Linear Algebra and Application(Peter.D.Lax著)答案 英文版，很好懂的！
19. 15.3 Decomposition of a finite dimensional linear vector space under a linear operator 50 19.15. 4 Computation of elementary divisors and invariant factors 19.15.5 Examples 2 19.16Numerical range 53 docn豆丁 www.docin.com This is a solution manual for the textbook Linear Algebra and Its Applications, 2nd Edition, by Peter Lax (John Wiley Sons, 2007). This version omits the following problems: Exercise 2, 9 of Chapter 8 Exercise 3 of Appendix 3; Exercise problems of Appendix 4, 5, 8 and 11 1 Fundamentals Proof. Suppose 0 and 0 are two zeros of vector addition, then by the definition of zero and commutativity we have0=0+0=0+0′=0. Po0 : For any a=(x1,…,xn)∈K", we have +0=( )+(0,…,0)=(x1+0 So0=(0,.,0) is zero element of classical vector addition 3. Proof. The isomorphism T can be defined as T((al, .,an))=a1+a2+..+an-7-I Prof. Suppose S={s1,…,sn}. The isomorphism T can be defined as T(f)=(f(81),…,∫(Sn)),vf∈ 5. Proof For any p(a)=a1+a2+.+anz"-,we define 丁 T(p)=p(x), where p on the left side of the equation is regarded as a polynomial over R while p(r)on the right side of the equation is regarded as a function defined on S=[sl, ..., Sn. To prove T is an isomorphism, it suffices to prove T'is one-to-one. This is seen through the observation that p(82) om 1 and the vandermonde matrix S T-1 is invertible for distinct s1,s2,……,sn 6. Proof. For any 9,yEY, 2, tEZ and k E K, we have(by commutativity and associative law) (y+z)+(y+z)=(z+y)+(y/+2)=z+(y+(y+x)=z+(y+y/)+x)=z+(z+(y+y) (z+z)+(y+y)=(y+y)+(z+2)∈Y+Z, k(y+x)=ky+kz∈Y+Z So r+Z is a linear subspace of X if Y and Z are 7 Proof. For any c1,2∈Y∩z, since Y and Z are linear subspaces of X,m1+r2∈ Y and1+m2∈Z. Therefore,r1+x2∈Y∩z. For any k E K and a∈Ynz, since Y and Z are linear subspaces of X,kx∈ and kr E Z. Therefore, k EYnZ. Combined, we conclude Y n Z is a linear subspace of X Proof. By definition of zero vector, 0+0=0E 0. For any k E K, k0= k(0+0)=k0+k0. So k0=0E Combined, we can conclude 10 is a linear subspace of X 9. Pmoo. Define y={k11+…+k;:k…,k;∈K}. Then clearly a1=1m1+0x2+…+0r;∈Y Similarly, we can show a2,…,x∈￥. Since for any k,…,k,H,…,k∈K, (h121+…+kx)+(k11+…+k)=(k1+)x1+…+(k+k)∈Y for any k1,…,kj,k∈K, k(k1x1+…+k)=(kk1)1+…+(k);∈Y, we can conclude Y is a linear subspace of X containing 1, ..,Ii. Finally, if Z is any linear subspace of X containing 1, .. Cj, it is clear that Y c Z as Z must be closed under scalar multiplication and vector ddition. Combined, we have proven Y is the smallest linear subspace of X containing 1, .. j 10. Proof. We prove by contradiction. Without loss of generality, assume 21=0. Then 11+022+..+0zj 0. This shows 31,..., a, are linearly dependent, a contradiction. So r1 #0. We can similarly prove 2,…,x≠0 Po0. Suppose Yi has a basis yi,…, yn:. Then it suffices to prove y1,…,3hm;…,班,…, ymm form a basis of X. By definition of direct sum, these vectors span X, so we only need to show they are linearly independent In fact, if not, then 0 has two distinct representations: 0=0+...+0 and 0= >,(aiyi +...