Table Of ContentMA251 Algebra I – Advanced Linear Algebra
Daan Krammer
November 27, 2014
Contents
1 Review of Some Linear Algebra 2
1.1 The matrix of a linear map with respect to two bases . . . . . . . . . . 2
1.2 The column vector of a vector with respect to a basis . . . . . . . . . . 2
1.3 Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 The Jordan Canonical Form 3
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 The Cayley-Hamilton theorem . . . . . . . . . . . . . . . . . . . . . . . 5
2.3 The minimal polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.4 Jordan chains and Jordan blocks . . . . . . . . . . . . . . . . . . . . . 7
2.5 Jordan bases and the Jordan canonical form . . . . . . . . . . . . . . . 8
2.6 The JCF when n = 2 and 3 . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.7 The JCF for general n . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.8 Proof of theorem 26 (non-examinable) . . . . . . . . . . . . . . . . . . 14
2.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.10 Powers of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.11 Applications to difference equations . . . . . . . . . . . . . . . . . . . 18
2.12 Functions of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.13 Applications to differential equations . . . . . . . . . . . . . . . . . . . 20
3 Bilinear Maps and Quadratic Forms 21
3.1 Bilinear maps: definitions . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 Bilinear maps: change of basis . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 Quadratic forms: introduction . . . . . . . . . . . . . . . . . . . . . . . 23
3.4 Quadratic forms: definitions . . . . . . . . . . . . . . . . . . . . . . . . 25
3.5 Change of variable under the general linear group. . . . . . . . . . . . 26
3.6 Change of variable under the orthogonal group . . . . . . . . . . . . . 28
3.7 Applications of quadratic forms to geometry . . . . . . . . . . . . . . . 33
3.7.1 Reduction of the general second degree equation . . . . . . . . 33
3.7.2 The case n = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.7.3 The case n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.8 Unitary, hermitian and normal matrices . . . . . . . . . . . . . . . . . 38
3.9 Applications to quantum mechanics (non-examinable) . . . . . . . . . 40
4 Finitely Generated Abelian Groups 42
4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
4.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.3 Cosets and quotient groups . . . . . . . . . . . . . . . . . . . . . . . . 46
4.4 Homomorphisms and the first isomorphism theorem . . . . . . . . . . 48
4.5 Free abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.6 Unimodular elementary row and column operations and the Smith
normal form for integral matrices . . . . . . . . . . . . . . . . . . . . . 51
4.7 Subgroups of free abelian groups . . . . . . . . . . . . . . . . . . . . . 53
4.8 General finitely generated abelian groups . . . . . . . . . . . . . . . . 54
4.9 Finite abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.10 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.11 Hilbert’s Third problem . . . . . . . . . . . . . . . . . . . . . . . . . . 58
2 MA251AlgebraI November27,2014
4.12 Possible topics for the second year essays . . . . . . . . . . . . . . . . . 60
1 Review of Some Linear Algebra
StudentswillneedtobefamiliarwiththewholeofthecontentsofthefirstyearLinear
Algebra module (MA106). In this section, we shall review the material on matrices
of linear maps and change of basis. Other material will be reviewed as it arises.
Throughout these lecture notes, K is a field and all vector spaces and linear map
are over K. If K is not allowed to be an arbitrary field we will say so.
1.1 The matrix of a linear map with respect to two bases
Let T: V W be a linear map. Let E = (e ,...,e ) be a basis of V and F =
1 n
!
(f ,..., f ) of W. It is known that there exist unique scalars a for 1 j n and
1 m ij
1 i m such that
m
T(e ) = Â a f
j ij i
i=1
for all j. The matrix
a11 ··· a1n
(a ) = .. ..
ij ij 0 . . 1
Bam1 ··· amnC
@ A
is written [F,T,E] or [FTE] and called the matrix of T with respect to E and F.
The set (or vector space) of linear maps V W is written Hom(V,W). For fixed
!
V,W,E,Fasabove,theassociationT [FTE]definesabijectivemapHom(V,W)
7! !
