3501 handouts

MRQ 2014

School of Mathematics and Statistics

MT3501 Linear Mathematics

Handout 0: Course Information

Lecturer: Martyn Quick, Room 326.

Prerequisite: MT2001 Mathematics

Lectures: Mon (even), Tue & Thu 12 noon, Purdie Theatre B.

Tutorials: These begin in Week 2:

Times: Mon 2pm (Maths Theatre B), Mon 3pm (Maths Theatre B),Thu 2pm (Maths Theatre B), Fri 3pm (Maths Theatre B).

Assessment: 10% Continuous assessment, 90% examContinuous assessment questions will be distributed later in the semester. (Full

details t.b.a.)

Recommended Texts:

T. S. Blyth & E. F. Robertson, Basic Linear Algebra, Second Edition, SpringerUndergraduate Mathematics Series (Springer-Verlag 2002)

T. S. Blyth & E. F. Robertson, Further Linear Algebra, Springer UndergraduateMathematics Series (Springer-Verlag 2002)

R. Kaye & R. Wilson, Linear Algebra, Oxford Science Publications (OUP 1998)

Webpage: All my handouts, slides, problem sheets and solutions will be availablein PDF format from MMS and in due course on my webpage. Solutions will be postedonly in MMS and only after all relevant tutorials have happened. The versions on mywebpage will be as single PDFs so will be posted later in the semester (the currentversions are from last year, though there wont be a huge number of changes).

The lecture notes and handouts will contain a number (currently a small number!)of extra examples. I dont have time to cover these in the lectures themselves, butthey are intended to be useful supplementary material.

1

Course Content:

Vector spaces: subspaces, spanning sets, linear independence, bases

Linear transformations: rank, nullity, matrix of a linear transformation,change of basis

Direct sums: projection maps

Diagonalisation: eigenvectors & eigenvalues, characteristic polynomial, mini-mum polynomial

Jordan normal form

Inner product spaces: orthogonality, orthonormal bases, GramSchmidt pro-cess

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MRQ 2014



Handout I: Vector spaces

1 Vector spaces

Definition 1.1 A field is a set F together with two binary operations

F F F F F F

(, ) 7 + (, ) 7

called addition and multiplication, respectively, such that

(i) + = + for all , F ;

(ii) (+ ) + = + ( + ) for all , , F ;

(iii) there exists an element 0 in F such that + 0 = for all F ;

(iv) for each F , there exists an element in F such that + () = 0;

(v) = for all , F ;

(vi) () = () for all , , F ;

(vii) ( + ) = + for all , , F ;

(viii) there exists an element 1 in F such that 1 6= 0 and 1 = for all F ;

(ix) for each F with 6= 0, there exists an element 1 (or 1/) in F such that1 = 1.

Example 1.2 The following are examples of fields: (i) Q; (ii) R; (iii) C, all with theusual addition and multiplication;

(iv) Z/pZ = {0, 1, . . . , p 1} (where p is a prime number) with addition andmultiplication being performed modulo p.

1

Definition 1.3 Let F be a field. A vector space over F is a set V together with thefollowing operations

V V V F V V

(u, v) 7 u+ v (, v) 7 v,

called addition and scalar multiplication, respectively, such that

(i) u+ v = v + u for all u, v V ;

(ii) (u+ v) + w = u+ (v + w) for all u, v, w V ;

(iii) there exists a vector 0 in V such that v + 0 = v for all v V ;

(iv) for each v V , there exists a vector v in V such that v + (v) = 0;

(v) (u+ v) = u+ v for all u, v V and F ;

(vi) (+ )v = v + v for all v V and , F ;

(vii) ()v = (v) for all v V and , F ;

(viii) 1v = v for all v V .

We shall use the term real vector space to refer to a vector space over the field Rand complex vector space to refer to one over the field C.

2

Example 1.4 (i) Let n be a positive integer and let

Fn =

x1x2...xn

x1, x2, . . . , xn F

.

This is a vector space over F with addition given byx1x2...xn

+

y1y2...yn

=

x1 + y1x2 + y2

...xn + yn

,

scalar multiplication by

x1x2...xn

=

x1x2...

xn

,

zero vector

0 =

00...0

,

and negatives given by

x1x2...xn

=

x1x2...

xn

.

(ii) C forms a vector space over R.

(iii) A polynomial over a field F is an expression of the form

f(x) = a0 + a1x+ a2x2 + + amx

m

for somem > 0, where a0, a1, . . . , am F . The set F [x] of all polynomials over Fforms a vector space over F . Addition is given by

i

aixi +i

bixi =i

(ai + bi)xi

and scalar multiplication by

i

aixi =i

(ai)xi.

3

(iv) Let FR denote the set of all functions f : R R. Define

(f + g)(x) = f(x) + g(x) (for x R)

and(f)(x) = f(x) (for x R).

Then FR forms a vector space over R with respect to this addition and scalarmultiplication. In this vector space negatives are given by

(f)(x) = f(x)

and the zero vector is 0 : x 7 0 for all x R.

Basic properties of vector spaces

Proposition 1.5 Let V be a vector space over a field F . Let v V and F . Then

(i) 0 = 0;

(ii) 0v = 0;

(iii) if v = 0, then either = 0 or v = 0;

(iv) ()v = v = (v).

