Linear Algebra: A Modern Introduction, 3rd Edition ... · Maple In order to use Maple to do linear algebra,you must ﬁrst load the linear algebra pack-age. This is done with the

1

Although most of the examples and exercises in this book can be done without theaid of technology, there are some that are intended to be done with a matrix-capable calculator or a computer algebra system. These are clearly marked withthe symbol . If you have access to such technology, it is also useful for checkinghand calculations and for doing computations once you have mastered the basictechniques.

In this section, you will see what selected examples from each chapter look likewhen done using the computer algebra systems Maple, Mathematica, and MATLAB.By imitating these examples (and making use of the extensive built-in help facilitiesof each of the systems), you should be able to exploit the power of these systems tohelp you with the exercises throughout this book.

CASAny sufficiently advanced technologyis indistinguishable from magic.

—Arthur C. Clarke

“Technology and the Future,” inReport on Planet Three and OtherSpeculations

Harper & Row, 1972

Maple In order to use Maple to do linear algebra, you must first load the linear algebra pack-age. This is done with the command

> with (LinearAlgebra):

The package must be loaded for each new Maple session and each new worksheet.

Chapter 1: VectorsA calculation that results in a vector or matrix must be performed using the command evalm. Observe that ending an input statement with a colon sup-presses the output, while ending an input statement with a semicolon displays theoutput.

Example 1.6

> u:=vector([1,0,-1]):> v:=vector([2,-3,1]):> w:=vector([5,-4,0]):> evalm(3*u+2*v-w);

Example 1.15

> u:=vector([1,2,-3]): v:=vector([-3,5,2]):> dotprod(u,v);

1

32, �2, �1 4

Technology Bytes

Exercises E.qxd 1/1/70 5:51 AM Page 1

2 Appendix E

If v is a vector, the Maple command norm(v,2) gives its (Euclidean) length.

Example 1.17

(> norm(vector([2,3]),2);

The cross product of vectors in �3 is computed using the command crossprod.

> u:=vector([1,2,3]): v:=vector([-1,1,5]):> crossprod(u,v);

All linear algebra commands can be performed symbolically as well as numerically.

> u:=vector([u1,u2,u3]): v:=vector([v1,v2,v3]):> evalm(u+3*v);

(> dotprod(u,v);

What are the bars? Maple is assuming that the vectors may have entries that arecomplex numbers, in which case this is the version of the dot product that is needed.(See Exploration: Vectors and Matrices with Complex Entries in Chapter 7) Use thefollowing command to obtain the real dot product symbolically:

> dotprod(u,v,'orthogonal');

> norm(u,2);

> crossprod(u,v);

Maple can also perform modular arithmetic.

Example 1.12

> 3548 mod 3;

2

Example 1.14

> u:=vector([2,2,0,1,2]): v:=vector([1,2,2,2,1]):> dotprod(u,v) mod 3;

1

3u2 v3 � u3 v2, u3 v1 � u1 v3, u1 v2 � u2 v1 4

2�u1 �2 � �u2 �2 � �u3 �2

u1 v1 � u2 v2 � u3 v3

u11v1 2 � u21v2 2 � u31v3 2

3u1 � 3 v1, u2 � 3 v2, u3 � 3 v3 4

37, �8, 3 4

113


Example 1.40

> c:=vector([3,1,3,1,3,1,3,1,3,1,3,1]):> v:=vector([0,7,4,9,2,7,0,2,0,9,4,6]):> dotprod(c,v) mod 10;

0

To perform modular arithmetic calculations where the result is a vector or matrix,you need to first define a procedure called matmod.

> matmod:=proc(A,p):> map('mod',evalm(A),p):> end;

matmod : � proc(A, p)map(‘mod ’, evalm(A), p) end proc

> u:=vector([2,2,0,1,2]): v:=vector([1,2,2,2,1]):> matmod(u+v,3);

Chapter 2: Systems of Linear Equations

Example 2.6

> A:=matrix(3,3,[1,-1,-1,3,-3,2,2,-1,1]);

> b:=vector([2,16,9]);

> linsolve(A,b);

Elementary row operations can be performed individually. To construct an aug-mented matrix, use the augment command.

> Ab:=augment(A,b);

> addrow(A,1,2,-3);

C1 �1 �1 2

0 0 5 10

2 �1 1 9

S

Ab : � C1 �1 �1 2

3 �3 2 16

2 �1 1 9

S

33, �1, 2 4

b : � 32, 16, 9 4

A: � C1 �1 �1

3 �3 2

2 �1 1

S

30, 1, 2, 0, 0 4

Appendix E 3


This command adds �3 times row 1 to row 2 (putting the result in row 2).

> swaprow(%,2,3);

This command swaps rows 2 and 3 of the previous output (denoted by %).

> mulrow(%,3,1/5);

This command multiplies row 3 of the last result by .

There are two other commands that can be useful in solving a linear system di-rectly: rref and gaussjord.

Examples 2.11 and 2.13

> A:=matrix(3,4,[1,-1,-1,2,2,-2,-1,3,-1,1,-1,0]):> b:=vector([1,3,-3]):> Ab:=augment(A,b);

> rref(Ab);

> gaussjord(Ab);

Applying linsolve to this example produces the following:

> linsolve(A,b);

�_t2, _t2 �3 � _t1, _t1, �1 � _t1�

C1 �1 0 1 2

0 0 1 �1 1

0 0 0 0 0

S

C1 �1 0 1 2

0 0 1 �1 1

0 0 0 0 0

S

C 1 �1 �1 2 1

2 �2 �1 3 3

�1 1 �1 0 �3

S

15

C1 �1 �1 2

2 �1 1 9

0 0 1 2

S

C1 �1 �1 2

2 �1 1 9

0 0 5 10

S

4 Appendix E


Although this solution looks different from the solution to Example 2.11, thesubstitutions

show that they are equivalent.You can combine the procedure matmod with the above techniques to handle

systems over �p.

Example 2.16

> A:=matrix(3,3,[1,2,1,1,0,1,0,1,2]): b:=vector([0,2,1]):> Ab:=augment(A,b);

> matmod(rref(Ab),3);

> matmod(linsolve(A,b),3);

Chapter 3: MatricesAll of the basic matrix operations are built into Maple, as the following examplesillustrate.

> A:=matrix(2,2,[2,-1,3,4]); B:=matrix(2,2,[1,2,0,1]);

> evalm(A-2*B);

To evaluate the matrix A2, use the command

> evalm(A^2);

c1 �6

18 13d

c0 �5

3 2d

B : � c1 2

0 1d

A: � c2 �1

3 4d

31, 2, 1 4

C1 0 0 1

0 1 0 2

0 0 1 1

S

Ab : � C1 2 1 0

1 0 1 2

0 1 2 1

S

s � _t2 � 3 � _t1 and t � �1 � _t1

Appendix E 5


Use the command &* for matrix multiplication.

> evalm(A&*B);

> transpose(A);

> inverse(A)

> det(A);

11

Like vector operations, matrix operations can be performed symbolically inMaple.

> A:=matrix(2,2,[a,b,c,d]);

> evalm(A^2);

> det(A);

> inverse(A);

The matmod procedure, as defined for the examples in Chapter 1, is used for matrixcalculations involving modular arithmetic.

Dd

ad � bc�

b

ad � bc

� c

ad � bc

a

ad � bc

T

ad � bc

ca2 � bc ab � bd

ca � dc bc � d2 d

A: � ca b

c dd

D4

11

1

11

�3

11

2

11

T

c2 3

�1 4d

c2 3

3 10d

6 Appendix E


> A:=matrix(2,2,[2,-1,3,4]):> A3:=matmod(A,3);

> matmod(inverse(A3),3);

So the matrix A3 is its own inverse modulo 3.Another built-in function of Maple will determine the rank of a matrix:

> rank(A);

2

Maple can produce bases for the row, column, and null spaces of a matrix.