Km,n.
Theorem 1. Let S:U V and T: V W be linear maps. Let (respectively) E,F,G
! !
be bases of (respectively)U,V,W. Then
[G,TS,E] = [G,T,F][F,S,E]. ⇤
1.2 The column vector of a vector with respect to a basis
Let E = (e ,...,e ) be a basis of a vector space V and let x V. It is known that
1 n
2
there are unique scalars a for 1 i n such that
i
n
v = Â a e .
i i
i=1
The column vector
a1
.. (2)
0 . 1
BanC
@ A
is written [E,v] or [Ev] and is called the column vector of v with respect to E. The a
i
are known as the coordinates of v with respect to E.
For typographical reasons we often denote the column vector (2) by (a ,...,a )T
1 n
(T denotes transpose).
For fixed V,E as above, the association v [E,v] defines a bijective map V
7! !
Kn,1.
Theorem 3. Let T: V W be a linear map and let v V. Let (respectively) E,F be
! 2
bases of (respectively) V,W. Then
[F,T(v)] = [F,T,E][E,v]. ⇤
November27,2014 MA251AlgebraI 3
If E = (e ,...,e ) is the standard basis of Kn and v Kn then [E,v] = v. If
1 n
2
F = (f ,..., f ) is another basis of Kn then [E,1,F]e = f for all i because
1 n i i
[E,1,F]e = [E,1,F][F, f ] = [E, f ] = f .
i i i i
1.3 Change of basis
LetUbeavectorspace. TheidentitymapU Uisdefinedbyx xandisvariously
! 7!
written I, id, 1, I , id , 1 .
U U U
An easy consequence of theorem 3 is:
Corollary 4. Let T: V W be a linear map. Let (respectively) E ,E ,F ,F be bases
1 2 1 2
!
of (respectively) V,V,W,W. Then
[F2,T,E2] = [F2,1,F1][F1,T,E1][E1,1,E2]. ⇤
The foregoing corollary explains why matrices of the form [F,1,E] are called
change of base matrices.
Definition5. Twomatrices A,B Km,n aresaidtobeequivalent ifthereexistinvert-
2
ible matrices P Km,m, Q Kn,n such that B = PAQ.
2 2
Theorem 6. A,B Km,n. Then the following are equivalent:
2
(a) The matrices A,B are equivalent.
(b) The matrices A,B represent the same linear map with respect to possibly different
bases.
(c) The matrices A,B have the same rank. ⇤
2 The Jordan Canonical Form
2.1 Introduction
Throughout this section V will be a vector space of dimension n over a field K, T :
V V will be a linear operator1, and A will be the matrix of T with respect to a
!
fixed basis e ,...,e of V. Our aim is to find a new basis e ,...,e for V, such that
1 n 01 0n
the matrix of T with respect to the new basis is as simple as possible. Equivalently
(by corollary 4), we want to find an invertible matrix P (the associated basis change
matrix) such that P 1AP is a simple as possible.
�
Our preferred form of matrix is a diagonal matrix, but we saw in MA106 that
the matrix (1 1), for example, is not similar to a diagonal matrix. We shall generally
0 1
assumethatK = C. Thisistoensurethatthecharacteristicpolynomialof Afactorises
into linear factors. Under this assumption, it can be proved that A is always similar
toamatrix B = (� )ofacertaintype(calledtheJordancanonicalformorsometimes
ij
Jordan normal form of the matrix), which is not far off being diagonal. In fact � is
ij
zero except when j = i or j = i+1, and� is either 0 or 1.
i,i+1
We start by summarising some definitions and results from MA106. We shall use
0 both for the zero vector in V and the zero n n matrix. The zero linear operator
⇥
0 : V V corresponds to the zero matrix 0, and the identity linear operator I :
V V
!
V V corresponds to the identity n n matrix I .
n
! ⇥
Becauseofthecorrespondencebetweenlinearmapsandmatrices,whichrespects
addition and multiplication, all statements about A can be rephrased as equivalent
statements about T. For example, if p(x) is a polynomial equation in a variable x,
then p(A) = 0 p(T) = 0 .