Subspaces

Definition 1.6 Let V be a vector space over a field F . A subspace W of V is anon-empty subset such that

(i) if u, v W , then u+ v W , and

(ii) if v W and F , then v W .

Lemma 1.7 Let V be a vector space and let W be a subspace of V . Then

(i) 0 W ;

(ii) if v W , then v W .

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Definition 1.9 Let V be a vector space and let U and W be subspaces of V .

(i) The intersection of U and W is

U W = { v | v U and v W }.

(ii) The sum of U and W is

U +W = {u+ w | u U, w W }.

Proposition 1.10 Let V be a vector space and let U and W be subspaces of V . Then

(i) U W is a subspace of V ;

(ii) U +W is a subspace of V .

Corollary 1.11 Let V be a vector space and let U1, U2, . . . , Uk be subspaces of V .Then

U1 + U2 + + Uk = {u1 + u2 + + uk | ui Ui for each i }

is a subspace of V .

5

Spanning sets

Definition 1.12 Let V be a vector space over a field F and suppose that A ={v1, v2, . . . , vk} is a set of vectors in V . A linear combination of these vectors is avector of the form

1v1 + 2v2 + + kvk

for some 1, 2, . . . , k F . The set of all such linear combinations is called the spanof the vectors v1, v2, . . . , vk and is denoted by Span(v1, v2, . . . , vk) or by Span(A ).

Proposition 1.13 Let A be a set of vectors in the vector space V . Then Span(A ) isa subspace of V .

Definition 1.14 A spanning set for a subspace W is a set A of vectors such thatSpan(A ) = W .

Linear independent elements and bases

Definition 1.16 Let V be a vector space over a field F . A set A = {v1, v2, . . . , vk}is called linearly independent if the only solution to the equation

ki=1

ivi = 0

(with i F ) is 1 = 2 = = k = 0.If A is not linearly independent, we shall call it linearly dependent.

6

Example 1A Determine whether the set {x+x2, 12x2, 3+6x} is linearly independentin the vector space P of all real polynomials.

Solution: We solve

(x+ x2) + (1 2x2) + (3 + 6x) = 0;

that is,( + 3) + (+ 6)x + ( 2)x2 = 0. (1)

Equating coefficients yields the system of equations

+ 3 = 0

+ 6 = 0

2 = 0;

that is, 0 1 31 0 61 2 0

=

000

.

A sequence of row operations (Check!) converts this to1 2 00 1 30 0 0

=

000

.

Hence the original equation (1) is equivalent to

2 = 0

+ 3 = 0.

Since there are fewer equations remaining than the number of variables, we have enoughfreedom to produce a non-zero solution. For example, if we set = 1, then = 3 =3 and = 2 = 6. Hence the set {x+ x2, 1 2x2, 3 + 6x} is linearly dependent.

Lemma 1.17 Let A be a set of vectors in the vector space V . Then A is linearlyindependent if and only if no vector in A can be expressed as a linear combination ofthe others.

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Lemma 1.18 Let A be a linearly dependent set of vectors belonging to a vectorspace V . Then there exists some vector v in A such that A \ {v} spans the samesubspace as A .

If we start with a finite set and repeatedly remove vectors which are linear combi-nations of the others in the set, then we must eventually stop and produce a linearlyindependent set. This establishes:

Theorem 1.19 Let V be a vector space (over some field). If A is a finite subset of Vand W = Span(A ), then there exists a linearly independent subset B with B Aand Span(B) = W .

Definition 1.20 Let V be a vector space over the field F . A basis for V is a linearlyindependent spanning set. We say that V is finite-dimensional if it possesses a finitespanning set; that is, if V possesses a finite basis. The dimension of V is the size ofany basis for V and is denote by dimV .

Lemma 1.22 Let V be a vector space of dimension n (over some field) and supposeB = {v1, v2, . . . , vn} is a basis for V . Then every vector in V can be expressed as alinear combination of the vectors in B in a unique way.

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Theorem 1.23 Let V be a finite-dimensional vector space. Suppose {v1, v2, . . . , vm}is a linearly independent set of vectors and {w1, w2, . . . , wn} is a spanning set for V .Then

m 6 n.

Corollary 1.24 Let V be a finite-dimensional vector space. Then any two bases for Vhave the same size and consequently dimV is uniquely determined.

Proposition 1.26 Let V be a finite-dimensional vector space. Then every linearlyindependent set of vectors in V can be extended to a basis for V by adjoining a finitenumber of vectors.

Corollary 1.27 Let V be a vector space of finite dimension n. If A is a linearlyindependent set containing n vectors, then A is a basis for V .

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MRQ 2014



Handout II: Linear transformations

2 Linear transformations

Definition 2.1 Let V andW be vector spaces over the same field F . A linear mapping(also called a linear transformation) from V to W is a function T : V W such that

(i) T (u+ v) = T (u) + T (v) for all u, v V , and

(ii) T (v) = T (v) for all v V and F .

Comment: Sometimes we shall write Tv for the image of the vector v under thelinear transformation T (instead of T (v)).

