哈工大选修课 LINEAR ALGEBRA 试卷及答案

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 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 matrix

361171

22312

4

5

8

4A ---??

??=--????--??

5. Find a basis for Col A of the matrix

1

332-9-2-22-82230

7

13

4

-111

-8A ?????

?=??????

6. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is

222

2

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 : 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 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 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 ?????

?????????????????→

?????

??????110003.000

03

.01y x y x scale

????

?

?????????

?

?????????

??????-??

→?11000

3.000

03

.01000

010

10

y x R o t a t e

????

?

????????????????????

?????-?????

???

??-??

?→?11000

3.000

03

.01000

010

10

125.00

100

01

y x T r a n s l a t e

The matrix for the composite transformation is ????

?

??????????

?????-??????????-1000

3.00003

.01000

01010

125.00

10001

?????

?????????

?

?????--=10003.000

03

.012

5.00

01010

????

??????--=125.00

03

.00

3.00

4. Find a basis for the null space of the matrix 3

61171

22312

4

5

8

4A ---??

??=--????--??

Solution:

First, write the solution of A X=0 in parametric vector form: A ~ ????

?

?????---0000

230

210

100

020

1

, x 1-2x 2 -x 4+3x 5=0 x 3+2x 4-2x 5=0 0=0

The 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. ????

???

?

????????-+????????????????-+????????????????=??

?

???

?

?????????+--+=???????

?

????????102030120100012223254254542

54254321x 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 A of the matrix 1332-9-2-22-82230

7

13

4

-111

-8A ??????=??????

Solution: A ~ ?????

????

???---074

90

0120

0230

0130

001, so the rank of A is 3. Then we have a basis for Col A of the matrix: U = ????????????0001, v = ????????????0013and w = ?????

????

???--074

9

6. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is

222

2

1x 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=??

?

?

??b a 00

, because if u= ??????21u u , x=???

???21x x , and x = Au, then u 1 =

a

x 1 and u 2 =

b

x 2

It follows that u is in the unit disk, with 12

22

1≤+u u , if and only if x is in E , with

1)()(2

221≤+b 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 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 span

n

.

i. The linear transformation x → Ax maps n

onto

n

.

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 ()()T

11

T 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 ()()L

11

L A A --= ( L is a natural number )

s. ()*

1

n *

A K)

(KA --=-

t. If 0A ≠, then *

1

A 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 n

, define

11x v = 1112222v v v v x x v ??-

=

22

22311

11333v v v v x v v v v x x v ??-

??-=

. .

.

1p 1

p 1p 1p p 22

22p 11

11p 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 ≤≤1 PROOF

For 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 . Define

1k w

1k 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. PROOF

Let v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say, 1λand 2λ. T o show that 0v v 21=?, compute

2T 12T 11211v )(A v v )v (λv v λ==? Since v 1 is an eigenvector ()

()2

T

12T T

1Av

v v A v ==

)(221v v T

λ=

2122T

12v 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 c

11111++=+???+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 have

0)()(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.