最优化导论第四版答案

A are linearly dependent, and Ay is a linear combination of the columns of A). Let a be a solution to Ac=b. Then clearly a+yt is also a solution. This contradicts the uniqueness of the solution. Hence rank A +:By Theorem 2.1, a solution exists. It remains to prove that it is unique. For this, let a and y be solutions, i.e.,A= b and Ay=b. Subtracting, we get A(a-y)=0. Since rank A= n and A has n columns, then -y=0 and hence =y, which shows that the solution is unique 2.3 Consider the vectors ai=1,a∈Rn+1,讠=1,...,k. Since k≥n+2, then the vectors a1,, ak must be linearly independent in Rn+I. llence, there exist a1, .. ak, not all zero, such that 0 The first component of the above vector equation is 2i_i ai=0, while the last n components have the form i-l diai=0, completing the proof 2.4 We first postmultiply M by the matrix Mm-kk Im-k to obtain 7-h,k Mkk O Mm-k, k Im-kMkk O Note that the determinant of the postmultiplying matrix is 1. Next we postmultiply the resulting product m-k O Mkk O O O ME, k otice that NM=(O)(.6 where det The above easily follows from the fact that the determinant changes its sign if we interchange columns, as discusscd in Scction 2. 2. Morcovcr, det (I k)det(mk, k) Mk,k) o Mkk dctM=± dct Mk b. We can see this on the following examples. We assume, without loss of generality that Mm-kk=O and let kb=2. Thus k=1. First consider the case when m= 2. Then we have k, ke O Us det m=-2= det(-M k, k) Next consider the case when m=3. Then m-h =2≠det(-Mk,k) Mkk O Therefore, in general detM≠det(-Mk,k) However, when k= m/2, that is, when all sub-matrices are square and of the same dimension then it is true that etm= det (Mn, k) Sc[121] 2.5 Let A B C D and suppose that each block is k k. John R. Silvester [121] showed that if at least one of the blocks is equal too(zero matrix), thell the desired formula holds. Indeed, if a row or coluMn block is zero, then the determinant, is equal to zero as follows from the determina.nt,'s properties discussed Section 2.2. That, is, if A=B=O, or A=C=O, and so on, then obviously dct M=0. This includes the casc when any thrcc or all four block matrices are zero matrices IfB= or c=o then A B det m= det det (ad) C D The only case left to analyze is when a=O or D=O. We will show that in either case det M =det(-BC) Without loss of generality suppose that D=O. Following arguments of John R. Silvester [121], we premul tiply M by the product of three matrices whose determinants are unity: Ik IkA B o I OI A B He ence a B de A B det(c)det B det(-1k)detC det B Thus we have detaB C O det(bc)=det(CB) 2.6 We represent the given system of equations in the form Aa=b, where 1 and b 1-20-1 Using elementary row operations yield 1121 4 ane A,b 1-20-1-2 from which rank A=2 and rankA, b= 2. Therefore, by Theorem 2.1, the system has a solution We next represent the system of equations as 2+4 Assigning arbitrary values to cs and 4(3=d3, a4=d4), we get 1 1[1-23 4 2+x2 2-11-2 4 2+x ds -od 1-2d3 Therefore, a general solution is d3-寺d4 d 3 d4+ 3 1 0 where d and d4 are arbitrary values 2.7 1. Apply the definition of -a if 0 if-a if >= if a >0 2. If>0, then a=a. If a 0>a. Hence a>u On the other hand, -u2-a (by the above). Hence, a2--a=-la(by property 1) 3. We have four cases to consider. First, if a, b20, then a+b20. Ilence, a+b=a+b-a+bl Second, if a, b20, then a+bs0. Hence a+b=-(a+b) b=a+ b Third. if a>o and b