Lemma 2.2 Let T : V W be a linear mapping between two vector spaces over thefield F . Then

(i) T (0) = 0;

(ii) T (v) = T (v) for all v V ;

(iii) if v1, v2, . . . , vk V and 1, 2, . . . , k F , then

T

( ki=1

ivi

)=

ki=1

i T (vi).

1

Definition 2.3 Let T : V W be a linear transformation between vector spaces overa field F .

(i) The image of T isT (V ) = imT = {T (v) | v V }.

(ii) The kernel or null space of T is

ker T = { v V | T (v) = 0W }.

Proposition 2.4 Let T : V W be a linear transformation between vector spacesV and W over the field F . The image and kernel of T are subspaces of W and V ,respectively.

Definition 2.5 Let T : V W be a linear transformation between vector spaces overthe field F .

(i) The rank of T , which we shall denote by rankT , is the dimension of the imageof T .

(ii) The nullity of T , which we shall denote by nullT , is the dimension of the kernelof T .

Theorem 2.6 (Rank-Nullity Theorem) Let V and W be vector spaces over thefield F with V finite-dimensional and let T : V W be a linear transformation. Then

rankT + nullT = dimV.

Comment: [For those who have done MT2002.] This can be viewed as an ana-logue of the First Isomorphism Theorem for groups within the world of vector spaces.Rearranging gives

dimV dimker T = dim imT

and since (as we shall see on Problem Sheet II) dimension essentially determines vectorspaces we conclude

V/ker T = imT.

2

Constructing linear transformations

Proposition 2.7 Let V be a finite-dimensional vector space over the field F withbasis {v1, v2, . . . , vn} and let W be any vector space over F . If y1, y2, . . . , yn arearbitrary vectors in W , there is a unique linear transformation T : V W such that

T (vi) = yi for i = 1, 2, . . . , n.

This transformation T is given by

T

( ni=1

ivi

)=

ni=1

iyi

for an arbitrary linear combinationn

i=1 ivi in V .

Proposition 2.9 Let V be a finite-dimensional vector space over the field F withbasis {v1, v2, . . . , vn} and let W be a vector space over F . Fix vectors y1, y2, . . . , ynin W and let T : V W be the unique linear transformation given by T (vi) = yifor i = 1, 2, . . . , n. Then

(i) imT = Span(y1, y2, . . . , yn).

(ii) kerT = {0} if and only if {y1, y2, . . . , yn} is a linearly independent set.

3

Example 2A Define a linear transformation T : R3 R3 in terms of the standardbasis B = {e1, e2, e3} by

T (e1) = y1 =

211

, T (e2) = y2 =

10

2

, T (e3) = y3 =

01

4

.

Show that ker T = {0} and imT = R3.

Solution: We check whether {y1,y2,y3} is linearly independent. Solve

y1 + y2 + y3 = 0;

that is,

2 = 0

= 0

+ 2 + 4 = 0.

The second equation tells us that = while the first says = 2. Substituting for and in the third equation gives

+ 4 + 4 = 7 = 0.

Hence = 0 and consequently = = 0.This shows {y1,y2,y3} is linearly independent. Consequently, ker T = {0} by

Proposition 2.9. The Rank-Nullity Theorem now says

dim imT = dimR3 dimker T = 3 0 = 3.

Therefore im T = R3 as it has the same dimension.[Alternatively, since dimR3 = 3 and {y1,y2,y3} is linearly independent, this set

must be a basis for R3 (see Corollary 1.24). Therefore, by Proposition 2.9(i),

imT = Span(y1,y2,y3) = R3,

once again.]

4

The matrix of a linear transformation

Definition 2.11 Let V andW be finite-dimensional vector spaces over the field F andlet B = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be bases for V andW , respectively.If T : V W is a linear transformation, let

T (vj) =

mi=1

ijwi

express the image of the vector vj under T as a linear combination of the basis C(for j = 1, 2, . . . , n). The m n matrix [ij ] is called the matrix of T with respect tothe bases B and C . We shall denote this by Mat(T ) or, when we wish to be explicitabout the dependence upon the bases B and C , by MatB,C (T ).

Note that the entries of the jth column of the matrix of T are:

1j2j...

mj

i.e., the jth column specifies the image of T (vj) by listing the coefficients when it isexpressed as a linear combination of the vectors in C .

Informal description of what the matrix of linear transformation does: Sup-pose that V and W are finite-dimensional vector spaces of dimension n and m, respec-tively. Firstly V andW look like Fn and Fm. Moreover, from this viewpoint, a lineartransformation T : V W then looks like multiplication by the matrix Mat(T ).

5

Change of basis

The matrix of a linear transformation depends heavily on the choice of bases for thetwo vector spaces involved. The following theorem describes how these matrices fora linear transformation T : V V are linked when calculated with respect to twodifferent bases for V . Similar results apply when using different bases for vectorsspaces V and W for a linear transformation T : V W .

Theorem 2.13 Let V be a vector space of dimension n over a field F and let T : V V be a linear transformation. Let B = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wn} bebases for V and let A and B be the matrices of T with respect to B and C , respectively.Then there is an invertible matrix P such that

B = P1AP.

Specifically, the (i, j)th entry of P is the coefficient of vi when wj is expressed as alinear combination of the basis vectors in B.

6

Example 2B Let

B =

011

, 101

, 21

0

.