Examples 3.45, 3.47, and 3.48

> A:=matrix(4,5,[1,1,3,1,6,2,-1,0,1,-1,-3,2,1,-2,1,4,1,6,1,3]):> rowspace(A);

> colspace(A);

> nullspace(A);

The bases given above for the column and null spaces of A are not the same as in thetext. Can you show that they are equivalent?

You can define linear transformations in Maple.


> T:=(x,y)->[x,2*x-y,3*x+4*y]:> T(1,-1);

> S:=(u,v,w)->[2*u+w,3*v-w,u-v,u+v+w]:> T(x,y);

To compute S(T(x,y)), you have to pick out the components of the previousoutput. These are known to Maple as %[1], %[2], and %[3].

> S(%[1],%[2],%[3]);

35x � 4y, 3x � 7y, �x � y, 6x � 3y 4

3x, 2x � y, 3x � 4y 4

31, 3, �1 4

5 30, �5, 1, �4, 1 4 , 3 1, �3, 0,�4, 1 4 6

5 31, 0, 0, 1 4 , 3 0, 1, 0, 6 4 , 3 0, 0, 1, 3 4 6

5 30, 0, 0, 1, 4 4 , 3 0, 1, 2, 0, 3 4 , 3 1, 0, 1, 0, �1 4 6

c2 2

0 1d

A3 : � c2 2

0 1d

Appendix E 7


Maple has a built-in command for the LU factorization of a matrix.

Example 3.33

> A:=matrix(3,3,[2,1,3,4,-1,3,-2,5,5]);

> LUdecomp(A,L='L',U='U'):> L:=evalm(L); U:=evalm(U);

To check that this result is correct, compute

> evalm(L&*U);

which is A, as expected.

Chapter 4: Eigenvalues and EigenvectorsEigenvalues, eigenvectors, and the characteristic polynomial are all included in theLinearAlgebra package.

Example 4.18

> A:=matrix(3,3,[0,1,0,0,0,1,2,-5,4]):

The characteristic polynomial of A can be produced using the command

> c:=charpoly(A,lambda);

(Replacing lambda by xwould give the characteristic polynomial as a function of x.)You can now factor this polynomial to find its zeros (the eigenvalues of A):

> factor(c);

Or, you can solve the characteristic equation directly, if you like:

> solve(c=0,lambda);

2, 1, 1

1l � 2 2 1l � 1 2 2

l3 � 4l2 � 5l � 2

C 2 1 3

4 �1 3

�2 5 5

S

U : � C2 1 3

0 �3 �3

0 0 2

SL : � C 1 0 0

2 1 0

�1 �2 1

S

A: � C 2 1 3

4 �1 3

�2 5 5

S

8 Appendix E


If all you require are the eigenvalues of A, you can use

> eigenvals(A);

2, 1, 1

The command eigenvects gives the eigenvalues of a matrix, their algebraic multi-plicities, and a basis for each corresponding eigenspace.

> eigenvects(A);

For a diagonalizable matrix A, you can find a matrix P such that P�1 AP is diagonal.

Example 4.26

> A:=matrix(3,3,[-1,0,1,3,0,-3,1,0,-1]):> eig:=eigenvects(A);

Now you need to extract the three eigenvectors.

> p1:=eig[1,3,1];

This command tells Maple that the first vector you want is inside the first set of brack-ets, in the third entry (which is {[0, 1, 0], [1, 0, 1]}), and is the first vector in this set.The commands for the second and third eigenvectors are similar:

> p2:=eig[1,3,2]; p3:=eig[2,3,1];

Next, you build the matrix P with these three eigenvectors as its columns:

> P:=augment(p1,p2,p3);

To check that this matrix is correct, compute

> evalm(inverse(P)&*A&*P);

which is, as expected, a diagonal matrix displaying the eigenvalues of A.When approximating eigenvalues (or other numerical results) in Maple, use the

floating-point evaluation command, evalf.

C0 0 0

0 0 0

0 0 �2

S

P : � C1 1 �1

1 0 3

0 1 1

S

p3 : � 3�1, 3, 1 4

p2 : � 31, 0, 1 4

p1: � 30, 1, 0 4

eig : � 30, 2, 5 30, 1, 0 4 , 31, 0, 1 4 6 4 , 3�2, 1, 5 3�1, 3, 1 4 6 4

31, 2, 5 31, 1, 1 4 6 4 , 32, 1, 5 31, 2, 4 4 6 4

Appendix E 9


> A:=matrix(2,2,[1,2,3,4]):> eigenvals(A);

> evalf(eigenvals(A));

Maple can also handle matrix exponentials.

Example 4.47

> A:=matrix(2,2,[1,2,3,2]):> exponential(A);

Chapter 5: OrthogonalityMaple includes the Gram-Schmidt Process as a built-in procedure. It can be used toconvert a basis for a subspace of �n into an orthogonal or an orthonormal basis forthat subspace.

Example 5.13

> x1:=vector([1,-1,-1,1]):> x2:=vector([2,1,0,1]):> x3:=vector([2,2,1,2]):> GramSchmidt([x1,x2,x3]);

To produce an orthonormal basis, add the option normalized.

> GramSchmidt([x1,x2,x3], normalized);

The QR factorization of a matrix is performed in Maple using the command QRdecomp, which returns the upper triangular matrix R. To obtain theorthogonal matrix Q, you must instruct Maple to produce it. The command

c�1

61312, 0,

1

61312,

1

31312 d d

c c1

2,

�1

2,

�1

2,

1

2d , c

3

1015,

3

1015,

1

1015,

1

1015 d ,

c 31, �1, �1, 1 4 , c3

2,

3

2,

1

2,

1

2d , c

�1

2, 0,

1

2, 1 d d

D3

5e1�12 �

2

5e4 2

5e4 �

2

5e1�12

3

5e4 �

3

5e1�12 2

5e1�12 �

3

5e4

T

5.372281323, �.372281323

5

2�

1

2133,

5

2�

1

2133

10 Appendix E


fullspan=false ensures that the dimensions of Q agree with those of A andthat R is a square matrix.

Example 5.15

> A:=matrix(4,3,[1,2,2,-1,1,2,-1,0,1,1,1,2]):> R:=QRdecomp(A, Q='Q', fullspan=false);

> Q:=evalm(Q);

You can check that this result is correct by verifying that A � QR:

> evalm(Q&*R);

Chapter 6: Vector SpacesYou can adapt Maple’s matrix commands for use in finding bases for finite-dimensionalvector spaces or testing vectors for linear independence. The key is to use the isomor-phism between an n-dimensional vector space and �n.

Example 6.26

To test the set {1 � x, x � x2, 1 � x2} for linear independence in �2, use the standardisomorphism of �2 with �3. This gives the correspondence

1 � x 4 C1

1

0

S , x � x 2 4 C0

1

1

S , 1 � x 2 4 C1

0

1

S

D1 2 2

�1 1 2

�1 0 1

1 1 2

T

Q : � H1

2

3

1015 �

1

616

�1

2

3

1015 0

�1

2

1

1015

1

616

1

2

1

1015

1

316

X

R : � F2 1

1

2

0 153

215

0 01

216

V

Appendix E 11


Now build a matrix with these vectors as its columns:

> A:=matrix(3,3,[1,0,1,1,1,0,0,1,1]);

> rank(A);

3

Since rank(A) � 3, the columns of A are linearly independent. Hence, the set {1 � x,x � x 2, 1 � x 2} is also linearly independent, by Theorem 6.7.

