(完整版)哈工大选修课LINEARALGEBRA试卷及答案,推荐文档

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 matrix that corresponds to the composite transformation of a scaling by 0.3,

33?a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.

90?4. Find a basis for the null space of the matrix 361171223124584A ---????=--????--??

5. Find a basis for Col of the matrix

A 1332-9-2-22-822307134-111-8A ??????=??????

6. Let and be positive numbers. Find the area of the region bounded by the ellipse

a b whose equation is

22

22

1x y a b +=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 : R n → R m is said to be onto R m if each b in R m is the image of at least one x in R n.

A mapping T : R n → R m is said to be one-to-one if each b in R m is the image of at most one x in R n.

(5) Linearly independence:

An indexed set of vectors {V1, . . . ,V p} in R n is said to be linearly independent if the vector equation

x 1v 1+x 2v 2+ . . . +x p v p = 0

Has 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 = 0

2. 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 matrix that corresponds to the composite transformation of a scaling by 0.3,

33?a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.

90?Solution:

If ψ=π/2, then sin ψ=1 and cos ψ=0. Then we have

??????????????????????→?????

??????110003.00003.01y x y x scale ????

??????????????????????????-??→?110003.00003.010*******y x Rotate

????

??????????????????????????-??????????-???→?110003.00003.0100001010125.0010001y x Translate The 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 ~ , ????

??????---000023021010002001x 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 x

u 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 of the matrix A 1332-9-2-22-822307134-111-8A ??????=??????

Solution:

A ~ , so the rank of A is 3.????????????---074900120023

00130001Then we have a basis for Col of the matrix:

A U = , v = and w = ????????????0001????????????0013????

?

???????--07496. Let and be positive numbers. Find the area of the region bounded by the ellipse

a b whose equation is

22

221x y a b

+=Solution:

We claim that E is the image of the unit disk D under the linear transformation T

determined by the matrix A=, because if u= , x=, and x = Au, then ??????b a 00??????21u u ??

????21x x u 1 =

and u 2 = a x 1b

x 2It follows that u is in the unit disk, with , if and only if x is in E , with 1222

1≤+u u

1)()(2221≤+x a x . Then we have

{area of ellipse} = {area of T (D )}

= |det A| {area of D} = ab π(1)2 = πab 7. Provide twenty statements for the invertible matrix theorem.

Let A be a square matrix. Then the following statements are equivalent. That is, for a

n n ?given A, the statements are either all true or false.

a. A is an invertible matrix.

b. A is row equivalent to the identity matrix.

n n ?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 R n .

h. The columns of A span R n .

i. The linear transformation x Ax maps R n onto R n .

→j. There is an matrix C such that CA = I.

n n ?k. There is an matrix D such that AD = I.

n n ?l. A T is an invertible matrix.m. If , then 0A ≠()()T

11T A A --=n. If A, B are all invertible, then (AB)* = B *A *

o. T

**T )(A )(A =p. If , then 0A ≠()()*11*A A

--=q. ()*

1n *A 1)(A --=-r. If , then ( L is a natural number )0A ≠()()L 11L A A

--=s. ()*

1n *A K)(KA --=-t. If , then 0A ≠*1A A

1A =-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 R n , define

11x v = 11

12222v v v v x x v ??-= 22

2231111333v v v v x v v v v x x v ??-??-

=

. ..1

p 1p 1p 1p p 2222p 1111

p 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 addition

Span {v 1, . . . , v p } = {x 1, . . . , x p } for p

k ≤≤1PROOF

For , let W k = Span {v 1, . . . , v p }. Set , so that Span {v 1} = Span

p k ≤≤111x v ={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 . Define

1

k w 1k 1k x proj 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, because x k+1 is not in W k = Span {x 1, . . . ,

0v 1k ≠+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.

PROOF

Let v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say, and . To 1λ2λshow that , compute

0v v 21=? Since v 1 is an eigenvector

2T 12T 11211v )(Av v )v (λv v λ==?

()()2T 12T T 1Av v v A v ==

)(221v v T

λ= 2122T 12v v λv v λ?==Hence , but , so ()0v v λλ2121=?-()0λλ21≠-0

v 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

(1)

1p p p 11v v c v c +=+???+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 c (2)

11111++=+???+p p p p p v v c v c λλλMultiplying both sides of (1) by and subtracting the result from (2), we have

1+p λ (3)

0)()(11111=-+???+-++p p p p c v c λλλλSince {v 1, . . . , v p } is linearly independent, the weights in (3) are all zero. But none of the

factors are zero, because the eigenvalues are distinct. Hence for i = 1, . . . , p. 1+-p i λλ0=i

c

But when (1) says that , which is impossible. Hence {v 1, . . . , v r } cannot be linearly

01=+p v dependent and therefore must be linearly independent.