+an, 3 for r some a an,,.,ar,.,arm, where not all a; are zero. This is contradictory with the definition of direct sum. So we must have linear independence, which imply ,3hn,…,m,…, g form a basis of X. Consequently, dimX=>dimY, 12. Proof. Fix a basis 1, ...,n of X, any element a E X can be uniquely represented as 2isr ai(a)ri for some aa(x)∈K,i=1,……,n. We define the isomorphism as a→(a1(x),……,an(x). Clearly this isomorphism depends on the basis and by varying the choice of basis, we can have different isomorphisms 13. Poof. For any al,x2∈X,ifr1≡m2,ie.x1-m2∈Y, then a2-x1=-(x1-x2)∈Y,i.e.m2≡m1.This is symmetry. For any a∈X,m-m=0∈Y.Sox≡x. This is reflexivity. Finally,ix1≡r2,m2≡x3,then 1-3=(1- 2)+2-a3EY,1.e 31=23. This is transitivity 14 Proof. For any 1,m2∈X, we can find y∈{xn}n{2} if and only if a1-w∈ Y and o-y∈Y.Then 1 (x1-y)-(x2-3)∈Y So ina)#g if and only if 1=fa2. 15 Proof. If a)= and y)=y], then x-a,y-y'eY So(ar +y)-(a'+y)=(a-a)+(y-yEY This shows a+y=fa'+y. Also, for any k E K, k-k '=k(a -rEy. so kfa=ka=tka'y k{x}. 16 Proof. By theory of polynomials, we have y=a(II(t-t): q(t)is a polynomial of degree <n Then it's easy to see dimy=n-i and dim X/Y=dim X-dim y= j 17 Proof. By Theorem 6, dim X/y=dim X-dim Y=0, which implies X/Y= 0. So X=Y 18 Proof. Define y={(x,0):x∈X1,0∈X2}andY2={(0,x):0∈X1,x∈X2}. Then Yi and y2 are linear subspaces of X10 X2. It is easy to see Yi is isomorphic to X1, Y2 is isomorphic to X2, and Y1nY2=1(0, 0) So by Theorem 7, dim X1X2=dimI+dimY2-dim (YinY2)=dimX1+dimX2-0=dim X1+dimX2 19 Proof. By Exercise 18 and Theorem 6, dim(reX/y=dimY +dim(X/y=dimY +dimX-dimy=dim X Since linear spaces of same finite dimension are isomorphic(by one-to-one mapping between their bases) YOX/Y is isomorphic to X 20 Proof. (a) is not since a: a1 >0 is not closed under the scalar multiplication by -1.(b)is.(c) is not since 1 +12+1=0 and x+22+1=0 imply (a1+a1+(a2+32)+1=-1(d)is(e)is not since r being an integer does not guarantee ra1 is an integer for any rE R Proof. See the textbook's solution, page 279 uality Proof. Suppose el,..., en is a basis of X and suppose 1=2ii aiei. If the underlying field is R, we define a linear function I by setting l(ei)=ai(i=l, .. n) and extending its definition to X via linear combination. If the underlying field is C, we define I similarly by setting l(ei=ai, where ai is the complex conjugate of ai(i= 1, .. n). In either case, we have l(a1)=c12+0, where. is the Euclidean norm of r" or ct To generalize the above result to a general linear vector space X over a field K, we clearly need some notion of norm. This is exactly the starting point of Hahn-Banach Theorem, which claims a similar result for general linear vector spaces, not necessarily finite-dimensional (see Lax [6). So this exercise problem needs extra conditions if we need to go beyond K=R or K=C Proof For any I and l2EY, we have(11+l2) For any k E K,(kl)(y)=k(l(g))=k0=0 for any y E Y. So kL E Y. Combined, we conclude Yis ubspace of X 3. Po0. Since scy,Y-cS-.For“→”,letr1,…, Tm be a maximal linearly independent subset of s. Then S=span(x1,…,xm)andY={∑10:a1,…,am∈K} by Exercise9 of Chapter1. By the definition of annihilator, for any l∈S-andy=∑1a;∈Y, we have )=∑a(x;) SoLEY. By the arbitrariness of L, SCY. Combined, we have S=Y 丁 Proof. Suppose three linearly independent polynomials p1, P2 and P3 are applied to