Example 6.44

Extend the set {1 � x, 1 � x} to S � {1 � x, 1 � x, 1, x, x2} and use the standard iso-morphism of �2 with �3 to obtain

Form a matrix with these vectors as its columns and find its reduced echelon form:

> A:=matrix(3,5,[1,1,1,0,0,1,-1,0,1,0,0,0,0,0,1]);

> rref(A);

Since the leading 1s are in columns 1, 2, and 5, these are the vectors to use for a basis.Accordingly, {1 � x, 1 � x, x2} is a basis for �2 that contains 1 � x and 1 � x.

Many questions about a linear transformation between finite-dimensional vectorsspaces can be handled by using the matrix of the linear transformation.

Example 6.68

The matrix of with respect to the

basis B of W and the standard basis C � {1, x, x 2}

of �2 is

� e c1 0

0 0d , c

0 1

1 0d , c

0 0

0 1d f

1a � b 2 � 1b � c 2x � 1c � a 2x2T ca b

b cd �

E1 0

1

2

1

20

0 11

2

�1

20

0 0 0 0 1

U

A: � C1 1 1 0 0

1 �1 0 1 0

0 0 0 0 1

S

1 � x 4 C1

1

0

S , 1 � x 4 C 1

�1

0

S , 1 4 C1

0

0

S , x 4 C0

1

0

S , x 2 4 C0

0

1

S

A: � C1 0 1

1 1 0

0 1 1

S

12 Appendix E


(Check this.) By Exercises 40 and 41 in Section 6.6, the rank and nullity of T are thesame as those of A.

> A:=matrix(3,3,[1,-1,0,0,1,-1,-1,0,1]):> rank(A);

2

It now follows from the Rank Theorem that nullity (T ) � 3 � rank(T ) � 3 � 2 � 1.If you want to construct bases for the kernel and range of T, you can proceed as

follows:

> nullspace(A);

This is the coordinate vector of a matrix with respect to the basis B of W. Hence, thematrix is

so a basis for ker(T) is .

> colspace(A);

This time you have coordinate vectors with respect to the basis C of �2, so a basis forrange(T ) is {x � x 2, 1 � x2}.

Chapter 7: Distance and Approximation

Computations involving the inner product in �[a, b] are easy

to do in Maple.

Example 7.6

> f(x):=x:g(x):=3*x-2:> int(f(x)^2, x=0..1);

> int((f(x)-g(x))^2, x=0..1);

4

3

1

3

8 f, g 9 � �b

a

f 1x 2g 1x 2 dx

5 30, 1, �1 4 , 31, 0, �1 4 6

e c1 1

1 1d f

1 # c 1 0

0 0d � 1 # c 0 1

1 0d � 1 # c 0 0

0 1d � c

1 1

1 1d

5 31, 1, 1 4 6

A � C 1 �1 0

0 1 �1

�1 0 1

S

Appendix E 13


> int(f(x)*g(x),x=0..1);

0

The various norms discussed in Section 7.2 are built into Maple. The format of thecommand is norm(v,n), where v is a vector and n is a positive integer or the wordinfinity. Use n=1 for the sum norm,n=2 for the Euclidean norm, and n=infinityfor the max norm.

> v:=vector([1,(2,5]):> norm(v,1);

8

> norm(v,2);

> norm(v,infinity);

5

If the command norm is applied to a matrix, you get the induced operator norm.

Example 7.19

> A:=matrix(3,3,[1,-3,2,4,-1,-2,-5,1,3]);> norm(A,1);

10

> norm(A,infinity);

9

The condition number of a square matrix A can be obtained with the commandcond(A,n), where n is one of 1,2,infinity, or frobenius.

Example 7.21

> A:=matrix(2,2,[1,1,1,1.0005]):> cond(A,1); cond(A,2); cond(A,infinity); cond(A,frobenius);

8004.0005008002.0003738004.0005008002.000501

Maple can perform least squares approximation in one step.

Example 7.26

> A:=matrix(3,2,[1,1,1,2,1,3]): b:=vector([2,2,4]):> leastsqrs(A,b);

c2

3, 1 d

130

14 Appendix E


If A does not have linearly independent columns, then there is not a unique leastsquares solution of Ax � b. The additional argument optimize tells Maple to pro-duce the least squares solution of minimal (Euclidean) length.

Example 7.40

> A:=matrix(2,2,[1,1,1,1]): b:=vector([0,1]):> leastsqrs(A,b);

> leastsqrs(A,b,optimize);

The singular value decomposition is also built into Maple.

Example 7.34(b)

> A:=matrix(3,2,[1,1,1,0,0,1]):> singularvals(transpose(A));

This command produces the exact singular values of A. Their numerical approxima-tions are given by

> evalf(sv);

The same result is obtained with

> sv:=evalf(Svd(A,U,V));

(Note that evalf must be used.) This time, the orthogonal matrices U and V arecomputed (numerically) as well.

> U:=evalm(U); V:=evalm(V);

(Note that the signs of some of the entries differ from those in the solution in the text. This is not surprising, since the SVD is not unique. Also notice that u12 ��0.3513817914 � 10�16 0.) You can easily verify that you have an SVD of A by computing .©

�

V : � c�.7071067812 �.7071067812

�.7071067812 .707067812d

U : � C�.8164965809 �.351381791410�16 �.5773502692

�.4082482905 �.7071067812 .5773502692

�.4082482905 .7071067812 .5773502692

S

sv : � 31.732050808, 1.000000000 4

31.732050808, 1. 4

sv : � 313, 1 4

c1

4,

1

4d

c1

2� _t1, _t1 d

Appendix E 15


> evalm(transpose(U)&*A&*V);

Correct!The pseudoinverse is not a built-in function in Maple, but it is not hard to construct.

Example 7.39(b)

> A:=matrix(3,2,[1,1,1,0,0,1]: sv:=evalf(Svd(A,U,V)):> d:=diag(sv[1],sv[2]);

> Splus:=augment(inverse(d),[0,0]);

> Aplus:=evalm(V&*Splus&*transpose(U));

A plus: � c.3333333332 .6666666666 �.3333333334

.3333333332 �.3333333334 .6666666666d

S plus : � c.5773502690 0 0

0 1.000000000 0d

d: � c1.732050808 0

0 1.000000000d

C1.732050807 0

0 1.000000000

0 0

S

16 Appendix E

Mathematica In order to use Mathematica to do linear algebra, you must first load the appropriatelinear algebra package. For basic vector and matrix operations, this is done with thecommand

<LinearAlgebra 'MatrixManipulation'

The package must be loaded for each new Mathematica session and each new notebook.

Chapter 1: VectorsVectors are entered as lists, delimited by braces. A semicolon is used to suppress theoutput of an input statement.

Example 1.6

In[1]:=

u={1,0,-1}; v={2,-3,1}; w={5,-4,0};

In[2]:=

3*u+2*v-w

Out[2]=

{2,-2,-1}

Example 1.15

In[1]:=

u={1,2,-3}; v={-3,5,2}


In[2]:=

u.v

Out[2]=

1

To find the (Euclidean) length of a vector, use the function Norm.

Example 1.17

In[1]:=

Norm[{2,3}]

Out[1]=

The Cross product of vectors in �3 is computed using the command Cross.

In[1]:=

u={1,2,3}; v={-1,1,5};

In[2]:=Cross[u,v];Out[2]=

[7,-8,3]

All linear algebra commands can be performed symbolically as well as numerically.

In[1]:=

u={u1,u2,u3}; v={v1,v2,v3};

In[2]:=u+3*v;Out[2]=

{u1+3 v1, u2+3 v2, u3+3 v3}

In[1]:=u.vOut[1]=

u1 v1+u2 v2+u3 v3

In[1]:=

Norm[u]

Out[1]=

In[1]:=

Cross[u,v];

Out[1]=

{-u3 v2+u2 v3, u3 v1-u1 v3, -u2 v1+u1 v2}

Mathematica can also perform modular arithmetic.