(i) Show that B is a basis for R3.

(ii) Write down the change of basis matrix from the standard basis E = {e1, e2, e3}to B.

(iii) Let

A =

2 2 31 1 21 2 2

and view A as a linear transformation R3 R3. Find the matrix of A withrespect to the basis B.

Solution: (i) We first establish that B is linearly independent. Solve

011

+

101

+

21

0

=

000

;

that is,

+ 2 = 0

= 0

= 0.

Thus = and the first equation yields 2 + = 0. Adding the third equation nowgives = 0 and hence = = 0. This show B is linearly independent and it istherefore a basis for R3 since dimR3 = 3 = |B|.

(ii) We write each vector in B in terms of the standard basis 011

= e2 e3

101

= e1 e3

21

0

= 2e1 e2

and write the coefficients appearing down the columns of the change of basis matrix:

P =

0 1 21 0 11 1 0

.

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(iii) Theorem 2.13 says MatB,B(A) = P1AP (as the matrix of A with respect to

the standard basis is A itself). We first calculate the inverse of P via the usual rowoperation method:

0 1 21 0 11 1 0

1 0 00 1 00 0 1

0 1 21 0 10 1 1

1 0 00 1 00 1 1

r3 7 r3 + r1

1 0 10 1 20 1 1

0 1 01 0 00 1 1

r1 r2

1 0 10 1 20 0 1

0 1 01 0 01 1 1

r3 7 r3 + r2

1 0 00 1 00 0 1

1 2 11 2 21 1 1

r1 7 r1 + r3

r2 7 r2 2r3

Hence

P1 =

1 2 11 2 2

1 1 1

and so

MatB,B(A) = P1AP

=

1 2 11 2 2

1 1 1

2 2 31 1 21 2 2

0 1 21 0 11 1 0

=

1 2 12 4 32 3 3

0 1 21 0 11 1 0

=

1 0 01 1 0

0 1 1

.

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MRQ 2014



Handout III: Direct sums

3 Direct sums

Definition 3.1 Let V be a vector space over a field F . We say that V is the directsum of two subspaces U1 and U2, written V = U1 U2 if every vector v in V can beexpressed uniquely in the form v = u1 + u2 where u1 U1 and u2 U2.

Proposition 3.2 Let V be a vector space and U1 and U2 be subspaces of V . ThenV = U1 U2 if and only if the following conditions hold:

(i) V = U1 + U2,

(ii) U1 U2 = {0}.

Comment: Many authors use the two conditions to define what is meant by a directsum and then show it is equivalent to our unique expression definition.

Proposition 3.4 Let V = U1 U2 be a finite-dimensional vector space expressed asa direct sum of two subspaces. If B1 and B2 are bases for U1 and U2, respectively,then B1 B2 is a basis for V .

Corollary 3.5 If V = U1 U2 is a finite-dimensional vector space expressed as adirect sum of two subspaces, then

dimV = dimU1 + dimU2.

1

Projection maps

Definition 3.6 Let V = U1 U2 be a vector space expressed as a direct sum of twosubspaces. The two projection maps P1 : V V and P2 : V V onto U1 and U2,respectively, corresponding to this decomposition are defined as follows:

if v V , express v uniquely as v = u1 + u2 where u1 U1 and u2 U2,then

P1(v) = u1 and P2(v) = u2.

Lemma 3.7 Let V = U1 U2 be a direct sum of subspaces with projection mapsP1 : V V and P2 : V V . Then

(i) P1 and P2 are linear transformations;

(ii) P1(u) = u for all u U1 and P1(w) = 0 for all w U2;

(iii) P2(u) = 0 for all u U1 and P2(w) = w for all w U2;

(iv) kerP1 = U2 and imP1 = U1;

(v) kerP2 = U1 and imP2 = U2.

Proposition 3.8 Let P : V V be a projection corresponding to some direct sumdecomposition of the vector space V . Then

(i) P 2 = P ;

(ii) V = kerP imP ;

(iii) I P is also a projection;

(iv) V = kerP ker(I P ).

Here I : V V denotes the identity transformation I : v 7 v for v V .

2

Example 3A Let V = R3 and U = Span(v1), where

v1 =

31

2

.

(i) Find a subspace W such that V = U W .

(ii) Let P : V V be the associated projection onto W . Calculate P (u) where

u =

444

.

Solution: (i) We first extend {v1} to a basis for R3. We claim that

B =

31

2

,100

,010

is a basis for R3. We solve

31

2

+

100

+

010

=

000

;

that is,3+ = + = 2 = 0.

Hence = 0, so = 3 = 0 and = = 0. Thus B is linearly independent. SincedimV = 3 and |B| = 3, we conclude that B is a basis for R3.

Let W = Span(v2,v3) where

v2 =

100

and v3 =

010

.

Since B = {v1,v2,v3} is a basis for V , if v V , then there exist 1, 2, 3 R suchthat

v = (1v1) + (2v2 + 3v3) U +W.

Hence V = U +W .If v U W , then there exist , 1, 2 R such that

v = v1 = 1v2 + 2v3.

Thereforev1 + (1)v2 + (2)v3 = 0.

Since B is linearly independent, we conclude = 1 = 2 = 0, so v = v1 = 0.Thus U W = {0} and so

V = U W.