V
,
1i.e.alinearmapfromaspacetoitself
4 MA251AlgebraI November27,2014
If Tv = �v for � K and 0 = v V, or equivalently, if Av = �v, then � is an
2 6 2
eigenvalue, and v a corresponding eigenvector of T and A. The eigenvalues can be
computed as the roots of the characteristic polynomial c (x) = det(A xI ) of A.
A n
�
The eigenvectors corresponding to � are the non-zero elements in the nullspace
(= kernel) of the linear operator T �I This nullspace is called the eigenspace of T
V
�
with respect to the eigenvalue �. In other words, the eigenspace is equal to v V
{ 2 |
T(v) = �v , which is equal to the set of eigenvectors together with 0.
}
Thedimensionoftheeigenspace,whichiscalledthenullityofT �I istherefore
V
�
equal to the number of linearly independent eigenvectors corresponding to �. This
numberplaysanimportantroleinthetheoryoftheJordancanonicalform. Fromthe
Dimension Theorem, proved in MA106, we know that
rank(T �I )+nullity(T �I ) = n,
V V
� �
where rank(T �I ) is equal to the dimension of the image of T �I .
V V
� �
For the sake of completeness, we shall now repeat the results proved in MA106
about the diagonalisability of matrices. We shall use the theorem that a set of n
linearlyindependentvectorsofV formabasisofV withoutfurtherexplicitreference.
Theorem 7. Let T : V V bealinearoperator. Thenthematrixof T isdiagonalwith
!
respect to some basis of V if and only if V has a basis consisting of eigenvectors of T.
Proof. Suppose that the matrix A = (↵ ) of T is diagonal with respect to the basis
ij
e ,...,e of V. Recall that the image of the i-th basis vector of V is represented by
1 n
the i-th column of A. But since A is diagonal, this column has the single non-zero
entry↵ . Hence T(e ) =↵ e , and so each basis vector e is an eigenvector of A.
ii i ii i i
Conversely, suppose that e ,...,e is a basis of V consisting entirely of eigenvec-
1 n
torsof T. Then,foreachi,wehave T(e ) = � e forsome� K. Butthenthematrix
i i i i
2
of A with respect to this basis is the diagonal matrix A = (↵ ) with↵ = � for each
ij ii i
i. ⇤
Theorem 8. Let � ,...,� be distinct eigenvalues of T : V V, and let v ,...,v
1 r 1 r
!
be corresponding eigenvectors. (So T(v ) = � v for 1 i r.) Then v ,...,v are
i i i 1 r
linearly independent.
Proof. We prove this by induction on r. It is true for r = 1, because eigenvectors are
non-zero by definition. For r > 1, suppose that for some↵ ,...,↵ K we have
1 r
2
↵ v +↵ v + +↵ v = 0.
1 1 2 2 r r
···
Then, applying T to this equation gives
↵ � v +↵ � v + +↵ � v = 0.
1 1 1 2 2 2 r r r
···
Now, subtracting � times the first equation from the second gives
1
↵ (� � )v + +↵ (� � )v = 0.
2 2 1 2 r r 1 r
� ··· �
By inductive hypothesis, v ,...,v are linearly independent, so ↵ (� � ) = 0 for
2 r i i 1
�
2 i r. But, by assumption, � � = 0 for i > 1, so we must have ↵ = 0 for
i 1 i
� 6
i > 1. But then↵ v = 0, so↵ is also zero. Thus↵ = 0 for all i, which proves that
1 1 1 i
v1,...,vr are linearly independent. ⇤
Corollary9. Ifthelinearoperator T : V V (orequivalentlythen nmatrix A)has
! ⇥
n distinct eigenvalues, where n = dim(V), then T (or A) is diagonalisable.