Example 1.12

In[1]:=

Mod[3548, 3]

Out[1]=

2

1u12 � u22 � u33

113

Appendix E 17


Example 1.14

In[1]:=

u={2,2,0,1,2}; v={1,2,2,2,1};

In[2]:=

Mod[u.v, 3]

Out[2]=

1

Example 1.40

In[1]:=

c={3,1,3,1,3,1,3,1,3,1,3,1}; v={0,7,4,9,2,7,0,2,0,9,4,6};

In[2]:=

Mod[c.v, 10]

Out[2]=

0

The same syntax is used to perform modular arithmetic calculations in which theresult is a vector or matrix.

In[1]:=

u={2,2,0,1,2}; v={1,2,2,2,1};

In[2]:=

Mod[u+v, 3]

Out[2]=

{0,1,2,0,0}

Chapter 2: Systems of Linear Equations

Example 2.6

In[1]:=

Solve[{x-y-z==2,3*x-3*y+2*z==16,2*x-y+z==9},{x,y,z}]

Out[1]=

{{x�3, y�-1, z�2}}

Alternatively, you can have Mathematica perform the same calculations using matri-ces and vectors:

In[1]:=

A={{1,-1,-1},{3,-3,2},{2,-1,1}}; b={{2},{16},{9}};

In[2]:=LinearSolve[A,b]

{{3},{-1},{2}}

To perform elementary row operations individually, begin by entering the aug-mented matrix:

In[1]:=

A={{1,-1,-1,2},{3,-3,2,16},{2,-1,1,9}};

In[2]:=

A[[2]]=A[[2]]-3*A[[1]];

18 Appendix E


In[3]:=

MatrixForm[A]

Out[3]=

This command adds �3 times row 1 to row 2 (putting the result in row 2). The com-mand MatrixForm displays the result as a matrix, rather than as a “list of lists.”

In[1]:=

ReplacePart[A,A,3,2]; ReplacePart[%,A,2,3]//MatrixForm

Out[1]=

This command swaps rows 2 and 3 of the previous output (in two steps).

In[1]:=

A[[3]]=1/5*%[[3]]; MatrixForm[A]

This command multiplies row 3 of the last result by and stores the answer back in A.

Another command that can be useful in solving a linear system directly isRowReduce.


In[1]:=

Ab={{1,-1,-1,2,1},{2,-2,-1,3,3},{-1,1,-1,0,-3}};

In[2]:=

RowReduce[Ab]//MatrixForm

Out[2]=

To see the solutions explicitly, use Solve rather than LinearSolve:

In[1]:=

Solve[{w-x-y+2z==1,2*w-2*x-y+3*z==3,-w+x-y==-3},{w,x,y,z}]

Out[1]=

{{w�2+x-z,y�1+z}}

You can solve systems over �p by specifying the modulus.

C1 -1 0 1 2

0 0 1 -1 1

0 0 0 0 0

S

15

C1 -1 -1 2

2 -1 1 9

0 0 1 2

S

C1 -1 -1 2

2 -1 1 9

0 0 5 10

S

C1 -1 -1 2

0 0 5 10

2 -1 1 9

S

Appendix E 19


Example 2.16

In[1]:=

A={{1,2,1},{1,0,1},{0,1,2}}; b={{0},{2},{1}};

In[2]:=

LinearSolve[A,b,Modulus->3]

Out[2]=

{{1},{2},{1}}

Chapter 3: MatricesMathematica handles matrices as “lists of lists.” A matrix is input as a list whoseelements are themselves lists containing the entries of the rows of the matrix.To display the result of a matrix calculation as a matrix, use the command MatrixForm.

In[1]:=

A={{2,-1},{3,4}}; B={{1,2},{0,1}};

In[2]:=A-2*B//MatrixFormOut[2]=


In[3]:=

MatrixPower[A,2]//MatrixForm

(Do not use A^2; it squares all the entries of A individually.) Use a period for matrix multiplication. (The command A*B will give an “elementwise” product instead.)

In[4]:=

A.B//MatrixForm

Out[4]=

In[5]:=

Transpose[A]//MatrixForm

Out[5]=

In[6]:=

Inverse[A]//MatrixForm

c2 3

-1 4d

c2 3

3 10d

c1 -6

18 13d

c0 -5

3 2d

20 Appendix E


Out[6]=

In[7]:=

Det[A]

Out[7]=

11

Like vector operations, matrix operations can be performed symbolically inMathematica.

In[1]:=

A={{a,b},{c,d}};

In[2]:=

MatrixPower[A,2]//MatrixForm

Out[2]=

In[3]:=

Det[A]

Out[3]=

-bc+ad

In[4]:=

Inverse[A]//MatrixForm

Out[4]=

The Mod and Modulus commands are used for matrix calculations involvingmodular arithmetic.

In[1]:=

A={{2,-1},{3,4}};

In[2]:=

A3=Mod[A,3]

Out[2]=

{{2,2},{0,1}}

Do not use the MatrixForm command here, since you are going to performanother calculation on the result.

Dd

ad-bc-

b

ad-bc

-c

ad-bc

a

ad-bc

T

ca2 � bc ab � bd

ca � dc bc � d2d

D4

11

1

11

-3

11

2

11

T

Appendix E 21


In[3]:=

Inverse[A3, Modulus->3]

Out[3]=

{{2,2},{0,1}}

So the matrix A3 is its own inverse modulo 3.Mathematica can produce a basis for the null space of a matrix.

Examples 3.45, 3.47, and 3.48

In[1]:=

A={{1,1,3,1,6},{2,-1,0,1,-1},{-3,2,1,-2,1},{4,1,6,1,3}};

In[2]:=

NullSpace[A]

Out[2]=

{{1,-3,0,-4,1},{-1,-2,1,0,0}}

To find bases for the row and column spaces of A, use the RowReduce command.

In[3]:=

R=RowReduce[A]

Out[3]=

{{1,0,1,0,-1},{0,1,2,0,3},{0,0,0,1,4},{0,0,0,0,0}}

From this output you see that the nonzero rows form a basis for the row space of A,and the columns of A corresponding to the positions of the leading 1s (columns 1, 2,and 4) give a basis for the column space of A. You can extract these as follows:

In[4]:=

TakeRows[R,3]

Out[4]=

{{1,0,1,0,-1},{0,1,2,0,3},{0,0,0,1,4}}

In[5]:=

{TakeColumns[A,{1}], TakeColumns[A,{2}], TakeColumns[A,{4}]}

Out[5]=

{{{1},{2},{-3},{4}},{{1},{-1},{2},{1}},{{1},{1},{-2},{1}}}

You can define linear transformations in Mathematica.


In[1]:=

T[x_,y_]:={x,2*x-y,3*x+4*y}

In[2]:=

T[1,-1]

Out[2]=

{1,3,-1}

In[3]:=

S[u_,v_,w_]:={2*u+w,3*v-w,u-v,u+v+w}

In[4]:=

T[x,y]

22 Appendix E


Out[4]=

{x,2 x-y,3 x+4 y}

To compute S(T(x,y)), you have to pick out the components of the previous out-put. These are known to Mathematica as %[[1]], %[[2]], and %[[3]].

In[5]:=

S[%[[1]],%[[2]],%[[3]]]

Out[5]=

{5 x+4 y,-3 x+3 (2 x-y)-4 y,-x+y,6 x+3 y}

In[6]:=

Simplify[%]

Out[6]=

{5 x+4 y,3 x-7 y,-x+y,3 (2 x+y)}

Mathematica has a built-in command for the LU factorization of a matrix.