3

(ii) We write u as a linear combination of the basis B. Inspection shows

u =

444

= 2

31

2

2

100

+ 6

010

=

62

4

+

26

0

,

where the first term in the last line belongs to U and the second to W . Hence

P (u) =

26

0

(since this is the W -component of u).

Direct sums of more summands

Definition 3.10 Let V be a vector space. We say that V is the direct sum of subspacesU1, U2, . . . , Uk, written V = U1 U2 Uk, if every vector in V can be uniquelyexpressed in the form u1 + u2 + + uk where ui Ui for each i.

The following analogues of earlier results then hold:

Proposition 3.11 Let V be a vector space with subspaces U1, U2, . . . , Uk. ThenV = U1 U2 Uk if and only if the following conditions hold:

(i) V = U1 + U2 + + Uk;

(ii) Ui (U1 + + Ui1 + Ui+1 + + Uk) = {0} for each i.

Proposition 3.12 Let V = U1 U2 Uk be a direct sum of subspaces. If Bi isa basis for Ui for i = 1, 2, . . . , k, then B1 B2 Bk is a basis for V .

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MRQ 2014



Handout IV: Diagonalisation of linear transformations

4 Diagonalisation of linear transformations

Throughout fix a vector space V over a field F and a linear transformation T : V V .

Eigenvectors and eigenvalues

Definition 4.1 Let V be a vector space over a field F and let T : V V be a lineartransformation. A non-zero vector v is an eigenvector for T with eigenvalue (where F ) if

T (v) = v.

Definition 4.2 Let V be a vector space over a field F , let T : V V be a lineartransformation, and F . The eigenspace corresponding to the eigenvalue is thesubspace

E = ker(T I) = { v V | T (v) v = 0 }

= { v V | T (v) = v }.

Recall I denotes the identity transformation given by v 7 v.

From now on we strengthen out standing assumption to V being a finite-dimensionalvector space over F .

Definition 4.3 Let T : V V be a linear transformation of the finite-dimensionalvector space V (over F ) and let A be the matrix of T with respect to some basis. Thecharacteristic polynomial of T is

cT (x) = det(xI A)

where x is an indeterminate variable.

Lemma 4.4 Suppose that T : V V is a linear transformation of the finite-dimensionalvector space V over F . Then is an eigenvalue of T if and only if is a root of thecharacteristic polynomial of T .

1

Lemma 4.5 Let V be a finite-dimensional vector space V over F and T : V Vbe a linear transformation. The characteristic polynomial cT (x) is independent of thechoice of basis for V .

Diagonalisability

Definition 4.6 (i) Let T : V V be a linear transformation of a finite-dimensionalvector space V . We say that T is diagonalisable if there is a basis with respectto which T is represented by a diagonal matrix.

(ii) A square matrix A is diagonalisable if there is an invertible matrix P such thatP1AP is diagonal.

Proposition 4.7 Let V be a finite-dimensional vector space and T : V V be alinear transformation. Then T is diagonalisable if and only if there is a basis for Vconsisting of eigenvectors for T .

2

Algebraic and geometric multiplicities

Lemma 4.10 If the linear transformation T : V V is diagonalisable, then the char-acteristic polynomial of T is a product of linear factors.

Definition 4.12 Let V be a finite-dimensional vector space over the field F and letT : V V be a linear transformation of V . Let F .

(i) The algebraic multiplicity of (as an eigenvalue of T ) is the largest power k suchthat (x )k is a factor of the characteristic polynomial cT (x).

(ii) The geometric multiplicity of is the dimension of the eigenspace E correspond-ing to .

Proposition 4.13 Let T : V V be a linear transformation of a vector space V . Aset of eigenvectors of T corresponding to distinct eigenvalues is linearly independent.

Theorem 4.14 Let V be a finite-dimensional vector space over the field F and letT : V V be a linear transformation of V .

(i) If the characteristic polynomial cT (x) is a product of linear factors (as alwayshappens, for example, if F = C), then the sum of the algebraic multiplicitiesequals dimV .

(ii) Let F and let r be the algebraic multiplicity and n be the geometricmultiplicity of . Then

n 6 r.

(iii) The linear transformation T is diagonalisable if and only if cT (x) is a product oflinear factors and n = r for all eigenvalues .

3

Example 4A Let

A =

1 2 14 5 24 3 0

.

Show that A is not diagonalisable.

Solution: The characteristic polynomial of A is

cA(x) = det(xI A)

= det

x+ 1 2 14 x 5 2

4 3 x

= (x+ 1) det

(x 5 23 x

)+ 2det

(4 24 x

)+ det

(4 x 54 3

)= (x+ 1)

(x(x 5) + 6

)+ 2(4x 8) + (12 4x+ 20)

= (x+ 1)(x2 5x+ 6) + 8(x 2) 4x+ 8

= (x+ 1)(x 2)(x 3) + 8(x 2) 4(x 2)

= (x 2)((x+ 1)(x 3) + 8 4

)= (x 2)(x2 2x 3 + 4)

= (x 2)(x2 2x+ 1)

= (x 2)(x 1)2.