Proof. Under the hypothesis, there are n linearly independent eigenvectors, which
therefore form a basis of V. The result follows from theorem 7. ⇤
November27,2014 MA251AlgebraI 5
2.2 The Cayley-Hamilton theorem
This theorem says that a matrix satisfies its own characteristic equation. It is easy to
visualise with the following “non-proof”:
c (A) = det(A AI) = det(0) = 0.
A
�
Thisargumentisfaultybecauseyoucannotreallyplugthematrix Aintodet(A xI):
�
you must compute this polynomial first.
Theorem 10 (Cayley-Hamilton). Let c (x) bethecharacteristicpolynomialofthe n
A
⇥
n matrix A over an arbitrary field K. Then c (A) = 0.
A
Proof. Recall from MA106 that, for any n n matrix B, the adjoint adj(B) is the
⇥
n n matrix whose (j,i)th entry is the cofactor c = ( 1)i+jdet(B ), where B is
ij ij ij
⇥ �
the matrix obtained from B by deleting the i-th row and the j-th column of B. We
proved that Badj(B) = det(B)I .
n
By definition, c (x) = det(A xI ), and (A xI )adj(A xI ) = det(A
A n n n
� � � �
xI )I . Now det(A xI ) is a polynomial of degree n in x; that is det(A xI ) =
n n n n
� �
a x0 +a x1 + +a xn, with a K. Similarly, putting B = A xI in the last
0 1 n i n
··· 2 �
paragraph, we see that the (j,i)-th entry ( 1)i+jdet(B ) of adj(B) is a polynomial
ij
�
of degree at most n 1 in x. Hence adj(A xI ) is itself a polynomial of degree
n
� �
at most n 1 in x in which the coefficients are n n matrices over K. That is,
� ⇥
adj(A xI ) = B x0+B x+ +B xn 1, where each B is an n n matrix over
n 0 1 n 1 � i
� ··· � ⇥
K. So we have
(A xIn)(B0x0+B1x+ +Bn 1xn�1) = (a0x0+a1x1+ +anxn)In.
� ··· � ···
Since this is a polynomial identity, we can equate coefficients of the powers of x on
the left and right hand sides. In the list of equations below, the equations on the left
are the result of equating coefficients of xi for 0 i n, and those on right are
obtained by multiplying Ai by the corresponding left hand equation:
AB = a I AB = a I
0 0 n 0 0 n
AB B = a I A2B AB = a A
1 0 1 n 1 0 1
� �
AB B = a I A3B A2B = a A2
2 1 2 n 2 1 2
� �
··· ···
AB B = a I AnB An 1B = a An 1
n 1 n 2 n 1 n n 1 � n 2 n 1 �
� � � � � � � �
B = a I AnB = a An.
n 1 n n n 1 n
� � � �
Now summing all of the equations in the right hand column gives
0 = a0A0+a1A+...+an 1An�1+anAn
�
(remember A0 = In), which says exactly that cA(A) = 0. ⇤
By the correspondence between linear maps and matrices, we also have c (T) =
A
0.
2.3 The minimal polynomial
Westartthissectionwithabriefgeneraldiscussionofpolynomialsinasinglevariable
x with coefficients in a field K, such as p = p(x) = 2x2 3x+11. The set of all such
�
polynomialsisdenotedby K[x]. Therearetwobinaryoperationsonthisset: addition
and multiplication of polynomials. These operations turn K[x] into a ring, which will
be studied in great detail in Algebra-II.
6 MA251AlgebraI November27,2014
As a ring K[x] has a number of properties in common2 with the integers Z. The
notation a b mean a divides b. It can be applied to integers (for instance, 3 12), and
| |
also to polynomials (for instance, (x 3) (x2 4x+3)).
� | �
We can divide one polynomial p (with p = 0) into another polynomial q and get
6
aremainderwithdegreelessthan p. Forexample,ifq = x5 3, p = x2+x+1,then
�
we find q = sp+r with s = x3 x2+1 and r = x 4. For both Z and K[x], this is
� � �
known as the Euclidean Algorithm.