Example 3.33

In[1]:=

A={{2,1,3},{4,-1,3},{-2,5,5}};

In[2]:=

{LU,P,cond}=LUDecomposition[A]

Out[2]=

{{{2,1,3},{2,-3,-3},{-1,-2,2}},{1,2,3},1}

The term {{2,1,3},{2,-3,-3},{-1,-2,2}} contains the matrices Land U combined into a single matrix that leaves out the diagonal 1s of L. Mathe-matica’s help facility gives the procedures Upper and Lower, which will extract L and U :

In[3]:=

Upper[LU_?MatrixQ]:=LUTable[If[i ≤ j,1,0],{i,Length[LU]},{j,Length[LU]}]

In[4]:=

Lower[LU_?MatrixQ]:=LU-Upper[LU]+IdentityMatrix[Length[LU]]

In[5]:=

MatrixForm[L=Lower[LU]]

Out[5]=

In[6]:=

MatrixForm[U=Upper[LU]]

Out[6]=

C2 1 3

0 -3 -3

0 0 2

S

C 1 0 0

2 1 0

-1 -2 1

S

Appendix E 23



In[7]:=

MatrixForm[L.U]

Out[7]=


Chapter 4: Eigenvalues and EigenvectorsEigenvalues and eigenvectors are included in Mathematica.

Example 4.18

In[1]:=

A={{0,1,0},{0,0,1},{2,-5,4}};


In[2]:=

Eigenvalues[A]

Out[2]=

{1,1,2}

The command Eigenvectors gives the corresponding eigenvectors.

In[3]:=

Eigenvectors[A]

Out[3]=

{{1,1,1},{0,0,0},{1,2,4}}

Entering Eigensystem combines the two previous calculations.

In[4]:=

Eigensystem[A]

Out[4]=

{{1,1,2},{{1,1,1},{0,0,0},{1,2,4}}}

For a diagonalizable matrix A, you can find a matrix P such that P �1 AP is diagonal.

Example 4.26

In[1]:=

A={{-1,0,1},{3,0,-3},{1,0,-1}};

In[2]:=

{evalues,evectors}=Eigensystem[A]

Out[2]=

{{-2,0,0},{{-1,3,1},{1,0,1},{0,1,0}}}

Now you need to extract the three eigenvectors and put them into the columns of P.

In[3]:=

MatrixForm[P = Transpose[evectors]]

C 2 1 3

4 -1 3

-2 5 5

S

24 Appendix E


Out[3]=

The corresponding diagonal matrix is created with the command

In[4]:=

MatrixForm[Diag=DiagonalMatrix[evalues]]

Out[4]=

To check that the matrix is correct, compute

In[5]:=

P.Diag.Inverse[P]//MatrixForm

Out[5]=

As expected, this is A!When approximating eigenvalues (or other numerical results) in Mathematica,

use the floating-point evaluation command, N.

In[1]:=

A=N[{{1,2},{3,4}}]

Out[1]=

{{1.,2.},{3.,4.}}

In[2]:=

Eigenvalues[A]

Out[2]=

{5.37228,-0.372281}

Mathematica can also handle matrix exponentials.

Example 4.47

In[1]:=

A={{1,2},{3,2}};

In[2]:=

MatrixExp[A]//MatrixForm

D3 � 2E5

5E

21-1 � E5 2

5E31-1 � E5 2

5E

2 � 3E5

5E

T

C-1 0 1

3 0 -3

1 0 -1

S

C-2 0 0

0 0 0

0 0 0

S

C-1 1 0

3 0 1

1 1 0

S

Appendix E 25


Chapter 5: OrthogonalityMathematica includes the Gram-Schmidt Process as a built-in procedure. To use it,you must first load the orthogonalization package.

In[1]:=

<LinearAlgebra'Orthogonalization'

Example 5.13

In[2]:=

x1={1,-1,-1,1}; x2={2,1,0,1}; x3={2,2,1,2};

In[3]:=

GramSchmidt[{x1,x2,x3}]

Out[3]:=

The default is an orthonormal basis; to obtain an orthogonal basis only, use theoption Normalized->False.

In[4]:=

GramSchmidt[{x1,x2,x3}, Normalized->False]

Out[4]=

The QR factorization of a matrix is performed in Mathematica using the com-mand QRDecomposition, which returns an orthogonal matrix Q and an uppertriangular matrix R such that QTR � A.

Example 5.15

In[1]:=

A={{1,2,2},{-1,1,2},{-1,0,1},{1,1,2}};

In[2]:=

{Q,R}=QRDecomposition[A];

In[3]:=

Map[MatrixForm,{Q,R}]

You can check that this result is correct by verifying that QTR � A:

G2 1

1

2

0 15 3152

0 0 A3

2

0 0 0

W wg G1

2-1

2-1

2

1

2

3

215 3

215 1

215 1

215-116 0

116 A2

3

0 0 0 0

W,

e 51,-1,-1,16, e3

2,3

2,1

2,1

2f, e-

1

2,0,

1

2,1 ff

e1

2, -

1

2,-

1

2,1

2f, e

3

215, 3

215, 1

215, 1

215 f, e- 116,0, 116,A2

3f

26 Appendix E


In[4]:=

Transpose[Q].R//MatrixForm

Out[4]=

Chapter 6: Vector SpacesYou can adapt Mathematical matrix commands for use in finding bases for finite-dimensional vector spaces or testing vectors for linear independence. The key is to usethe isomorphism between an n-dimensional vector space and �n.

Example 6.26

To test the set {1 � x, x � x2, 1 � x2} for linear independence in �2, use the standardisomorphism of �2 with �3. This gives the correspondence


In[1]:=

A={{1,0,1},{1,1,0},{0,1,1}};

In[2]:=

RowReduce[A]

Out[2]=

{{1,0,0},{0,1,0},{0,0,1}}

Since the reduced row echelon form of A is I, you know that the columns of A are lin-early independent. Hence, the set {1 � x, x � x 2, 1 � x 2} is also linearly independent,by Theorem 6.7.

Example 6.44

Extend the set {1 � x, 1 � x} to S � {1 � x, 1 � x, 1, x, x 2} and use the standardisomorphism of �2 with �3 to obtain


In[1]:=

MatrixForm[A={{1,1,1,0,0},{1,-1,0,1,0},{0,0,0,0,1}}]

1 � x 4 C1

1

0

S , 1 � x 4 C 1

�1

0

S , 1 4 C1

0

0

S , x 4 C0

1

0

S , x2 4 C0

0

1

S

1 � x 4 C1

1

0

S , x � x2 4 C0

1

1

S , 1 � x2 4 C1

0

1

S

D1 2 2

-1 1 2

-1 0 1

1 1 2

T

Appendix E 27


Out[1]=

In[2]:=

RowReduce[A]//MatrixForm

Out[2]=

Since the leading 1s are in columns 1, 2, and 5, these are the vectors to use for a basis.Accordingly, {1 � x, 1 � x, x 2} is a basis for �2 that contains 1 � x and 1 � x.

Many questions about a linear transformation between finite-dimensional vectorspaces can be handled by using the matrix of the linear transformation.

Example 6.68

The matrix of with respect to the basis

B of W and the standard basis C� {1, x, x 2} of �2 is


In[1]:=

A={{1,-1,0},{0,1,-1},{-1,0,1}};

In[2]:=

RowReduce[A]

Out[2]:=

{{1,0,-1},{0,1,-1},{0,0,0}}

You can see that the rank is 2. It now follows from the Rank Theorem that nullity(T) � 3 � rank(T) � 3 � 2 � 1.