formula(9). Then ml m2 and m must satisfy the linear equations WW p(t)m1(t2)p1(t3)m1 p2(t1)P2(t2)p2(t3 3(t1)p3(t2)p3( 1=m We take p1(t)=1, p2(t)=t and p3(t)=t2. The above equation becomes 1111「m 0 111 」l- 202 a 0 m 9 Then it's easy to see that for a>v1/3, all three weights are positive To show formula(9) holds for all polynomials of degree 6 when a= V3 5, we note for any oddnEN e n dr=0, mip(-a)+m3p(a)=0 since m=m2 and p(-a)=-p(a), and m2p(0)=0 So( 9 ) holds for any a" of odd degree n. In particular, for p(a)=a and p(a)=as. For p(r)=x, we have ∧只之4d5’m1p(t1)+m2p(2)+m3p(t3)=2m4s2 So formula(9)holds for p(a)=x4 when a= v3/5. Combined, we conclude for a √3/5,(9) holds for all polynomials of degree 6 Remark 1. In this erercise problem and exercise 5 below. "Theorem 6 should be corrected to " Theorem Proof. We take pi(t)=l, P2(t)=t, p3(t)=t, and p4(t)=t. Then 1, m2, m3, and m4 solve the following equation: 2 6 6 0 62 b2 2/3 Then m22 a 6 b a 0 23 2b2 2/3 b3 63 0 dot 302 2a2-2b22a3+2ab2 2a2-2b2 2a2b+282a2+2b22a2b-2b 0 2a2-2b2 2a2b-2 2a2b+282 /3 -2a2+22-2a3+2ab2 2a3-2ab2 3 3 So the weights are positive if and only if one of the following two mutually exclusive cases hold 1)b2>,a2<b2,a2> 6. Proof.(from the textbook's solution)(a) Suppose there is a linear relation al1(P)+b2(p)+c23(P)=0 Set p=p(a=(a-52)(a-53). Then P($2)=P(E3=0, PI( 51)#0; so we get from the above relation that 0. Similarly 6=0, c=0 (b)Since dim P2=3, also dim P2=3. Since l1, l2, l3 are linearly independent, the span P? (cl) We define li by setting l1(e)= if j=1 ifj≠1 ind extending 1 to V by linear combination, i.e. l(2i-1ajej): =2israilej)=a1. l2 can be ly. If constructed similarly there exist al,..., an such that all1+..+anln=0, we have 0=a111ei)+.anIn(ei=aj, 3 So l1,..., In are linearly independent. Since dim V=dim V =n,(1,., In) is a basis of V (c2) We define -x2)(- 21=x2 aC-C1-. pi )( 2 7 Proof(from the textbook's solution) l(e has to be zero for =(1,0, -1, 2)and a =(2, 3, 1, 1). These yield two equations for c1,……,c4 c1-c3+2c4=0,2c1+3c2+c3+c4=0. We express CI and c2 in terms of c and c4. From the first equation, C1 C3-2c4. Setting this into the second equation gives C2=-C3+C4 3 Linear Mappings Proof. For any y,y’∈T(X), there exist,x′∈ X such that(x)= y and T(x)=y.Soy+y T(a)+T()=T(a +TET(X). For any k E K, ky= kT(a)=T(ka)E T(X). Combined, we conclude T(X) is a linear subspace of U (b) Proof. Suppose V is a linear subspace of U. For any a, a'ET-(v), there exist 3, y/E V such that T(e)=y andT(x)=y. Since 1(x+x)=(x)+T(x)=y+y′∈V,x+m∈T-1(V). or any k∈K, sInce T(kr)=kr(a)=kyeV, krEr-(V). Combined, we conclude T-(V)is a linear subspace of X 2. Proof.(from the textbook's solution) Suppose we drop the ith equation; if the remaining equations do not determine a uniquely, there is an a that is mapped into a vector whose components except the ith are zero If this were true for all i= 1, ..,m, the range of the mapping a -u would be m-dimensional; but according to Theorem 2, the dimension of the range is n m. Therefore one of the equations may be dropped without losing uniqueness; by induction m-n of the equations may be omitted Alternative solation: Uniqueness of the solution a implies the column vectors of the matrix T=(tii) are linearly independent. Since the column rank of a matrix equals its row rank( see Chapter 4), it is possible to select a subset of n of these equations which uniquely determine the solution Remark 2. The tectbook's solution is a proof that the column rank of a matric equals its row rank. 3. Proof.So(ax+by)=S(r(ar+by)=S(a(x)+b(y)=aS(T(x)+bS(T(y)=asoT(x)+bsoT(y) So SoT is also a linear mapping. Pmof.