In particular, the algebraic multiplicity of the eigenvalue 1 is 2.We now determine the eigenspace for eigenvalue 1. We solve (A I)v = 0; that is,

2 2 14 4 24 3 1

xyz

=

000

. (2)

We solve this by applying row operations:2 2 14 4 24 3 1

000

2 2 10 0 0

0 1 1

000

r2 7 r2 2r1

r3 7 r3 2r1

2 0 10 0 0

0 1 1

000

r1 7 r1 + 2r3

So Equation (2) is equivalent to

2x+ z = 0 = y + z.

Hence z = 2x and y = z = 2x. Therefore the eigenspace is

E1 =

x2x2x

x R

= Span

122

and we conclude dimE1 = 1. Thus the geometric multiplicity of 1 is not equal to thealgebraic multiplicity, so A is not diagonalisable.

4

Minimum polynomial

Note: The minimum polynomial is often also called the minimal polynomial.

Definition 4.15 Let T : V V be a linear transformation of a finite-dimensionalvector space over the field F . Theminimum polynomial mT (x) is themonic polynomialover F of smallest degree such that

mT (T ) = 0.

To say that a polynomial is monic is to say that its leading coefficient is 1. Ourdefinition of the characteristic polynomial ensures that cT (x) is also always a monicpolynomial.

The definition of the minimum polynomial (in particular, the assumption that it ismonic) ensures that it is unique.

Theorem 4.16 (CayleyHamilton Theorem) Let T : V V be a linear transfor-mation of a finite-dimensional vector space V . If cT (x) is the characteristic polynomialof T , then

cT (T ) = 0.

5

Facts about polynomials: Let F be a field and recall F [x] denotes the set ofpolynomials with coefficients from F :

f(x) = anxn + an1x

n1 + + a1x+ a0 (where ai F ).

Then F [x] is an example of what is known as a Euclidean domain (see MT4517 Ringsand Fields for full details). A summary of its main properties are:

We can add, multiply and subtract polynomials;

Euclidean Algorithm: if we attempt to divide f(x) by g(x) (where g(x) 6= 0), weobtain

f(x) = g(x)q(x) + r(x)

where either r(x) = 0 or the degree of r(x) satisfies deg r(x) < deg g(x) (i.e., wecan perform long-division with polynomials).

When the remainder is 0, that is, when f(x) = g(x)q(x) for some polynomial q(x),we say that g(x) divides f(x).

If f(x) and g(x) are non-zero polynomials, their greatest common divisor is thepolynomial d(x) of largest degree dividing them both. It is uniquely determinedup to multiplying by a scalar and can be expressed as

d(x) = a(x)f(x) + b(x)g(x)

for some polynomials a(x), b(x).

Those familiar with divisibility in the integers Z (particularly those who have at-tended MT1003 ) will recognise these facts as being standard properties of Z (which isalso a standard example of a Euclidean domain).

6

Proposition 4.18 Let V be a finite-dimensional vector space over a field F and letT : V V be a linear transformation. If f(x) is any polynomial (over F ) such thatf(T ) = 0, then the minimum polynomial mT (x) divides f(x).

Corollary 4.19 Suppose that T : V V is a linear transformation of a finite-dimensionalvector space V . Then the minimum polynomial mT (x) divides the characteristic poly-nomial cT (x).

Theorem 4.20 Let V be a finite-dimensional vector space over a field F and letT : V V be a linear transformation of V . Then the roots of the minimum polyno-mial mT (x) and the roots of the characteristic polynomial cT (x) coincide.

Theorem 4.21 Let V be a finite-dimensional vector space over the field F and letT : V V be a linear transformation. Then T is diagonalisable if and only if theminimum polynomial mT (x) is a product of distinct linear factors.

Lemma 4.22 Let T : V V be a linear transformation of a vector space over thefield F and let f(x) and g(x) be coprime polynomials over F . Then

ker f(T )g(T ) = ker f(T ) ker g(T ).

7

Example 4B Let

A =

0 2 11 5 31 2 0

.

Calculate the characteristic polynomial and the minimum polynomial of A. Hencedetermine whether A is diagonalisable.

Solution:

cA = det(xI A)

= det

x 2 11 x 5 3

1 2 x

= xdet

(x 5 32 x

) 2 det

(1 31 x

)+ det

(1 x 51 2

)= x(x(x 5) + 6

) 2(x+ 3) + (2 x+ 5)

= x(x2 5x+ 6) + 2(x 3) x+ 3

= x(x 3)(x 2) + 2(x 3) (x 3)

= (x 3)(x(x 2) + 2 1

)= (x 3)(x2 2x+ 1)

= (x 3)(x 1)2.

Since the minimum polynomial divides cA(x) and has the same roots, we deduce

mA(x) = (x 3)(x 1) or mA(x) = (x 3)(x 1)2.

We calculate

(A 3I)(A I) =

3 2 11 2 31 2 3

1 2 11 4 31 2 1

=

2 0 22 0 2

2 0 2

6= 0.

Hence mA(x) 6= (x 3)(x 1). We conclude

mA(x) = (x 3)(x 1)2.

This is not a product of distinct linear factors, so A is not diagonalisable.

8

MRQ 2014



Handout V: Jordan normal form

5 Jordan normal form

Definition 5.1 A Jordan block is an n n matrix of the form

Jn() =

1

1 0. . .

