上海交通大学：《矩阵论》课程教学资源（讲义）CHAPTER 05

MATRIX THEORY CHAPTER 5 FALL 2017 1.VECTOR NORMS (4④z+≤z+ (任，≥0 四在+=,)+(2, (⑤)）红，到=，或 Theorem 1(Cauchy-Schwarz inequality).If is an inner product on a vector space V over the field F CorR,then for all x,y∈V I红，P≤a,,以 Theorem2.f(,）is an inner product on a vector space V,then‖=√在，E)is a vector norm. Example 1.The IP norm on Cn is -( forl≤p≤In particular: z=+lz2+…+zn: l2=VP+2P+…+EnF Ilall=l. Theorem3.Suppose Virpace f finte dimension.Thenn.onV are equivalent in the sense that there erist constant C>such that 1/CIxl.≤lzl。≤Cl,VxeV 2.MATRIX NORMS Definition3.A function·l:Mn-→R is a Matrir norm if for any A,B∈Mn, (1)WAl≥0： (②)ⅢA=0 if and only if A=0:

MATRIX THEORY - CHAPTER 5 FALL 2017 1. Vector Norms Definition 1. Let V be a vector space over a field F (C or R). A function k · k : V −→ R is a vector norm if for any x, y ∈ V , (1) kxk ≥ 0; (2) kxk = 0 if and only if x = 0; (3) kcxk = |c|kxk; (4) kx + yk ≤ kxk + kyk. Definition 2. Let V be a vector space over a field F (C or R). A function h·, ·i : V × V −→ F is an inner product if for any x, y, z ∈ V , (1) hx, xi ≥ 0; (2) hx, xi = 0 if and only if x = 0; (3) hcx, yi = chx, yi for any c ∈ F; (4) hx + z, yi = hx, yi + hz, yi; (5) hx, yi = hy, xi. Theorem 1 (Cauchy-Schwarz inequality). If h·, ·i is an inner product on a vector space V over the field F (C or R), then for all x, y ∈ V |hx, yi|2 ≤ hx, xihy, yi. Theorem 2. If h·, ·i is an inner product on a vector space V , then kxk = p hx, xi is a vector norm. Example 1. The l p norm on C n is kxkp = Xn i=1 |xi | p !1 p for 1 ≤ p ≤ ∞ In particular: kxk1 = |x1| + |x2| + · · · + |xn|; kxk2 = p |x1| 2 + |x2| 2 + · · · + |xn| 2; kxk∞ = max 1≤i≤n |xi |. Theorem 3. Suppose V is a vector space of finite dimension. Then any two vector norms k · k?, k · k◦ on V are equivalent in the sense that there exist constant C > 0 such that 1/Ckxk? ≤ kxk◦ ≤ Ckxk?, ∀ x ∈ V. 2. Matrix Norms Definition 3. A function 9 · 9 : Mn −→ R is a Matrix norm if for any A, B ∈ Mn, (1) 9A9 ≥ 0; (2) 9A9 = 0 if and only if A = 0; (3) 9cA9 = |c| 9 A9; (4) 9A + B9 ≤ 9A 9 + 9 B9, (5) 9AB9 ≤ 9A 9 9B9. 1

2 FALL 2017 A:=罗件 Example 2.(1) 4=警件-（学空）广=网 (② 4=器快=器 小=a…,the c eor九 =器 A,there erists a matriz norm such that p(A)All p(A)+e. Proof.Fix A and e>0,we construct a matrix norm in the following way: (1)There exists a unitary matrix U such that UAU=T is an upper triangular matrix.Hencet=Ai, Thenl(A)+e.Fix D.. (3)Define a matr x norm for arbitrary BE Mn byB =D.UBUD Applying this norm to A we haveA≥pA) Theorem 7.For,limA=0 if and only if p(A)<1. Proof.:Suppose z is an eigenvector associated to eigenvalue A.Then Ar =r.Take koo.Since 0=limgAr,we have limg=0.Hence 1.Since A is an arbitrary eigenvalue, Theorem 8.For any AE Mn and any matriz norm,p(A)=limkA/