A polynomial r is said to be a greatest common divisor of p,q K[x] if r p, r q,
2 | |
and, for any polynomial r with r p, r q, we have r r. Any two polynomials p,q
0 0 0 0
| | | 2
K[x] have a greatest common divisor and a least common multiple (which is defined
similarly), but these are only determined up to multiplication by a constant. For
example, x 1 isagreatestcommondivisorof x2 2x+1 and x2 3x+2,butsois
� � �
1 x and 2x 2. To resolve this ambiguity, we make the following definition.
� �
Definition 11. A polynomial with coefficients in a field K is called monic if the coef-
ficient of the highest power of x is 1.
For example, x3 2x2+x+11 is monic, but 2x2 x 1 is not.
� � �
Now we can define gcd(p,q) to be the unique monic greatest common divisor of
p and q, and similarly for lcm(p,q).
As with the integers, we can use the Euclidean Algorithm to compute gcd(p,q).
Forexample,if p = x4 3x3+2x2, q = x3 2x2 x+2,then p = q(x 1)+r with
� � � �
r = x2 3x+2, and q = r(x+1), so gcd(p,q) = r.
�
Theorem 12. Let A be an n n matrix over K representing the linear operator T :
⇥
V V. The following hold:
!
(a) There is a unique monic non-zero polynomial p(x) with minimal degree and coef-
ficients in K such that p(A) = 0,
(b) If q(x) is any polynomial with q(A) = 0, then p q.
|
Proof. (a). If we have any polynomial p(x) with p(A) = 0, then we can make
p monic by multiplying it by a constant. By theorem 10, there exists such a p(x),
namely c (x). If we had two distinct monic polynomials p (x), p (x) of the same
A 1 2
minimal degree with p (A) = p (A) = 0, then p = p p would be a non-zero
1 2 1 2
�
polynomial of smaller degree with p(A) = 0, contradicting the minimality of the
degree, so p is unique.
(b). Let p(x)betheminimalmonicpolynomialin(a)andsupposethatq(A) = 0.
As we saw above, we can write q = sp+r where r has smaller degree than p. If r is
non-zero, then r(A) = q(A) s(A)p(A) = 0 contradicting the minimality of p, so
�
r = 0 and p q. ⇤
|
Definition13. Theuniquemonicpolynomialµ (x)ofminimaldegreewithµ (A) =
A A
0 is called the minimal polynomial of A or of the corresponding linear operator T.
(Note that p(A) = 0 p(T) = 0 for p K[x].)
() 2
By theorem 10 and theorem 12 (ii), we have:
Corollary 14. The minimal polynomial of a square matrix A divides its characteristic
polynomial. ⇤
Similar matrices A and B represent the same linear operator T, and so their min-
imal polynomial is the same as that of T. Hence we have
Proposition 15. Similar matrices have the same minimal polynomial. ⇤
For a vector v V, we can also define a relative minimal polynomial µ as
A,v
2
2Technicallyspeaking,theyarebothEuclideanDomainsthatisanimportanttopicinAlgebra-II.
November27,2014 MA251AlgebraI 7
the unique monic polynomial p of minimal degree for which p(T)(v) = 0 . Since
V
p(T) = 0 if and only if p(T)(v) = 0 for all v V,µ is the least common multiple
V A
2
of the polynomialsµ for all v V.
A,v
2
But p(T)(v) = 0 for all v V if and only if p(T)(b ) = 0 for all b in a basis
V i V i
2
b ,...,b of V (exercise), so µ is the least common multiple of the polynomials
1 n A
µ .
A,bi
This gives a method of calculating µ . For any v V, we can compute µ
A A,v
2
by calculating the sequence of vectors v, T(v), T2(v), T3(v) and stopping when
it becomes linearly dependent. In practice, we compute T(v) etc. as Av for the
corresponding column vector v Kn,1.
2
For example, let K = R and
3 1 0 1
�
1 1 0 1
A = 0 1.
