哈工大选修课 LINEAR ALGEBRA 试卷及答案

LINEAR ALGEBRAANDITS 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 33⨯ matrix 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 matrix361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦5. Find a basis for Col A of the matrix1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦6. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is22221x y ab+=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 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 rectangular 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 zeros.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 refer 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 two 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 independence:An indexed set of vectors {V1, . . . ,V p} in n is said to be linearly independent if the vector equationx 1v 1+x 2v 2+ . . . +x p v p = 0Has only the trivial solution. The set {V 1, . . . ,V p } is said to be linearly dependent if there exist weights c 1, . . . ,c p , not all zero, such that c 1v 1+c 2v 2+ . . . +c p v p = 02. Describe the row 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 pivot 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 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 33⨯ matrix 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. Solution:If ψ=π/2, then sin ψ=1 and cos ψ=0. Then we have ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡110003.00003.01y x y x scale⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−→−110003.00003.0100001010y x R o t a t e⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−−→−110003.00003.0100001010125.0010001y x T r a n s l a t eThe matrix for the composite transformation is ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-10003.00003.0100001010125.0010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=10003.00003.0125.0001010⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=125.0003.003.004. Find a basis for the null space of the matrix 361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦Solution:First, write the solution of A X=0 in parametric vector form: A ~ ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---00002302101000201, x 1-2x 2 -x 4+3x 5=0 x 3+2x 4-2x 5=0 0=0The general solution is x 1=2x 2+x 4-3x 5, x 3=-2x 4+2x 5, with x 2, x 4, and x 5 free. ⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+--+=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡10203012010001222325425454254254321x x x x x x x x x x x x x x x xu v w=x 2u+x 4v+x 5w (1)Equation (1) shows that Nul A coincides with the set of all linear conbinations of u, v and w . That is, {u, v, w}generates Nul A. In fact, this construction of u, v and w automatically makes them linearly independent, because (1) shows that 0=x 2u+x 4v+x 5w only if the weights x 2, x 4, and x 5 are all zero.So {u, v , w} is a basis for Nul A.5. Find a basis for Col A of the matrix 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦Solution: A ~ ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---07490012002300130001, so the rank of A is 3. Then we have a basis for Col A of the matrix: U = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0001, v = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0013and w = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--07496. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is22221x y ab+=Solution:We claim that E is the image of the unit disk D under the linear transformation Tdetermined by the matrix A=⎥⎦⎤⎢⎣⎡b a 00, because if u= ⎥⎦⎤⎢⎣⎡21u u , x=⎥⎦⎤⎢⎣⎡21x x , and x = Au, then u 1 =ax 1 and u 2 =bx 2It follows that u is in the unit disk, with 12221≤+u u , if and only if x is in E , with1)()(2221≤+b x a x . Then we have{area of ellipse} = {area of T (D )} = |det A| {area of D} = ab π(1)2= πab7. Provide twenty statements for the invertible matrix theorem.Let A be a square n n ⨯ matrix. Then the following statements are equivalent. That is, for a given A, the statements are either all true or false. a. A is an invertible matrix.b. A is row equivalent to the n n ⨯ 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 one solution for each b in n.h. The columns of A spann.i. The linear transformation x → Ax maps nonton.j. There is an n n ⨯ matrix C such that CA = I. k. There is an n n ⨯ matrix D such that AD = I. l. A T is an invertible matrix. m. If 0A ≠, then ()()T11T A A --=n. If A, B are all invertible, then (AB)* = B *A *o. T**T )(A )(A =p. If 0A ≠, then ()()*11*A A --=q. ()*1n *A 1)(A --=-r. If 0A ≠, then ()()L11L A A --= ( L is a natural number )s. ()*1n *A K)(KA --=-t. If 0A ≠, then *1A A1A =-8. Show and prove the Gram-Schmidt process.Solution:The Gram-Schmidt process:Given a basis {x 1, . . . , x p } for a subspace W of n, define11x v = 1112222v v v v x x v ⋅⋅-=222231111333v v v v x v v v v x x v ⋅⋅-⋅⋅-=. ..1p 1p 1p 1p p 2222p 1111p p p v v v v x v v v v x v v v v x x v ----⋅-⋅⋅⋅-⋅⋅-⋅⋅-=Then {v 1, . . . , v p } is an orthogonal basis for W. In additionSpan {v 1, . . . , v p } = {x 1, . . . , x p } for p k ≤≤1 PROOFFor p k ≤≤1, let W k = Span {v 1, . . . , v p }. Set 11x v =, so that Span {v 1} = Span {x 1}.Suppose, for some k < p, we have constructed v 1, . . . , v k so that {v 1, . . . , v k } is an orthogonal basis for W k . Define1k w1k 1k x p r o j x v k+++-= By the Orthogonal Decomposition Theorem, v k+1 is orthogonal to W k . Note that proj Wk x k+1 is in W k and hence also in W k+1. Since x k+1 is in W k+1, so is v k+1 (because W k+1 is a subspace and is closed under subtraction). Furthermore, 0v 1k ≠+ because x k+1 is not in W k = Span {x 1, . . . , x p }. Hence {v 1, . . . , v k } is an orthogonal set of nonzero vectors in the (k+1)-dismensional space W k+1. By the Basis Theorem, this set is an orthogonal basis for W k+1. Hence W k+1 = Span {v 1, . . . , v k+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 orthogonal. PROOFLet v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say, 1λand 2λ. T o show that 0v v 21=⋅, compute2T 12T 11211v )(A v v )v (λv v λ==⋅ Since v 1 is an eigenvector ()()2T12T T1Avv v A v ==)(221v v Tλ=2122T12v v λv v λ⋅==Hence ()0v v λλ2121=⋅-, but ()0λλ21≠-, so 0v v 21=⋅10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent. Solution:If v 1, . . . , v r are eigenvectors that correspond to distinct eignvalues λ1, . . . , λr of an n n ⨯ matrix A.Suppose {v 1, . . . , v r } is linearly dependent. Since v 1 is nonzero, Theorem, Characterization of Linearly Dependent Sets, says that one of the vectors in the set is linear combination of the preceding vectors. Let p be the least index such that v p +1 is a linear combination of he preceding (linearly independent) vectors. Then there exist scalars c 1, . . . ,c p such that 1p p p 11v v c v c +=+⋅⋅⋅+ (1) Multiplying both sides of (1) by A and using the fact that Av k = λk v k for each k, we obtain 111+=+⋅⋅⋅+p p p Av Av c Av c11111++=+⋅⋅⋅+p p p p p v v c v c λλλ (2) Multiplying both sides of (1) by 1+p λ and subtracting the result from (2), we have0)()(11111=-+⋅⋅⋅+-++p p p p c v c λλλλ (3) Since {v 1, . . . , v p } is linearly independent, the weights in (3) are all zero. But none of the factors 1+-p i λλ are zero, because the eigenvalues are distinct. Hence 0=i c for i = 1, . . . ,p. But when (1) says that 01=+p v , which is impossible. Hence {v 1, . . . , v r } cannot be linearly dependent and therefore must be linearly independent.。