. . .

0 . . . 1

for some positive integer n and some scalar .A linear transformation T : V V (of a vector space V ) has Jordan normal form A

if there exists a basis for V with respect to which the matrix of T is

Mat(T ) = A =

Jn1(1) 0 00 Jn2(2) 0 0... 0

. . .. . .

......

.... . . 0

0 0 0 Jnk(k)

for some positive integers n1, n2, . . . , nk and scalars 1, 2, . . . , k. (The occurrencesof 0 here indicate zero matrices of appropriate sizes.)

Theorem 5.2 Let V be a finite-dimensional vector space and T : V V be a lineartransformation of V such that the characteristic polynomial cT (x) is a product of linearfactors with eigenvalues 1, 2, . . . , n, then there exist a basis for V with respect towhich Mat(T ) is in Jordan normal form where each Jordan block has the form Jm(i)for some m and some i.

Corollary 5.3 Let A be a square matrix over C. Then there exist an invertiblematrix P (over C) such that P1AP is in Jordan normal form.

1

Basic properties of Jordan normal form

Proposition 5.4 Let J = Jn() be an n n Jordan block. Then

(i) cJ(x) = (x )n;

(ii) mJ(x) = (x )n;

(iii) the eigenspace E of J has dimension 1.

Finding the Jordan normal form

Problem: Let V be a finite-dimensional vector space and let T : V V be a lineartransformation. If the characteristic polynomial cT (x) is a product of linear factors,find a basis B with respect to which T is in Jordan normal form and determine whatthis Jordan normal form is.

Observation 5.5 The algebraic multiplicity of as an eigenvalue of T equals the sumof the sizes of the Jordan blocks Jn() (associated to ) occurring in the Jordan normalform for T .

Observation 5.6 If is an eigenvalue of T , then the power of (x ) occurring inthe minimum polynomial mT (x) is (x )

m where m is the largest size of a Jordanblock associated to occurring in the Jordan normal form for T .

Observation 5.9 The geometric multiplicity of as an eigenvalue of T equals thenumber of Jordan blocks Jn() occurring in the Jordan normal form for T .

These observations are enough to determine the Jordan normal form for manysmall examples. We need to generalise Observation 5.9 in larger cases and considermore general spaces than the eigenspace E. Determining

dimker(T I)2, dimker(T I)3, . . .

would prove to be sufficient.

To find the basis B such that MatB,B(T ) is in Jordan normal form, boils down tofinding the Jordan normal form and then solving appropriate systems of linear equa-tions. (It is essentially a generalisation of finding the eigenvectors for a diagonalisablelinear transformation.)

Example 5A Omitted from the handout due to length. A document containing thissupplementary example can be found in MMS and on my webpage.

2

MRQ 2014



Handout VI: Inner product spaces

6 Inner product spaces

Throughout this section, our base field F will be either R or C. Recall that if z =x+ iy C, the complex conjugate of z is given by

z = x iy.

To save space and time, we shall use the complex conjugate even when F = R. Thus,when F = R and appears, we mean = for a scalar R.

Definition 6.1 Let F = R or C. An inner product space is a vector space V over Ftogether with an inner product

V V F

(v,w) 7 v,w

such that

(i) u+ v,w = u,w + v,w for all u, v, w V ,

(ii) v,w = v,w for all v,w V and F ,

(iii) v,w = w, v for all v,w V ,

(iv) v, v is a real number satisfying v, v > 0 for all v V ,

(v) v, v = 0 if and only if v = 0.

In the case when F = R, our inner product is symmetric in the sense that Condi-tion (iii) then becomes

v,w = w, v for all v,w V .

1

Example 6.2 (i) The vector space Rn is an inner product space with respect tothe usual dot product :

x1x2...xn

,y1y2...yn

=

x1x2...xn

y1y2...yn

=

ni=1

xiyi.

(ii) We can endow Cn with an inner product by

z1z2...zn

,w1w2...wn

=

ni=1

ziwi.

(iii) The set C[a, b] of continous functions f : [a, b] R is an inner product spacewhen we define

f, g =

baf(x)g(x) dx.

(iv) The space of real polynomials of degree at most n inherits the above inner productfrom the space of continous functions.

The space of complex polynomials

f(x) = nxn + n1x

n1 + + 1x+ 0 (with i C)

becomes an inner product space when we define

f, g =

1

0

f(x)g(x) dx

wheref(x) = nx

n + n1xn1 + + 1x+ 0.

2

Definition 6.3 Let V be an inner product space with inner product , . The normis the function : V R defined by

v =v, v.

(This makes sense since v, v > 0 for all v V .

Lemma 6.4 Let V be an inner product space with inner product , . Then

(i) v, w = v,w for all v,w V and F ;

(ii) v = || v for all v V and F ;

(iii) v > 0 whenever v 6= 0.

Theorem 6.5 (CauchySchwarz Inequality) Let V be an inner product spacewith inner product , . Then

|u, v| 6 u v

for all u, v V .

Corollary 6.6 (Triangle Inequality) Let V be an inner product space. Then

u+ v 6 u + v

for all u, v V .

3

Orthogonality and orthonormal bases

Definition 6.7 Let V be an inner product space.