0 0 1 0
B 0 0 0 1 C
B C
@ A
Using the standard basis b = (1 0 0 0)T, b = (0 1 0 0)T, b = (0 0 1 0)T,
1 2 1
b4 = (0 0 0 1)T of R4,1, we have:
Ab = (3 1 0 0)T, A2b = A(Ab ) = (8 4 0 0)T = 4Ab 4b , so (A2 4A+
1 1 1 1 1
� �
4)b = 0, and henceµ = x2 4x+4 = (x 2)2.
1 A,b1 � �
Ab = ( 1 1 0 0)T, A2b = ( 4 0 0 0)T = 4Ab 4b , soµ = x2 4x+4.
2 � 2 � 2� 2 A,b2 �
Ab = b , soµ = x 1.
3 3 A,b3 �
Ab = (1 1 0 1)T, A2b = (3 3 0 1)T = 3Ab 2b , so µ = x2 3x+2 =
4 4 4 � 4 A,b4 �
(x 2)(x 1).
� �
So we haveµ = lcm(µ ,µ ,µ ,µ ) = (x 2)2(x 1).
A A,b1 A,b2 A,b3 A,b4 � �
2.4 Jordan chains and Jordan blocks
The Cayley-Hamilton theorem and the theory of minimal polynomials are valid for
any matrix over an arbitrary field K, but the theory of Jordan forms will require an
additional assumption that the characteristic polynomial c (x) is split in K[x], i.e. it
A
factorises into linear factors. If the field K = C then all polynomials in K[x] factorise
into linear factors by the Fundamental Theorem of Algebra and JCF works for any
matrix.
Definition 16. A Jordan chain of length k is a sequence of non-zero vectors v , ...,
1
v Kn,1 that satisfies
k
2
Av = �v , Av = �v +v , 2 i k,
1 1 i i i 1
�
for some eigenvalue � of A.
Exercise 17. Prove that every Jordan chain is independent.
Example 18. Let V be the vector space of functions in the form f(z)e�z where f(z)
is the polynomial of degree less than k. Consider the derivative, that is, the linear
operator T : V V given by T(�(z)) = � (z). The vectors v = zi 1e�z/(i 1)!
0 i �
! �
formaJordanchainfor T andabasisofV. Inparticular,thematrixof T inthisbasis
is the Jordan block defined below.
Definition 19. Let T: V V be linear, � K and n > 0. The kernel of the linear
! 2
map (T �I )i is called a generalised eigenspace. Likewise for matrices.
V
�
Note that ker(T �I ) is an eigenspace; this is the case i = 1. The nonzero
V
�
elements of generalised eigenspaces are called generalised eigenvectors.
Noticethat v V isaneigenvector witheigenvalue � ifandonly ifµ = x �.
A,v
2 �
Similarly, generalised eigenvectors are characterised by the propertyµ = (x �)i
A,v
�
for some � and i.
8 MA251AlgebraI November27,2014
Example 20. Let’s find the generalised eigenspaces and a Jordan chain of
3 1 0
A = 0 3 1 .
0 1
0 0 3
@ A
Firstly, ker(A �I ) = 0 if � = 3. Furthermore
3
� 6
0 1 0 0 0 1
A 3I3 = 0 0 1 , (A 3I3)2 = 0 0 0 , (A 3I3)3 = 0.
� 0 1 � 0 1 �
0 0 0 0 0 0
@ A @ A
So the remaining generalised eigenspaces are
ker(A 3I ) = b , ker(A 3I )2 = b ,b , ker(A 3I )3 = b ,b ,b
3 1 3 1 2 3 1 2 3
� h i � h i � h i
where b denotes the standard base vectors.
i
Also Ab = 3b , Ab = 3b +b , Ab = 3b +b , so b ,b ,b is a Jordan chain
1 1 2 2 1 3 3 2 1 2 3
of length 3 for the eigenvalue 3 of A. The generalised eigenspaces of index 1, 2, and
3 are respectively b , b ,b , and b ,b ,b .
