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1、 LINEAR ALGEBRA AND ITS APPLICATIONS 姓名：易 学号：成绩：1.Definitions(1)Pivot position in a matrix;(2)Echelon Form;(3)Elementary operations;(4)Onto mapping and one-to-one mapping;(5)Linearly independence.2.Describe the row reduction algorithm which produces a matrix in reduced echelon form.3.Find the 3 3 ma
2、trix that corresponds to the composite transformation of a scaling by 0.3,a rotation of 90,and finally a translation that adds(-0.5,2)to each point of a figure.4.Find a basis for the null space of the matrix 361171223124584A 5.Find a basis for ColA of the matrix 1332-9-2-22-822307134-1 11-8A 6.Let a
3、 and b be positive numbers.Find the area of the region bounded by the ellipse whose equation is 22221xyab 7.Provide twenty statements for the invertible matrix theorem.8.Show and prove the Gram-Schmidt process.9.Show and prove the diagonalization theorem.10.Prove that the eigenvectors corresponding
4、to distinct eigenvalues are linearly independent.Answers:1.Definitions(1)Pivot position in a matrix:A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A.A pivot column is a column of A that contains a pivot position.(2)Echelon Form:A rect
5、angular matrix is in echelon form(or row echelon form)if it has the following three properties:1.All nonzero rows are above any rows of all zeros.2.Each leading entry of a row is in a column to the right of the leading entry of the row above it.3.All entries in a column below a leading entry are zer
6、os.If a matrix in a echelon form satisfies the following additional conditions,then it is in reduced echelon form(or reduced row echelon form):4.The leading entry in each nonzero row is 1.5.Each leading 1 is the only nonzero entry in its column.(3)Elementary operations:Elementary operations can refe
7、r to elementary row operations or elementary column operations.There are three types of elementary matrices,which correspond to three types of row operations(respectively,column operations):1.(Replacement)Replace one row by the sum of itself anda multiple of another row.2.(Interchange)Interchange tw
8、o rows.3.(scaling)Multiply all entries in a row by a nonzero constant.(4)Onto mapping and one-to-one mapping:A mapping T:n m is said to be onto m if each b in m is the image of at least one x in n.A mapping T:n m is said to be one-to-one if each b in m is the image of at most one x in n.(5)Linearly
9、independence:An indexed set of vectors V1,.,Vp in n is said to be linearly independent if the vector equation x1v1+x2v2+.+xpvp=0 Has only the trivial solution.The set V1,.,Vp is said to be linearly dependent if there exist weights c1,.,cp,not all zero,such that c1v1+c2v2+.+cpvp=0 2.Describe the row
10、reduction algorithm which produces a matrix in reduced echelon form.Solution:Step 1:Begin with the leftmost nonzero column.This is a pivot column.The pivot position is at the top.Step 2:Select a nonzero entry in the pivot column as a pivot.If necessary,interchange rows to move this entry into the pi
11、vot position.Step 3:Use row replacement operations to create zeros in all positions below the pivot.Step 4:Cover(or ignore)the row containing the pivot position and cover all rows,if any,above it.Apply steps 1-3 to the submatrix that remains.Repeat the process until there all no more nonzero rows to
12、 modify.Step 5:Beginning with the rightmost pivot and working upward and to the left,create zeros above each pivot.If a pivot is not 1,make it 1 by scaling operation.3.Find the 3 3 matrix that corresponds to the composite transformation of a scaling by 0.3,a rotation of 90,and finally a translation
13、that adds(-0.5,2)to each point of a figure.Solution:If=/2,then sin=1 and cos=0.Then we have 110003.00003.01yxyxscale 110003.00003.0100001010yxRotate 110003.00003.0100001010125.0010001yxTranslate The matrix for the composite transformation is 10003.00003.0100001010125.0010001 10003.00003.0125.0001010
14、 125.0003.003.00 4.Find a basis for the null space of the matrix 361171223124584A Solution:First,write the solution of AX=0 in parametric vector form:A 000023021010002001,x1-2x2 -x4+3x5=0 x3+2x4-2x5=0 0=0 The general solution is x1=2x2+x4-3x5,x3=-2x4+2x5,with x2,x4,and x5 free.1020301201000122232542