(i) Two vectors v and w are said to be orthogonal if v,w = 0.

(ii) A set A of vectors is orthogonal if every pair of vectors within it are orthogonal.

(iii) A set A of vectors is orthonormal if it is orthogonal and every vector in A hasunit norm.

Thus the set A = {v1, v2, . . . , vk} is orthonormal if

vi, vj = ij =

{0 if i 6= j

1 if i = j.

An orthonormal basis for an inner product space V is a basis which is itself anorthonormal set.

Theorem 6.9 An orthogonal set of non-zero vectors is linearly independent.

Problem: Given a (finite-dimensional) inner product space V , how do we find anorthonormal basis?

Theorem 6.10 (GramSchmidt Process) Suppose that V is a finite-dimensionalinner product space with basis {v1, v2, . . . , vn}. The following procedure constructs anorthonormal basis {e1, e2, . . . , en} for V .

Step 1: Define e1 =1

v1v1.

Step k: Suppose {e1, e2, . . . , ek1} have been constructed. Define

wk = vk

k1i=1

vk, eiei

and

ek =1

wkwk.

4

Example 6.12 (Laguerre polynomials) We can define an inner product on thespace P of real polynomials f(x) by

f, g =

0

f(x)g(x)ex dx.

If we apply the GramSchmidt process to the standard basis of monomials

{1, x, x2, x3, . . . },

then we produce an orthonormal basis for P consisting of the Laguerre polynomials

{Ln(x) | n = 0, 1, 2, . . . }.

Example 6.13 Define an inner product on the space P of real polynomials by

f, g =

1

1f(x)g(x) dx.

Applying the GramSchmidt process to the monomials {1, x, x2, x3, . . . } produces anorthonormal basis (with respect to this inner product). The polynomials produced arescalar multiples of the Legendre polynomials:

P0(x) = 1

P1(x) = x

P2(x) =1

2(3x2 1)

...

The set {Pn(x) | n = 0, 1, 2, . . . } of Legendre polynomials is orthogonal, but notorthonormal. This is the reason why the GramSchmidt process only produces ascalar multiple of them: they do not have unit norm.

The Hermite polynomials form an orthogonal set in the space P when we endow itwith the following inner product

f, g =

f(x)g(x)ex2/2 dx.

The orthonormal basis produced by applying the GramSchmidt process to the mono-mials are scalar multiples of the Hermite polynomials.

5

Orthogonal complements

Definition 6.14 Let V be an inner product space. If U is a subspace of V , theorthogonal complement to U is

U = { v V | v, u = 0 for all u U }.

Lemma 6.15 Let V be an inner product space and U be a subspace of V . Then

(i) U is a subspace of V , and

(ii) U U = {0}.

Theorem 6.16 Let V be a finite-dimensional inner product space and U be a subspaceof V . Then

V = U U.

We can consider the projection map PU : V V onto U associated to the decom-position V = U U. This is given by

PU (v) = u

where v = u+ w is the unique decomposition of v with u U and w U.

Theorem 6.17 Let V be a finite-dimensional inner product space and U be a subspaceof V . Let PU : V V be the projection map onto U associated to the direct sumdecomposition V = U U. If v V , then PU (v) is the vector in U closest to v.

Example 6A Omitted from the handout due to length. A document containing thissupplementary example can be found in MMS and on my webpage.

6

MRQ 2014



Handout VII: The adjoint of a transformation and self-adjoint

transformations

7 The adjoint of a transformation and self-adjoint trans-

formations

Throughout this section, V is a finite-dimensional inner product space over a field F(where, as before, F = R or C) with inner product , .

Definition 7.1 Let T : V V be a linear transformation. The adjoint of T is a mapT : V V such that

T (v), w = v, T (w) for all v,w V .

Remark: More generally, if T : V W is a linear map between inner productspaces, the adjoint T : W V is a map satisfying the above equation for all v Vand w W . Appropriate parts of what we describe here can be done in this moregeneral setting.

Lemma 7.2 Let V be a finite-dimensional inner product space and let T : V V bea linear transformation. Then there is a unique adjoint T for T and, moreover, T isa linear transformation.

We also record what was observed in the course of this proof:

If A = MatB,B(T ) is the matrix of T with respect to an orthonormal basis,then

MatB,B(T) = AT

(the conjugate transpose of A).

1

Definition 7.3 A linear transformation T : V V is self-adjoint if T = T .

Lemma 7.4 (i) A real matrix A defines a self-adjoint transformation if and only ifit is symmetric: AT = A.

(ii) A complex matrix A defines a self-adjoint transformation if and only if it isHermitian: AT = A.

Theorem 7.5 A self-adjoint transformation of a finite-dimensional inner product spaceis diagonalisable.

Corollary 7.6 (i) A real symmetric matrix is diagonalisable.

(ii) A Hermitian matrix is diagonalisable.

Lemma 7.7 Let V be a finite-dimensional inner product space and T : V V be aself-adjoint transformation. Then the characteristic polynomial is a product of linearfactors and every eigenvalue of T is real.

Lemma 7.8 Let V be an inner product space and T : V V be a linear map. If U isa subspace of V such that T (U) U (i.e., U is T -invariant), then T (U) U

(i.e., U is T -invariant).

2

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