Final Exam Linear Algebra

8/2/2019 Final Exam Linear Algebra

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1

Final Exam

1. Consider the following matrix

=

21000

1-0100

20021

A

(a). State the Rank and nullity of matrixA.

=

21000

1-0100

20021

A MatrixA is already in row reduced form

BecauseA has three nonzero row, the rank ofA is 3. Also, the number of columns of A

is n = 5, which implies that the nullity of A is n - rank = 5 - 3 = 2.

(b). Write a basis for the solution spaceAx = 0.

The system of equation corresponding to the reduced row-echelon form is

02

0

022

54

53

521

=+

=

=++

xx

xx

xxx

We choose2x and 5x as free variables to represent the solutions in this parametric form

tx

tx

tx

sx

tsx

=

=

=

=

=

5

4

3

2

1

2

22

This means that the solution space of 0x =A consists of all solution vector x of the form

shown below

Darcel FordMAT-22043-SP-2011

Linear AlgebraProfessor: Dr. Timothy Lo

March 1 2011


2/11

2

=x

=

=

=

12

1

0

2

00

0

1

2

2

22

5

4

3

2

1

ts

tt

t

s

ts

xx

x

x

x

A basis for the solution space ofA consist of the vectors

0

0

0

1

2

and

1

2

1

0

2

2. Let basis B = {(1, 2), (2, 3)} be given for R2. Find the coordinate matrix for

=

5

2x

We write x as a linear combination of the vectors inB.

)3,2()2,1()5,2(21

+== ccx

Equating corresponding components yields the following system of linear equations.

532

22

)3,2()2,1()5,2(

21

21

21

=

=+

+==

cc

cc

ccx

The solution of this system is7

41 =c and .

7

92 =c

So, )3,2(7

9)2,1(

7

4)5,2( +==x and [ ]

=

7

97

4

Bx


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3. Consider two vectors in R2: u = (1, 3) and v =(-2, 1)

)3,1(=u and )1,2(=v

vu Find(a)

132)1)(3()2)(1( =+=+= vu

vu andFind(b)

10)3()1( 22 =+=u

5)1()2( 22 =+=v

(c) What is the angle (in radians) formed by vectors u and v?

The cosine of the angle between u and v is given by

10

2

25

1

50

1

510

1

cos ====

vu

vu

So, 1.42910

2cos 1-

(d) Find the projection of vector u onto vector v.

==+

+

=

= 5

1,

5

2)1,2(

5

1)1,2(

)1()2(

)1)(3()2)(1(proj 22vvv,

vu,uv

4. Suppose u = (u1, u2) and v = (v1, v2) are vectors inR2. Explain why the operation

21vuvu = cannot be an inner product.

Since ;212211

vuvuvuvu += 21

vu cannot be an inner product.


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5. Consider the basis B = {(3, 2), (1, -4)} for R2. Use the Gram-Schmidt process torewrite this basis as an orthonormalized basis.

)}4,1(),2,3{(

21

=B

vv

First, we orthogonalize each vector inB.

)2,3(11 == vw

=

=

=

+

+=

=

1342,

1328

13

10,

13

15)4,1()2,3(

13

5)4,1(

)2,3(23

)2)(4()3(1)4,1(

,

,221

12

1222 w

ww

wvvw

Then, we normalize the vectors.

=

+

==13

2,

13

3)2,3(

23

1

221

11

w

wu

=

+

==

2548

42,

2548

28

13

42,

13

28

13

42

13

28

1

222

22

w

wu

So, the orthonormal basis is

=

2548

42,

2548

28,

13

2,

13

3B


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5

6. Explain why the functionf(x) = ex

fromR toR is not a linear transformation.

f(x) = ex is not a linear transformation fromR toR because, in general

( ) 2121 xxxx eee ++

For instance, ( ) 3232 eee ++

7.

)5,2()0,1( =T

)3,1()1,0( =T

Find )7,4( T

(a) Because )7,4( can be written as

),1,0(7)0,1(4)7,4( = we can use property 4 of theorem 6.1 to write

)1,15(

)3,1(7)5,2(4

)1,0(7)0,1(4)7,4(

=

=

= TTT

(b) The system of equation for the transformation is

2

1

35

2

vba

vba

=+

=

Therefore,

135

32

=+

=

ba

ba

Solution to this system is 11

8

=a and 11

17

=b

So,

=

11

17,

11

8v


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8. Consider the linear transformation T: R3 R

2defined by xx

=

421

312)(T

(a) Find the kernel ofT.

The kernel of T is the set of all x = (x1, x2, x3) in R3 such that

T(x1, x2, x3) = 0

From this equation, we can write the homogeneous system

421

312

=

0

0

3

2

1

x

x

x

0421

032

321

321

=++

=++

xxx

xxx

Writing the augmented matrix of this system in reduced row-echelon form produces

05

1110

05

201

32

31

5

11

5

2

xx

xx

=

=

Using the parameter t=x3 produces the family of solutions

=

=

15

115

2

5

115

2

3

2

1

t

t

t

t

x

x

x

Using the parameter t=x3 produces the family of solutions

So, the kernel of T is represented by

= numberrealais:1,

5

11,

5

2ker(T) tt

(b). State the rank and nullity ofT.