15、5454254254321xxxxxxxxxxxxxxxx u v w=x2u+x4v+x5w (1)Equation(1)shows that Nul A coincides with the set of all linear conbinations of u,v and w.That is,u,v,wgenerates Nul A.In fact,this construction of u,v and w automatically makes them linearly independent,because(1)shows that 0=x2u+x4v+x5w only if t
16、he weights x2,x4,and x5 are all zero.So u,v,w is a basis for Nul A.5.Find a basis for ColA of the matrix 1332-9-2-22-822307134-1 11-8A Solution:A 07490012002300130001,so the rank of A is 3.Then we have a basis for ColA of the matrix:U=0001,v=0013and w=0749 6.Let a and b be positive numbers.Find the
17、area of the region bounded by the ellipse whose equation is 22221xyab Solution:We claim that E is the image of the unit disk D under the linear transformation T determined by the matrix A=ba00,because if u=21uu,x=21xx,and x=Au,then u1=ax1 and u2=bx2 It follows that u is in the unit disk,with 12221 u
18、u,if and only if x is in E,with 1)()(2221bxax.Then we have area of ellipse=area of T(D)=|det A|area of D =ab(1)2=ab 7.Provide twenty statements for the invertible matrix theorem.Let A be a square nn matrix.Then the following statements are equivalent.That is,for a given A,the statements are either a
19、ll true or false.a.A is an invertible matrix.b.A is row equivalent to the nn identity matrix.c.A has n pivot positions.d.The equation Ax=0 has only the trivial solution.e.The columns of A form a linearly independent set.f.The linear transformation x Ax is one-to-one.g.The equation Ax=b has at least
20、one solution for each b in n.h.The columns of A span n.i.The linear transformation x Ax maps n onto n.j.There is an nn matrix C such that CA=I.k.There is an nn matrix D such that AD=I.l.AT is an invertible matrix.m.If 0A,thenT11TAA n.If A,B are all invertible,then(AB)*=B*A*o.T*T)(A)(A p.If 0A,then*1
21、1*AA q.*1n*A1)(A r.If 0A,thenL11LAA(L is a natural number)s.*1n*AK)(KA t.If 0A,then*1AA1A 8.Show and prove the Gram-Schmidt process.Solution:The Gram-Schmidt process:Given a basis x1,.,xp for a subspace W of n,define 11xv 1112222vvvvxxv 222231111333vvvvxvvvvxxv .1p1p1p1pp2222p1111pppvvvvxvvvvxvvvvxx
22、v Then v1,.,vp is an orthogonal basis for W.In addition Span v1,.,vp=x1,.,xp for pk 1 PROOF For pk 1,let Wk=Span v1,.,vp.Set 11xv,so that Span v1=Span x1.Suppose,for some k p,we have constructed v1,.,vk so that v1,.,vk is an orthogonal basis for Wk.Define 1kw1k1kxprojxvk By the Orthogonal Decomposit
23、ion Theorem,vk+1 is orthogonal to Wk.Note that projWkxk+1 is in Wk and hence also in Wk+1.Since xk+1 is in Wk+1,so is vk+1(because Wk+1 is a subspace and is closed under subtraction).Furthermore,0v1k because xk+1 is not in Wk=Span x1,.,xp.Hence v1,.,vk is an orthogonal set of nonzero vectors in the(
24、k+1)-dismensional space Wk+1.By the Basis Theorem,this set is an orthogonal basis for Wk+1.Hence Wk+1=Span v1,.,vk+1.When k+1=p,the process stops.9.Show and prove the diagonalization theorem.Solution:diagonalization theorem:If A is symmetric,then any two eigenvectors from different eigenspaces are o
25、rthogonal.PROOF Let v1 and v2 be eigenvectors that correspond to distinct eigenvalues,say,1and2.To show that 0vv21,compute 2T12T11211v)(Avv)v(vv Since v1 is an eigenvector 2T12TT1AvvvAv )(221vvT 2122T12vvvv Hence 0vv2121,but 021,so 0vv21 10.Prove that the eigenvectors corresponding to distinct eigen
26、values are linearly independent.Solution:If v1,.,vr are eigenvectors that correspond to distinct eignvalues 1,.,r of an nn matrix A.Suppose v1,.,vr is linearly dependent.Since v1 is nonzero,Theorem,Characterization of Linearly Dependent Sets,says that one of the vectors in the set is linear combinat
27、ion of the preceding vectors.Let p be the least index such that vp+1 is a linear combination of he preceding(linearly independent)vectors.Then there exist scalars c1,.,cp such that 1ppp11vvcvc (1)Multiplying both sides of(1)by A and using the fact that Avk=kvk for each k,we obtain 111pppAvAvcAvc 111
28、11pppppvvcvc (2)Multiplying both sides of(1)by1p and subtracting the result from(2),we have 0)()(11111ppppcvc (3)Since v1,.,vp is linearly independent,the weights in(3)are all zero.But none of the factors 1pi are zero,because the eigenvalues are distinct.Hence 0ic for i=1,.,p.But when(1)says that 01pv,which is impossible.Hence v1,.,vr cannot be linearly dependent and therefore must be linearly independent.