(R+S)oT(x)=(R+S)(T(x)=R(T(x)+S(T(x)=(RoT+Som)(x)andSo(T+P)(x)= S(+P)(x)=S(T(x)+P(x)=S(T(x)+S(P(x))=(SoT+SoP)(x) 4. Proof. Linearity of s and T is easy to see. For non-commutativity, consider the polynomial 8. Then TS(s)=T(82)=2≠s=S(1)=ST(s).SoST≠TS. Proof. For any a =(31, 2, I3)E X, S(a)=(1, 53,-22) and T(a)=(a3, 2,-21). So it's easy to see S and T are linear. For non-commutativity, note ST(r)= S(3, 2, -1)=(a3,-301, -.2)and Ts(a) T(1, T3, -12)=(2, 23, -11). So ST+TS in general Remark 3. Note the problem does not specify the direction of the rotation, so it is also possible that s() (31,-3, I2)and T(a)=(3, 32, T1). There are total of four choices of(S,T). But the corresponding proofs are similar to the one presented here 5 Proof. TT-I()=T(T-())=a by definition. So TT-=id 6 Proo∫. Suppose T':X→ u is invertible. Then for any y,y'∈U, there exist a unique I∈ X and a unique E X such that T(a)=y and T(a=y. So T(a +a)=T(a)+r(a=y+y' and by the injectivity of T,T-y+y=r+a'=T-y)+r-(y). For any k E K, since T(k )=kr(r)=ky, injectivity of T implies T-(kg)=ka=kr-I(g). Combined, we conclude T-is linear Proof. Suppose T:X-U and S: U-V. First, by the definition of multiplication, ST is a linear map Second, if a E X is such that ST(a)=0EV, the injectivity of S implies T()=OEU and the injectivity of T further implies =0E X. So, ST is one-to-one. For any z E V, there exists y E U such that S(y)=z Also, we can find a E X such that T(a)=y. So ST()=s(y)=z. This shows ST is onto. Combined, we conclude sr is invertible By associativity, we S with T-I and T with S-, we also have(T-S-ST)= id x. Therefore, we can conclude(ST)- T-S Proof. Suppose T: X+ U and S: U-v are linear maps. Then for any given IE v, (ST)L, r) L, ST )=(SL, T)=(TSL, a), Vr E X. Therefore,(ST)'I=T'ST. Let I run through every element of V, we conclude(ST)=TS. Proof. Suppose T and R are both linear maps from X to U. For any given LEU, we have((T+ Ryl, a) (L,(T+R)x)=(, Tx+ Rr)=(L, Ta)+(L, Ra)=(TL, r)+(R'L, a)=((T'+R)L, r),Va E X. Therefore (T+R'L=(T+R'L. Let L run through every element of V, we conclude(T+r)'=T+R Proof. Suppose T is an isomorphism from X to U, then T- is a well-defined linear map. We first show T is an isomorphism from U to X. Indeed, if E U is such that Tl=0, then for any EEX,0= (T'L, a)=(L,Tr). As a varies and goes through every element of X, Ta goes through every element of U. By considering the identification of U with U, we conclude l=0. So Tv is one-to-one. For any given mEX, define I=mr-, then LE U. For any r E X, we have(m, r)=(m, T-I(Tr))=(L, Tr)=(T'L, a) Since s is arbitrary, m= TI and T is therefore onto. Combined, we conclude T is an isomorphism from Uto X' and(T)- is hence well-defined iaV part(),(T-1)Tv=(TT-)'=(idu)=idu and T'(T-I)=(T-IT)'=(idx)=idx". This shows B y=(T) 10