Because the row-echelon form ofThas two nonzero row, it has a rank of 2. The

nullity is dim(domain) - rank = 3 - 2 = 1.


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9. Describe the kernel of the linear transform T:P1R, defined by dxxppT =1

0)()(

where P1 is the set of continuous linear function on [0, 1], and R is the set of realnumbers.

A generic integral transform takes the form ,)(),()( dxxfxpKpT =

withp

being the being the transform parameter. The transform takes a function )(xf and

produces a new function ).(pT The function ),( xpK is called the kernel of the

transform. The kernel of an integral transform, along with the limits and ,

distinguish a particular integral transform from another.

In the given case: solving the equation ( ) 0)()(1

010

1

0=+== dxxaadxxppT

yields .02

1

0

=+a

a Let ;21

aa = then, ,00

= aa and .0

aa =

{ }= aaxaT :2)(ker

10. Isomorphism

(a) IfMm,n, the space of all m by n matrices, andMj,k, the space of allj by k

matrices, are assumed to be isomorphic, what is the strongest requirementthat can be inferred about the values m, n, j and k?

The strongest requirement that can be inferred about the values m, n, j and k?

is that mn =jk, because thenMm,n andMj,k, will have the same dimension; and thereforeare isomorphic.

(b) Explain why R2 and R3 are not isomorphic

R2 and R3 are not isomorphic because they have different dimensions.


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11. Consider the vector 3P , the set of all cubic polynomials and the vector space 2P , the

set of all quadratic polynomials. Let { }32 ,,,13

xxxBP = be a basis for 3P and

{ }2,,12

xxBP = be the basis for 2P . Define the differential operation23: PPDx such

that .)( ppDx = Determine a matrixA such .)( pAPpDx == . Assume

32)( dxcxbxaxp +++=

23: PPDx is the differential operator defined by ][)( fdx

dfDx =

We compute the matrix of3P

B in the basis { }32 ,,,1 xxx as follows:

3223

322

32

32

03003)(

00202)(

00011)(

0000)1(

xxxxxD

xxxxxD

xxxxD

xxxD

x

x

x

x

+++==

+++==

+++==

+++=

and

[ ]

=

0000

3000

0200

0010

xD

Let 32)( dxcxbxaxp +++= and so 232)( dxcxbpD ++=

Hence, relative to the basis { },,,,1 32 xxx

[ ]dcbap ,,,][ = and [ ]0,3,2, dcbp =

pd

c

b

d

c

b

a

pDx =

=

=

0

3

2

0000

3000

0200

0010

)(


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12. Let .

100

010

301

=B Determine the basis and dimension of the eigenspace ofB.

The characteristic polynomial ofB is

3)1(

100

010

301

=

=

BI

So, the characteristic equation is 0)1( 3 =

So, the only eigenvalue is .1= To find the eigenvectors of ,1= we solve the

homogeneous linear system represented by 0)( = xBI

=

0100

000

300

BI

This implies that .03 =x Using the parameters 1xs = and ,2xt= we can find that the

eigenvectors of 1= are of the form

,

0

1

0

0

0

1

03

2

1

+

=

=

= tst

s

x

x

x

x s and tnot both zero.

Because 1= has two linearly independent eigenvectors, the dimension of its eigenspaceis 2. A basis for the eigenspace corresponding to 1= is

{ })0,1,0(),0,0,1(1 =B


10/11

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13.

Given

=

33

54A

(a) Determine the eigenvalues and associated eigenvectors ofA.

The characteristic polynomial ofA is

)1)(2(

2

)5)(2()3)(4(

)5)(2()3)(4(

32

54

2

+=

=

+=

+=

+

=

AI

So, the characteristic equation is ,0)1)(2( =+ which gives 21 = and 12 = as

the eigenvalues of A.

To find the corresponding eigenvectors, we use Gauss-Jordan elimination to solve the

homogeneous linear system represented by 0x = )( AI twice: first for ,21 == and

then for .12 ==

For ,21 = the coefficient matrix is

=

=

52

52

)3(22

)5(422 AI

which row reduces to

002

51

showing that .02

521 = xx Letting ,2 tx = we can conclude that every eigenvector of

is of the form

.0,

12

5

2

5

2

1

=

=

= tt

t

t

x

xx

For ,12 = the coefficient matrix is


11/11

11

=

=

22

55

)3(12

)5(411 AI

which row reduces to

00

11

showing that .021 =xx Letting ,2 tx = we can conclude that every eigenvector of is

of the form

.0,1

1

2

1

=

=

= tt

t

t

x

xx

(b) Show that A is diagonalizable into a similar matrixB by finding P such that

APPB1

=

We form the matrix P whose columns are the eigenvectors from part (a)

=

11

12

5

P

The matrix P is invertible, and its inverse is ,

3

5

3

23

2

3

2

1

=

P which implies that the

eigenvectors are linearly independent and A is diagonalizable. It follows that

=

==

10

02

11

12

5

32

54

3

5

3

23

2

3

2

1APPB

,

11

12

5

=P ,

3

5

3

23

2

3

2

1

=

P and

=

10

02B

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Final Exam Linear Algebra