1 1 2 1 2 3
h i h i h i
Notice that the dimension of a generalised eigenspace of A is the nullity of (T
�
�I )i, which is a a function of the linear operator T associated with A. Since similar
V
matrices represent the same linear operator, we have
Proposition 21. Let A,B Kn,n be similar. Then
2
dimker(A �)i = dimker(B �)i
� �
for all � K and i > 0. In words, the dimensions of the generalised eigenspaces of A
2
and B are equal. ⇤
Definition 22. We define a Jordan block with eigenvalue � of degree k to be a k k
⇥
matrix J = (� ), such that � = � for 1 i k, � = 1 for 1 i < k, and
�,k ij ii i,i+1
� = 0 if j is not equal to i or i+1.
ij
So, for example,
3 i 1 0 0 1 0 0
J1,2 = 10 11 , J�,3 = 0 0�2 3�2i 1 1, and J0,4 = 0 00 00 10 01 1
✓ ◆ 0 0 3 i
�2 B 0 0 0 0 C
B C
@ A @ A
are Jordan blocks, where � = 3 i in the second example.
�2
It should be clear that the matrix of T with respect to the basis v ,...,v of Kn,1
1 n
is a Jordan block of degree n if and only if v ,...,v is a Jordan chain for A.
1 n
Note also that for A = J , µ = (x �)i, so µ = (x �)k. Since J is an
�,k A,vi � A � �,k
upper triangular matrix with entries � on the diagonal, we see that the characteristic
polynomial c of A is also equal to (� x)k.
A
�
Warning: Someauthorsputthe1’sbelowratherthanabovethemaindiagonalin
a Jordan block. This corresponds to either writing the Jordan chain in the reversed
order or using rows instead of columns for the standard vector space. However, if an
author does both (uses rows and reverses the order) then the 1’s will go back above
the diagonal.
2.5 Jordan bases and the Jordan canonical form
Definition 23. We denote the m n matrix in which all entries are 0 by 0 . If A
m,n
⇥
is an m m matrix and B an n n matrix, then we denote the (m+n) (m+n)
⇥ ⇥ ⇥
matrix with block form
A 0
m,n
0 B
✓ n,m ◆
by A B. We call A B the block sum of A and B.
� �
November27,2014 MA251AlgebraI 9
Definition 24. Let T: V V belinear. AJordanbasis for T andV isafinitebasis E
!
of V such that there exist Jordan blocks J , ..., J such that
1 k
[ETE] = J J .
1 k
�···�
Likewise for matrices instead of linear maps.
Note that a linear map admits Jordan basis if and only if it is similar to a block
sum of Jordan blocks.
Combining definitions 16, 22 and 24 we find the following explicit characterisa-
tion of Jordan bases in terms of Jordan chains.
Proposition 25. Let T: V V be linear and E = (e ,...,e ) a basis of V. Then E is
1 n
!
a Jordan basis for T and V if and only if there exists k 1 and integers i(0), ..., i(k)
�
such that 0 = i(0) < < i(k) = n and
···
e , e , ..., e
i(t)+1 i(t)+2 i(t+1)
� �
is a Jordan chain for all t with 0 t < k. ⇤
Wecannowstatewithoutproofthemaintheoremofthissection,whichsaysthat
Jordan bases exist.
Theorem26. Let Abeann nmatrixoverK suchthatc (x)splitsintolinearfactors
A
⇥
in K[x].
(a) Then there exists a Jordan basis for A, and hence A is similar to a matrix J which
is a block sum of Jordan blocks.
(b) The Jordan blocks occurring in J are uniquely determined by A. More precisely,
if A is similar to J J and L L and J and L are (nonempty)
1 j 1 ` i i
�···� �···�
Jordan blocks then j = ` and there exists a permutation s of 1,..., j such that
{ }
J = L for all i.
i s(i)
The matrix J in the theorem is said to be the Jordan canonical form (JCF) or
sometimes Jordan normal form of A.
We will prove the theorem later. First we derive some consequences and study
methodsforcalculatingtheJCFofamatrix. Aswehavediscussedbefore,polynomials
over C always split. The gives the following corollary.