If you want to construct bases for the kernel and range of T, you can proceed asfollows:

In[3]:=

NullSpace[A]

Out[3]=

{{1,1,1}}

A � C 1 �1 0

0 1 �1

�1 0 1

S� e c

1 0

0 0d , c

0 1

1 0d , c

0 0

0 1d f

T ca b

b cd � 1a � b 2 � 1b � c 2x � 1c � a 2x

2

C1 0 12

12 0

0 1 12

-12 0

0 0 0 0 1

S

C1 1 1 0 0

1 -1 0 1 0

0 0 0 0 1

S

28 Appendix E



so a basis for ker(T) is . The first two columns of A form a basis for the

column space of A, since the leading 1s are in those columns. These correspond to abasis for the range of T. This time you have coordinate vectors with respect to thebasis C of �2, so a basis for range(T) is {1 � x 2, x � x 2}.


Computations involving the inner product in �[a, b] are easy

to do in Mathematica.

Example 7.6

In[1]:=

f=x; g=3*x-2;

In[2]:=

Integrate[f^2,{x,0,1}]

Out[2]=

In[3]:=

Integrate[(f-g)^2,{x,0,1}]

Out[3]=

In[4]:=

Integrate[f*g,{x,0,1}]

Out[4]=

0

For vector norms, Mathematica has the following:

In[1]:=

u={1,-2,5};

In[2]:=

Norm[u]

Out[2]=

In[3]:=

SumNorm[x_]:=Apply[Plus,Map[Abs,x]]

130

4

3

1

3

8 f, g 9 � �b

a

f 1x 2g 1x 2 dx

e c1 1

1 1d f

1 # c1 0

0 0d � 1 # c0 1

1 0d � 1 # c0 0

0 1d � c

1 1

1 1d

Appendix E 29


In[4]:=

SumNorm[u]

Out[4]=

8

In[5]:=

MaxNorm[x_]:=Apply[Max,Map[Abs,x]]

In[6]:=

MaxNorm[u]

Out[6]=

5

You can perform least squares approximation by having Mathematica solve thenormal equations.

Example 7.26

In[1]:=

A={{1,1},{1,2},{1,3}}; b={{2},{2},{4}};

In[2]:=

LinearSolve[Transpose[A].A,Transpose[A].b]

Out[2]=

If A does not have linearly independent columns, then there is not a unique leastsquares solution of Ax � b. In this case, the version of least squares that uses thepseudoinverse should be used.

Example 7.40

In[1]:=

A={{1,1},{1,1}}; b={{0},{1}};

In[2]:=

PseudoInverse[A].b

Out[2]=

The singular value decomposition is also built into Mathematica. It can be used onlyon numerical matrices. One way to ensure that a matrix is treated numerically is tomake sure that some of its entries include a decimal point.

Example 7.34(b)

In[1]:=

A={{1.,1.},{1.,0},{0,1.}};

In[2]:=

{U,S,V}=SingularValueDecomposition[A]

Out[2]=

{{{0.816497,0.408248,0.408248},{2.33321 x 10-16,0.707107,-0.707107}}, {1.73205,1.},{{0.707107,0.707107}, {0.707107,-0.707107}}}

ee1

4f, e

1

4ff

e e2

3f,1 f

30 Appendix E


This command gives a full SVD of A: The first entry in the list of lists is the matrix U,which is followed by a list containing the singular values, and finally V. To see these asmatrices, use the command

In[3]:=

Map[MatrixForm,{U,DiagonalMatrix[S],V}]

Out[3]=

Notice that Mathematica always returns the singular values in a square diagonalmatrix and that the matrices U and V are not necessarily square. This is differentfrom the way the text presents these values. However, the factorization has the sameproperty:

In[4]:=

MatrixForm[Transpose[U].DiagonalMatrix[S].V]

Out[4]=

which, as expected, is A. [The (2, 2) and (3, 1) entries are effectively zeros.]As we have already noted, the pseudoinverse is a built-in function in Mathematica.

Example 7.39(b)

In[1]:=

A={{1,1},{1,0},{0,1}};

In[2]:=

PseudoInverse[A]

Out[2]=

ee1

3,2

3,-

1

3f, e

1

3,-

1

3,2

3ff

D 1 # 1 #1 # -3.17341x10�16

7.60364x10�16 1 #T

c1.73205 0

0 1.d, c

0.707107 0.707107

0.707107 -0.707107d f

e c0.816497 0.408248 0.408248

2.33321x10-16 0.707107 -0.707107d,

Appendix E 31

MATLAB MATLAB stands for “Matrix Laboratory.” As its name suggests, it is designed to han-dle matrix (and vector) calculations, which it does with ease.

Chapter 1: VectorsMATLAB distinguishes between row and column vectors. The entries of a row vectorare separated by spaces or commas, while the entries of a column vector are separatedby semicolons:


»v=[1 2 3], w=[1;2;3]v=

1 2 3w=

123

Example 1.6

»u=[1 0 -1]; v=[2 -3 1]; w=[5 -4 0];»3*u+2*v-wans =

2 -2 -1

A semicolon after an input statement tells MATLAB to suppress the output. Thus, inthe above example, the vectors u, v, and w were not displayed.

Example 1.15

»u=[1 2 -3]; v=[-3 5 2];»dot(u,v)ans =

1

Example 1.17

»norm([2 3])ans =

3.6056

If MATLAB’s symbolic math toolbox is installed on your system, you can also ob-tain exact answers.

»sym(norm([2 3]))ans =

sqrt(13)

The cross product of vectors in �3 is computed using the command cross.

»u=[1 2 3]; v=[-1 1 5]; »cross(u, v) ans =

7 -8 3

MATLAB can also perform modular arithmetic.

Example 1.12

»mod(3548, 3) ans =

2

32 Appendix E


Example 1.14

»u=[2 2 0 1 2]; v=[1 2 2 2 1]; »mod(dot(u, v), 3) ans =

1

Example 1.40

»c=[3 1 3 1 3 1 3 1 3 1 3 1]; v=[0 7 4 9 2 7 0 2 0 9 4 6];»mod(dot(c, v), 10) ans =

0

The same syntax is used to perform modular arithmetic calculations in which theresult is a vector or matrix.

»u=[2 2 0 1 2]; v=[1 2 2 2 1];»mod(u+v, 3)ans =

0 1 2 0 0

Chapter 2: Systems of Linear EquationsMatrices are entered in MATLAB the way they are written by hand: row by row andcolumn by column. The entries in each row are separated by spaces or commas, anda semicolon is used to indicate the end of a row.

Example 2.6

»A=[1 -1 -1; 3 -3 2; 2 -1 1] A =

1 -1 -1

3 -3 2

2 -1 1

» b=[2; 16; 9]

b =

2

16

9

»linsolve(A, b) ans=

[ 3]

[-1]

[ 2]

Appendix E 33


The following syntax does the same thing as that above:

»A\bans =

3.0000

-1.0000

2.0000

You can also obtain the solution of a system from the reduced row echelon formof its augmented matrix. (This is necessary if the system has infinitely many or no so-lutions, since MATLAB cannot handle these cases.)

»Ab=[1 -1 -1 2; 3 -3 2 16; 2 -1 1 9];»rref(Ab) ans =

1 0 0 3

0 1 0 -1

0 0 1 2

If you like, you can ask MATLAB to walk you through the row reduction process.(Keep in mind, though, that MATLAB uses partial pivoting strategies.) This is donewith the command

rrefmovie(Ab)

Use MATLAB’s mod command to handle systems over �p.

Example 2.16

»A=[1 2 1; 1 0 1; 0 1 2]; b=[0; 2; 1];»mod(A\b, 3) ans =

1

2

1

This approach sometimes needs a little help, however. Suppose you need to solve thesystem Ax � b, where

over �5. You obtain

»A=[1 1 2;3 0 1;2 2 3]; b=[3;3;2];»sol=mod(A\b,5)sol =

14/3

1/3

4

This is not what you want, since the solution is not a vector with entries in �5. In thiscase, you need to use the following theorem, due to Euler:

A � C1 1 2

3 0 1

2 2 3

S and b � C3

3

2

S

34 Appendix E


Appendix E 35

Theorem E.1 If p is prime, then for all k 0 in �p,

kp�1 � 1

�

In this example, p � 5, so 3 5�1 � 34 � 1 in �5. Therefore, multiplying the solutionby 34 � 1 will not change it but will get rid of the fractions.

