4
The Jordan Canonical Form
The following subspaces are central for our treatment of the Jordan and
rational canonical forms of a linear transformation T : V ! V .
DEFINITION 4.1
With mT = pb11 . . . pbt t as before and p = pi , b = bi for brevity, we def
THEOREM 2.2
Let f be irreducible. Then if f6 | g, gcd(f, g) = 1 and 9u, v 2 F [x] such
that
uf + vg = 1.
PROOF Suppose f is irreducible and f6 | g. Let d = gcd(f, g) so
d|f
and d | g.
Then either d = cf for some constant c, or d = 1. But if d = cf then
f
and
g = x ) x = c1 p1 + + ct pt .
Hence with Ti = pi (T ),
IV
T
= T1 + + Tt
= c1 T1 + + ct Tt .
Next
mT = (x
c1 ) . . . (x
) (pi pj )(T ) = 0V
ct ) | pi pj
if i 6= j
if i 6= j
) pi (T )pj (T ) = 0V
or Ti Tj = 0V
Then Ti2 = Ti (T1 + + Tt ) = Ti IV = Ti .
N
= g + ( q2 )(f + ( q1 )g)
= g + ( q2 )f + (q1 q2 )g
.
.
= ( q2 )f + (1 + q1 q2 )g
rn = (. . .) f + (. . .) g.
|cfw_z
|cfw_z
u
v
In general, rk = sk f + tk g for
r
1
= f, r0 = g, s
1
1 k n, where
= 1, s0 = 0, t
1
= 0, t0 = 1
and
sk =
qk sk
1
+ sk
2,
tk =
q
and the result is proven.
Note:
Observe that (T (uj ) = Aj , the jth column of A. So Im T is mapped
under
into C(A). Also Ker T is mapped by
into N (A). Consequently
we get bases for Im T and Ker T from bases for C(A) and N (A), respectively.
(u 2 Ker T ,
If V is a vector space of all infinitely dierentiable functions on R, then
T (f ) = a0 Dn f + a1 Dn
1
f + + an
1 Df
+ an f
defines a linear transformation T : V 7! V .
The set of f such that T (f ) = 0 (i.e. the kernel of T ) is important.
Let T : U 7! V
1. f1 | f, . . . ft | f , and
2. f1 | e, . . . ft | e ) f | e.
This uniquely defines the lcm up to a constant multiple and so we set the
lcm to be the monic lcm .
EXAMPLES 2.1
If f g 6= 0, lcm (f, g) | f g .
(Recursive property)
lcm (f1 , . . . , ft+1 ) =
4.10.2 Determining the real Jordan form . . . . . . . . . . . 95
4.10.3 A real algorithm for finding the real Jordan form . . . 100
5 The
5.1
5.2
5.3
Rational Canonical Form
Uniqueness of the Rational Canonical Form . .
Deductions from the Rational Canoni
This proves the theorem for n = p, a prime.
But what if k is not prime? Equation (5) also tells us that
qk
Now let k
kNk .
2. Then
q k = kNk +
X
nNn
X
qn
n|k
n6=k
kNk +
kNk +
< kNk +
(as nNn q n )
n|k
n6=k
bk/2c
X
qn
n=1
bk/2c
X
(adding 1)
qn
n=0
q bk/2
Existence of factorization: If f 2 F [x] is not a constant polynomial, then
f being irreducible implies the result.
Otherwise, f = f1 F1 , with 0 < deg f1 , deg F1 < deg f . If f1 and F1 are
irreducible, stop. Otherwise, keep going.
Eventually we end with
where q, r 2 Zp [x] and deg r < deg g. So let
r = r 0 + r 1 x + + rn
where r0 , . . . , rn
1
n 1
1x
2 Zp . Then
g(A) = f (A)q(A) + r(A)
= 0q(A) + r(A)
= r(A)
= r0 In + r1 A + + rn
1A
n 1
Secondly, linear independence over Zp : Suppose that
r0 In + r1 A +
3
Invariant subspaces
DEFINITIONS 3.1
Subspaces V1 , . . . , Vt of V are called independent if
v1 + + vt = 0 ) v1 = 0, . . . , vt = 0
8v1 2 V1 , . . . , vt 2 Vt .
We say that V is the (internal) direct sum of the subspaces V1 , . . . , Vt if
(a) V1 , . .
U (refer to last years notes to show that this can be done).
Then T (ur+1 ), . . . , T (un ) span Im T . For
Im T
= hT (u1 ), . . . , T (ur ), T (ur+1 ), . . . , T (un )i
= h0, . . . , 0, T (ur+1 ), . . . , T (un )i
= hT (ur+1 ), . . . , T (un )i
So assum
note for the last step that terms will be of form
s
1
pa11 paRR
up to some prime pR , with ai
factorizations
0 8i = 1, . . . , R. and as R ! 1, the prime
pa11 paRR
map onto the natural numbers, N.