Corollary 27. Let Abeann nmatrixoverC. ThenthereexistsaJordanbasisfor A.
⇥
⇤
The proof of the following corollary requires algebraic techniques beyond the
scope of this course. You can try to prove yourself after you have done Algebra-II3.
Thetrickistofindafieldextension F K suchthat c (x) splitsin F[x]. Forexample,
A
�
consider the rotation by 90 degrees matrix A = ( 0 1). Since c (x) = x2 +1, its
1 0 A
eigenvalues are imaginary numbers i and i. Hen�ce, it admits no JCF over R but
�
over complex numbers it has JCF (i 0 ).
0 i
�
Corollary 28. Let A be an n n matrix over K. Then there exists a field extension
⇥
F K and a Jordan basis for A in Fn,1. ⇤
�
Thenexttwocorollariesareimmediate4 consequencesoftheorem26buttheyare
worth stating because of their computational significance.
Corollary 29. Let A be an n n matrix over K that admits a Jordan basis. If P is the
⇥
matrix having a Jordan basis as columns, then P 1AP is the JCF of A.
�
3OryoucantakeGaloisTheorynextyearandthisshouldbecomeobvious.
4ThismeansIamnotprovingthemherebutIexpectyoutobeabletoprovethem
10 MA251AlgebraI November27,2014
Notice that a Jordan basis is not, in general, unique. Thus, there exists multiple
matrices P such that J = P 1AP is the JCF of A.
�
The final corollary follows from an explicit calculation5 for J because both mini-
mal and characteristic polynomials of J and A are the same.
Corollary 30. Suppose that the eigenvalues of A are � ,...,� , and that the Jordan
1 t
blocks in J for the eigenvalue � are J ,...,J , where k k k .
i �i,ki,1 �i,ki,ji i,1 � i,2 �···� i,ji
Then the characteristic polynomial is cA(x) = ’it=1(�i �x)ki, where ki = ki,1 +
···+ki,ji for 1 i t. The minimal polynomial isµA(x) = ’it=1(x��i)ki,1.
2.6 The JCF when n = 2 and 3
Figure 1: All Jordan matrices of size 2 or 3 up to reordering the blocks. A dot means
zero; �,µ,⇡ K are distinct.
2
Jordan matrix c µ
A A
�
· (� x)(µ x) (x �)(x µ)
µ � � � �
✓· ◆
� 1 (� x)2 (x �)2
� � �
✓· ◆
� · (� x)2 (x �)
� � �
✓· ◆
�
µ· · (� x)(µ x)(⇡ x) (x �)(x µ)(x ⇡)
0· ·1 � � � � � �
⇡
· ·
@� 1 A
� · (� x)2(µ x) (x �)2(x µ)
0· ·1 � � � �
µ
· ·
@� A
�· · (� x)2(µ x) (x �)(x µ)
0· ·1 � � � �
µ
· ·
@� 1 A
� 1· (� x)3 (x �)3
0· 1 � �
�
· ·
@� 1 A
� · (� x)3 (x �)2
0· ·1 � �
�
· ·
@� A
�· · (� x)3 (x �)
0· ·1 � �
�
· ·
@ A
In figure 1 you can find a full list of the Jordan matrices of size 2 or 3 up to
reordering the blocks. For each we give the minimal and characteristic polynomials.
Corollary31. Let A Kn,n withn 2,3 admitaJCF.ThenaJCFof Aisdetermined
2 2{ }
by the minimal and characteristic polynomials of A.
Proof. Immediate from figure 1. ⇤
In the rest of this section we shall show examples where A Kn,n is given with
2
n 2,3 and one is asked to find a Jordan basis.
2{ }
Example 32. A = (1 4). We calculate c (x) = x2 2x 3 = (x 3)(x+1), so
1 1 A � � �
5The characteristic polynomial of J is the product of the characteristic polynomials of the Jordan
blocks and the minimal polynomial of J is the least common multiple of characteristic polynomials of
theJordanblocks