2 FALL 2017 Theorem 4. Let k· k? is a vector norm on C n. Then we can define a corresponding matrix norm on Mn(C) by 9A9? = max x6=0 kAxk? kxk? . Example 2. (1) 9A92 = max x6=0 kAxk2 kxk2 =  max x6=0 x ∗A∗Ax x ∗x 1 2 = p λmax(A∗A). (2) 9A91 = max x6=0 kAxk1 kxk1 = max 1≤j≤n kαjk1 where A = [α1, · · · , αn], i.e. αj is the j-th column vector of A. (3) 9A9∞ = max x6=0 kAxk∞ kxk∞ = max 1≤i≤n kβik1 where A =    β T 1 . . . β T n   , i.e. β T i is the i-th row vector of A. Remark 1. Since Mn(C) is also a vector space of dimension n 2 . So we have the vector norms for A = [aij ] as follows: kAk1 = Xn i,j=1 |aij |, kAk2 =   Xn i,j=1 |aij | 2   1 2 , kAk∞ = max 1≤i,j≤n |aij |. Theorem 5. Let ρ(A) be the spectrum radius of A, i.e. ρ(A) = max{|λ| : λ ∈ σ(A)}. Then for any matrix norm 9 · 9 on Mn(C), ρ(A) ≤ 9A 9 . Proof. Let λ be any eigenvalue of A. So there is some eigenvector x 6= 0 such that Ax = λx. We construct a square matrix X such that each column vector of X is equal to x. Then |λ| 9 X9 = 9λX9 = 9AX9 ≤ 9A 9 9X9 Since 9X9 6= 0, we have 9A9 ≥ |λ|. Since λ is arbitrary eigenvalue, we get 9A9 ≥ ρ(A). Theorem 6. Given A ∈ Mn, for any  > 0, there exists a matrix norm 9·9 such that ρ(A) ≤ 9A9 ≤ ρ(A)+. Proof. Fix A and  > 0,we construct a matrix norm in the following way: (1) There exists a unitary matrix U such that U ∗AU = T is an upper triangular matrix. Hence tii = λi , tij = 0 for i > j. Fix U. (2) Consider the matrix H = [hij ] = DsT D−1 s where Ds = diag{s 1 , s2 , · · · , sn}. Then H is also upper triangular with hii = tii = λi and hij = s i−j tij for i 0 such that ρ(A) +  < 1. By last theorem, there exists a matrix norm such that 9A9 < 1. Hence limk→ ∞ 9 Ak9 = 0. Theorem 8. For any A ∈ Mn and any matrix norm 9 · 9, ρ(A) = limk→ 9Ak9 1/k .

MATRIX THEORY-CHAPTER 5 Proof.For any 1>c>0,there exists a matrix norm.such that p(A)l.p(A)+e.Then p(A)=p(A)≤川Al.≤((A)+e) For the given norm.there exists a constant C such that C-1IB.≤B≤CIB: Hence p(A=p(A)≤川4Ⅲ≤C(P(A)+e)' p(A)≤IA/≤Ck((A)+) Then there exists ko such that when k>ko,1C1+e and 0≤I4/k-p(A)≤(p(A)+2 We finish the proof. ◇ Theorem9.ForA∈Mn,∑e1ak,Ak convergesif∑2，la l for some matrit norm·l Corollary 1.For A=[f,then A is invertible. Proof.We can show that in this case-D-Al<1 where D is the diagonal of A.Hence D-A and A are invertible

MATRIX THEORY - CHAPTER 5 3 Proof. For any 1 >  > 0, there exists a matrix norm 9 · 9?such that ρ(A) ≤ 9A9? ≤ ρ(A) + . Then ρ(A) k = ρ(A k ) ≤ 9A9? ≤ (ρ(A) + ) k For the given norm 9 · 9, there exists a constant C such that C −1 9 B9? ≤ 9B9 ≤ C 9 B9? Hence ρ(A) k = ρ(A k ) ≤ 9A k9 ≤ C(ρ(A) + ) k ρ(A) ≤ 9A k9 1/k ≤ C 1/k(ρ(A) + ) Then there exists k0 such that when k > k0, 1 ≤ C 1/k ≤ 1 +  and 0 ≤ 9A k 9 1/k −ρ(A) ≤ (ρ(A) + 2) We finish the proof. Theorem 9. For A ∈ Mn, P∞ k=1 akAk converges if P∞ k=1 |ak| 9 A9 k for some matrix norm 9 · 9. Example 3. (1) e A = I + P∞ k=1 A k k! converges for all A ∈ Mn (2) (I − A) −1 = I + P∞ k=1 Ak converges if and only if ρ(A) P j6=i |aij |, then A is invertible. Proof. We can show that in this case 9I − D−1A9∞ < 1 where D is the diagonal of A. Hence D−1A and A are invertible.