»mod(sol*(3^4),5)ans =

3

2

4

Chapter 3: MatricesAll of the basic matrix operations are built into MATLAB, as the following examplesillustrate.

»A=[2 -1;3 4]; B=[1 2;0 1];»A-2*Bans =

0 -5

3 2


»A^2 ans =

1 -6

18 13

The same calculation is done by mpower(A,2).Use the command * for matrix multiplication.

»A*B ans =

2 3

3 10

»transpose(A)ans =

2 3

-1 4

A more compact way to take the transpose of a matrix A is to type A’.

»inv(A) ans =

4/11 1/11

-3/11 2/11


Entering A^((1) does the same thing.

»det(A)ans =

11

Like vector operations,matrix operations can be performed symbolically in MATLAB.

»A=sym('[a b; c d]')A =

[a, b]

[c, d]

»A^2ans =

[a^2+b*c, a*b+b*d]

[c*a+d*c, b*c+d^d]

»det(A)ans =

a*d-b*c

»inv(A)ans =

[d/(a*d-b*c), -b/(a*d-b*c)]

[-c/(a*d-b*c), a/(a*d-b*c)]

For matrix calculations involving modular arithmetic, proceed as with systems oflinear equations.

»A=[2 -1; 3 4];»A3=mod(A, 3) A3 =

2 2

0 1

»mod(inv(A3), 3)ans =

1/2 2

0 1

This is not a matrix over �3, but Euler’s Theorem tells us that 23�1 � 22 � 1. There-fore, multiplying the last answer by 22 mod 3 will give the inverse of A3 over �3:

»mod(ans*(2^2),3) ans =

2 2

0 1

So A3 is its own inverse modulo 3.Another built-in function of MATLAB will determine the rank of a matrix.

»rank(A)ans =

2

36 Appendix E


MATLAB can produce bases for the row, column, and null spaces of a matrix.

Examples 3.45, 3.47, and 3.48

»A=[1 1 3 1 6; 2 -1 0 1 -1; -3 2 1 -2 1; 4 1 6 1 3];»rref(A) ans =

1 0 1 0 -1

0 1 2 0 3

0 0 0 1 4

0 0 0 0 0

From this matrix you can obtain a basis for row(A) by taking the nonzero rows.

»colspace(sym(A))ans =[1, 0, 0][0, 1, 0][0, 0, 1][1, 6, 3]

The basis given above for the column space of A is not the same as in the text. Can youshow that the two are equivalent?

»null(sym(A))ans =[-1, 1][-2, -3][-1, 0][ 0, -4][ 0, 1]

The same result arises from

»null (A, 'r') ans =

-1 1

-2 -3

1 0

0 -4

0 1

The 'r' tells MATLAB to give the result in rational arithmetic. Without it, you get anumerical answer. [In fact, MATLAB gives an orthonormal basis for null(A).]

»null(A)ans =

-0.5413 0.1443

-0.5867 -0.6327

0.4421 0.0400

0.3968 -0.7371

-0.0992 0.1843

Appendix E 37


MATLAB has a built-in command for the LU factorization of a matrix.

Example 3.33

»A=[2 1 3; 4 -1 3; -2 5 5];»[L, U]=1u(A) L =

0.5000 0.3333 1.0000

1.0000 0 0

-0.5000 1.0000 0

U =

4.0000 -1.0000 3.0000

0 4.5000 6.5000

0 0 -0.6667


»L*U ans =

2.0000 1.0000 3.0000

4.0000 -1.0000 3.0000

-2.0000 5.0000 5.0000


Chapter 4: Eigenvalues and EigenvectorsMATLAB can handle eigenvalues, eigenvectors, and the characteristic polynomial.

Example 4.18

»A=[0 1 0; 0 0 1; 2 -5 4];

The characteristic polynomial of A can be produced using the command

»c=poly(A) c =

1.0000 -4.0000 5.0000 -2.0000

This command gives the coefficients of the characteristic polynomial in order ofdecreasing powers of x. To see the actual polynomial, convert this result into standardform:

»poly2sym(c)ans =x^3-4*x^2+5*x-2

To solve the characteristic equation (in order to obtain the eigenvalues of A), ask forthe roots of c.

38 Appendix E


»roots(c)ans =

2.0000

1.0000 + 0.0000i

1.0000 - 0.0000i


»eig(A) ans =

1.0000 + 0.0000i

1.0000 - 0.0000i

2.0000

To see both the eigenvalues and the bases for the corresponding eigenspaces, use thelonger form:

»[P,D]=eig(A,'nobalance')P=

0.5774 0.5774 -0.2182

0.5774 0.5774 -0.4364

0.5774 0.5774 -0.8729

D=

1.0000 0 0

0 1.0000 0

0 0 2.0000

Here, the columns of P are eigenvectors of A corresponding to the eigenvalues of A,which appear on the diagonal of D.

For a diagonalizable matrix A, use the same syntax to find a matrix P such thatP�1 AP is diagonal.

Example 4.26

»A=[-1 0 1; 3 0 -3; 1 0 -1];»[P,D]=eig(A,'nobalance') P =

0.3015 0.3015 -0.0134

0.9045 -0.9045 -0.9998

0.3015 -0.3015 -0.0134

D =

-0.0000 0 0

0 -2.0000 0

0 0 0.0000

There are three distinct eigenvectors, so A is diagonalizable. Thus, PDP �1 shouldequal A.

Appendix E 39


»p*D*inv(P) ans =

-1.0000 0.0000 1.0000

3.0000 0.0000 -3.0000

1.0000 -0.0000 -1.0000

Bingo!MATLAB can also handle matrix exponentials using the command expm.

Example 4.47

»A=[1 2; 3 2];»expm(A)ans =

22.0600 21.6921

32.5382 32.9060

If the symbolic math toolbox is installed, you can get this result exactly.

»expm(sym(A))ans =

[ 3/5*exp(-1)+2/5*exp(4), 2/5*exp(4)-2/5*exp(-1)]

[ 3/5*exp(4)-3/5*exp(-1), 2/5*exp(-1)+3/5*exp(4)]

Chapter 5: OrthogonalityMATLAB includes the Gram-Schmidt Process as a built-in procedure. It can be usedto convert a basis for a subspace of �n into an orthonormal basis for that sub-space.You accomplish this by putting the given basis vectors into the columns of a matrixand then using the command orth.

Example 5.13

»x1=[1; -1; -1; 1]; x2=[2; 1; 0; 1]; x3=[2; 2; 1; 2];»A=[x1 x2 x3]A =

1 2 2

-1 1 2

-1 0 1

1 1 2

Q =

0.6701 -0.3781 -0.5241

0.4806 0.6444 -0.2321

0.1609 0.6018 0.2805

0.5423 -0.2823 0.7700

You can check that these vectors are orthonormal by verifying that QTQ � I.

40 Appendix E


»Q'*Qans =

1.0000 0.0000 -0.0000

0.0000 1.0000 0.0000

-0.0000 0.0000 1.0000

MATLAB can also perform the QR factorization of a matrix. It returns the uppertriangular matrix R.

Example 5.15

»A=[1 2 2; -1 1 2; -1 0 1; 1 1 2];»qr(A)ans =

-2.0000 -1.0000 -0.5000

-0.5000 -2.2361 -3.3541

-0.0000 0.4472 -1.2247

0.5000 0 0.9526

To obtain the orthonormal matrix Q, you must use a longer form. The command[Q,R] = qr(A,0) ensures that the dimensions of Q agree with those of A and thatR is a square matrix.