We let Nm denote the number of monic irreducibles of degre
then
2
or, more explicitly,
3
2
y1
6
6 . 7
4 . 5 = P4
yn
3
x1
. 7,
. 5
xn
y1 = p11 x1 + + p1n xn
.
.
yn = pn1 x1 + + p1n xn .
THEOREM 1.16 (Eect of changing basis on matrices of LTs)
Let T : V 7! V be a LT with bases and . Then
[T ] = P
1
[T ] P
where
P =
THEOREM 3.3
Suppose that mT, v = (x
v, (T
c)k . Then the vectors
cIV )(v), . . . , (T
cIV )k
1
(v)
for W = CT, v which we call the elementary Jordan basis.
form a basis
Also
[TW ] = Jk (c).
More generally suppose mT,v = pk , where p is a monic irreducible
PROOF.
) Suppose T is 1-1.
Then Ker T = cfw_0 and we have to show that Im T = V .
rank T + nullity T
= dim U
) rank T + 0 = dim V
i.e. dim( Im T ) = dim V
) Im T
= V
as T V .
( Suppose T is onto.
Then Im T = V and we must show that Ker T = cfw_0. The abov
1
Linear Transformations
We will study mainly finite-dimensional vector spaces over an arbitrary field
F i.e. vector spaces with a basis. (Recall that the dimension of a vector
space V (dim V ) is the number of elements in a basis of V .)
DEFINITION 1.1
(
3.3
Primary Decomposition Theorem
THEOREM 3.5 (Primary Decomposition)
If T : V 7! V is a LT with mT = pb11 . . . pbt t , where p1 , . . . , pt are monic
irreducibles, then
V = Ker pb11 (T )
Ker pbt t (T ),
a direct sum of T -invariant subspaces. Moreover
EXAMPLE 1.2
a b
Let A =
c d
defined by
2 M22 (F ) and let T : M22 (F ) 7! M22 (F ) be
T (X) = AX
XA.
Then T is linear2 , and Ker T consists of all 2 2 matrices A where AX =
XA.
Take to be the basis E11 , E12 , E21 , and E22 , defined by
1 0
0 1
0 0
0 0
E1
that mA = (x
a)n where
2
a 0
1 a
0 1
. . .
.
.
0
6
6
6
6
A = Jn (a) = 6
.
6
.
6
4 0 0 a 0
0 0
1 a
3
7
7
7
7
7
7
7
5
(i.e. A is an n n matrix with as on the diagonal and 1s on the subdiagonal).
Note: Again, the minimum polynomial happens to equal the char
2.5
Construction of a field of pn elements
(where p is prime and n 2 N)
Let f be a monic irreducible polynomial of degree n in Zp [x]that is,
Fq = Zp here.
For instance,
n = 2, p = 2 ) x2 + x + 1 = f
n = 3, p = 2 ) x3 + x + 1 = f or x3 + x2 + 1 = f.
Let A
Hence
pb (T ) = (up2b )(T ) + v(T )mT (T )
= (up2b )(T )
and thus
pb w = (pb
1
)(pb+1 w) = pb
1
0=0
and w 2 Ker pb , as required.
(iii)
Ker ph (T ) = Ker ph+1 (T ) ) Ker ph+1 (T ) = Ker ph+2 (T ),
i.e.
Ker ph (T ) Ker ph+1 (T ) ) Ker ph+1 (T ) Ker ph+2 (T
Now, knowing AB = In ,
) A(BC) = A
(AB)C = A
In C = A
)C = A
) BA = In
DEFINITION 1.9
Another standard isomorphism: Let dim V = m, with basis
Then
: V 7! Vm (F ) is the isomorphism defined by
= v1 , . . . , vm .
(v) = [v]
THEOREM 1.14
rank T = rank [T ]
P
THEOREM 1.17
Let A 2 Mnn (F ) and suppose that v1 , . . . , vn 2 Vn (F ) form a basis
for Vn (F ). Then if P = [v1 | |vn ] we have
P
1
AP = [TA ] .
PROOF. Let be the standard basis for Vn (F ) consisting of the unit vectors
E1 , . . . , En and let : v1 ,
Generally, if U = hu1 , . . . , un i, then Im T = hT (u1 ), . . . , T (un )i.
Note: Even if u1 , . . . , un form a basis for U , T (u1 ), . . . , T (un ) may not
form a basis for Im T . I.e. it may happen that T (u1 ), . . . , T (un ) are linearly
depende
THEOREM 3.7 (Fittings lemma)
Suppose T : V ! V is a linear transformation over T and
Ker Ker T 2 Ker T n = Ker T n+1 =
Then V = Im T n
Ker T n .
COROLLARY 3.1
If T : V ! V is an indecomposable linear transformation (that is the
only T invariant subspaces