∼ A implies A ∼ B; • transitive: C ∼ B and B ∼ A imply C ∼ A
4 FALL 2017 Theorem 5.If B=S-1AS,then .pA(t)=pB(t),i.e.A and B have the same eigenvalues with the same algebraic multiplicities. Theorem 6.For AM(C),the following are equivalent: (a)A is diagonalisable (b)A has n linearly independent eigenvectors; (c)for A E o(A).its algebraic multiplicity equals to its geometric multiplicity. Theorem 7.Suppose C=where AEM(C)and BEMm(C).Then C is diagonalisable if and 3.MORE ABOUT EIGENVECTOR Definition 3.For AEMn(C), 。ifAr=入r for some x≠0，then r is called the right eigenvector associated with入. ·fyA=Ay'for somey≠0，then y is called the left eigenvector associated with入 Theorem9.Suppose is a right eigenvector of A associated with andy is a left eigenvector of A associated ith4.f入≠h,then y'x=0
4 FALL 2017 Theorem 5. If B = S −1AS, then • pA(t) = pB(t), i.e. A and B have the same eigenvalues with the same algebraic multiplicities. • If Ax = λx (x 6= 0), then letting y = S −1x, n 6= 0 and By = λy. So A and B have the same eigenvalues with the same geometric multiplicities. Theorem 6. For A ∈ Mn(C), the following are equivalent: (a) A is diagonalisable; (b) A has n linearly independent eigenvectors; (c) for λ ∈ σ(A), its algebraic multiplicity equals to its geometric multiplicity. Corollary 1. For A ∈ Mn(C), if A has n distinct eigenvalues, then A is diagonalisable. (The reverse is not true.) Theorem 7. Suppose C = A 0 0 B where A ∈ Mn(C) and B ∈ Mm(C). Then C is diagonalisable if and only if A and B are both diagonalisable. 3. More about eigenvector Definition 3. For A ∈ Mn(C), • if Ax = λx for some x 6= 0, then x is called the right eigenvector associated with λ. • if y ∗A = λy∗ for some y 6= 0, then y is called the left eigenvector associated with λ. Theorem 8. If λ1, ..., λk are k distinct eigenvalues of A and αi is an eigenvector associated with λi. Then {α1, ..., αk} is a set of linearly independent eigenvectors. Theorem 9. Suppose x is a right eigenvector of A associated with λ and y is a left eigenvector of A associated with µ. If λ 6= µ, then y ∗x = 0