»[Q,R]=qr(A,0)Q =

-0.5000 -0.6708 0.4082

0.5000 -0.6708 -0.0000

0.5000 -0.2236 -0.4082

-0.5000 -0.2236 -0.8165

R =

-2.0000 -1.0000 -0.5000

0 -2.2361 -3.3541

0 6 -1.2247

You can check that this result is correct by verifying that A � QR:

»Q*Rans =

1.0000 2.0000 2.0000

-1.0000 1.0000 2.0000

-1.0000 0.0000 1.0000

1.0000 1.0000 2.0000

(The command [Q,R] = qr(A) produces the modified QR factorization discussedin the text.)

Chapter 6: Vector SpacesYou can adapt MATLAB’s matrix commands for use in finding bases for finite-di-mensional vector spaces or testing vectors for linear independence. The key is to usethe isomorphism between an n-dimensional vector space and �n.

Appendix E 41


Example 6.26

To test the set {1 � x, x � x 2, 1 � x 2} for linear independence in �2 use the standardisomorphism of �2 with �3. This gives the correspondence


»A=[1 0 1;1 1 0;0 1 1] A =

1 0 1

1 1 0

0 1 1

»rank(A) ans =

3

Since rank(A) � 3, the columns of A are linearly independent. Hence, the set {1 � x,x � x 2, 1 � x 2} is also linearly independent, by Theorem 6.7.

Example 6.44

Extend the set {1 � x, 1 � x} to S � {1 � x, 1 � x, 1, x, x2} and use the standard iso-morphism of �2 with �3 to obtain


»A=[1 1 1 0 0;1 -1 0 1 0;0 0 0 0 1]A =

1 1 1 0 0

1 -1 0 1 0

0 0 0 0 1

»rref (A)ans =

1.0000 0 0.5000 0.5000 0

0 1.0000 0.5000 -0.5000 0

0 0 0 0 1.0000

Since the leading 1s are in columns 1, 2, and 5, these are the vectors to use for a basis.Accordingly, {1 � x, 1 � x, x 2} is a basis for �2 that contains 1 � x and 1 � x.

Many questions about a linear transformation between finite-dimensional vectorspaces can be handled by using the matrix of the linear transformation.

1 � x 4 C1

1

0

S , 1 � x 4 C 1

�1

0

S , 1 4 C1

0

0

S , x 4 C0

1

0

S , x2 4 C0

0

1

S

1 � x 4 C1

1

0

S , x � x2 4 C0

1

1

S , 1 � x2 4 C1

0

1

S

42 Appendix E


Example 6.68

The matrix of with respect to the basis

B of W and the standard basis C� {1, x, x 2} of �2 is


»A=[1 -1 0;0 1 -1;-1 0 1];»rank(A)ans =

2

It now follows from the Rank Theorem that nullity(T) � 3 � rank(T) � 3 � 2 � 1.If you want to construct bases for the kernel and range of T, you can proceed as

follows:

»null(A,'r') ans =

1

1

1


so a basis for ker(T ) is .

»A=[1 -1 0; 0 1 -1; -1 0 1];»rref(A)ans =

1 0 -1

0 1 -1

0 0 0

Since the leading 1s are in columns 1 and 2, these columns in the original matrix forma basis for col(A). That is,

col1A 2 � span £ C 1

0

�1

S , C�1

1

0

S ≥

e c1 1

1 1d f

1 # c 1 0

0 0d � 1 # c 0 1

1 0d � 1 # c 0 0

0 1d � c

1 1

1 1d

A � C 1 �1 0

0 1 �1

�1 0 1

S� e c

1 0

0 0d , c

0 1

1 0d , c

0 0

0 1d f

T ca b

b cd � 1a � b 2 � 1b � c 2x � 1c � a 2x

2

Appendix E 43


This time you have coordinate vectors with respect to the basis C of �2, so a basis forrange(T) is {x � x 2, �1 � x 2}.


Computations involving the inner product in �[a, b] can be

done using MATLAB’s symbolic math toolbox. Symbolic expressions are defined byplacing the expression in single quotes and using the sym command.

Example 7.6

»f=sym('x'); g=sym('3*x-2');

The command int(F,a,b) computes .

»int(f^2,0,1)ans =

1/3

»int((f-g)^2,0,1)ans =

4/3

»int(f*g,0,1)ans=

0

The various norms discussed in Section 7.2 are built into MATLAB. The format of thecommand is norm(v,n), where v is a vector and n is a positive integer or the ex-pression inf. Use n=1 for the sum norm, n=2 for the Euclidean norm, and n=inffor the max norm. Typing norm(v) is the same as typing norm(v,2).

»v=[1 -2 5];»norm(v,1) ans =

8

»norm(v) ans =

5.4772

»norm(v,inf) ans =

5

If the command norm is applied to a matrix, you get the induced operator norm.

Example 7.19

»A=[1 -3 2; 4 -1 -2; -5 1 3]»norm(A,1)ans =

10

»norm(A,inf)ans =

9

�b

a

F 1x 2dx

8 f, g 9 � �b

a

f 1x 2g 1x 2 dx

44 Appendix E


The condition number of a square matrix A can be obtained with the commandcond(A,n), where n is equal to 1,2, or inf.

Example 7.21

»A=[1 1; 1 1.0005];»c1=cond(A,1), c2=cond(A), cinf=cond(A,inf)c1 =

8.0040e+03

c2 =

8.0020e+03

cinf =

8.0040e+03

MATLAB can perform least squares approximation in one step. If the system Ax � b has a nonsquare coefficient matrix A, then the command A\b returns theleast squares solution of the system.

Example 7.26

»A=[1 1; 1 2; 1 3]; b=[2; 2; 4];»A\bans =

0.6667

1.0000

The singular value decomposition is also built into Maple.

Example 7.34(b)

»A=[1 1; 1 0; 0 1];»svd(A)ans =

1.7321

1.0000

This command produces a vector containing numerical approximations to the singu-lar values of A. If you want to see the matrices U, , and V, then you must enter

»[U,S,V] = svd(A)U =

0.8165 0.0000 -0.5774

0.4082 -0.7071 0.5774

0.4082 0.7071 0.5774

S =

1.7321 0

0 1.0000

0 0

V =

0.7071 -0.7071

0.7071 0.7071

©

Appendix E 45


You can easily verify that you have an SVD of A by computing U VT.

»U*S*V'ans =

1.0000 1.0000

1.0000 -0.0000

0.0000 1.0000

As expected, the result is A.The pseudoinverse is a built-in function in MATLAB.

Example 7.39(b)

»A=[1 1; 1 0; 0 1];»pinv(A)ans =

0.3333 0.6667 -0.3333

0.3333 -0.3333 0.6667

Using the SVD, you can find minimum length least squares solutions of linear systems.

Example 7.40

»A=[1 1; 1 1]; b=[0; 1];»pinv(A)*bans =

0.2500

0.2500

MATLAB can apply the SVD to achieve digital image compression. The followingprocedure will convert a grayscale image called file.ext (where file is the name ofthe file and ext is the file format—JPEG, GIF, TIFF, etc.) into a matrix and then usethe SVD to produce rank k approximations to the original for any value of k up to therank of the matrix. The following example uses k � 12.

»A = imread('file.ext'); »A = double(A); »figure(1); »colormap(gray); »imagesc(A);

»[U,S,V] = svd(A); »k = 12;»Ak = U(:, 1:k)*S(1:k, 1:k)*V(:, 1:k)';»figure(2);»colormap(gray);»imagesc(Ak);

©

46 Appendix E


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