PROOF.
Let A = [T ] , where
and
is the basis u1 , . . . , un ,
T (uj ) =
m
X
is the basis v1 , . . . , vm ,
aij vi .
i=1
2
3
x1
6 . 7
Also let [u] = 4 . 5.
xn
P
Then u = nj=1 xj uj , so
T (u) =
=
n
X
j=1
n
X
j=1
=
m
X
i=1
) [T (u)]
2
xj T (uj )
xj
m
X
aij
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 +
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
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
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 ,
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
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
(
2. Let Y 2 Vn (F ) : then,
T (A
1
Y ) = A(A
1
Y)
= In Y = Y
Im TA = Vn (F )
so
THEOREM 1.10
If T is an isomorphism between U and V , then
dim U = dim V
PROOF.
Let u1 , . . . , un be a basis for U . Then
T (u1 ), . . . , T (un )
is a basis for V (i.e. hui
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
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 ) =
2.6
Characteristic and Minimum Polynomial of a Transformation
DEFINITION 2.8
(Characteristic polynomial of T : V 7! V )
Let be a basis for V and A = [T ] .
Then we define chT = chA . This polynomial is independent of the basis
:
PROOF ( chT is independent
DEFINITION 2.10
Let T : V ! V be a linear transformation over F . Then any polynomial
of least positive degree such that
f (T ) = 0V
is called a minimum polynomial of T .
We have corresponding results for polynomials in a transformation T to
those for pol
1.
(f + g)v = f v + gv
8f, g 2 F [x], v 2 V ;
f (v + w) = f v + f w
8f 2 F [x], v, w 2 V ;
2.
3.
(f g)v = f (gv)
8f, g 2 F [x], v 2 V ;
4.
1v = v
8v 2 V.
These axioms, together with the four axioms for addition on V , turn V into
what is called a left F [
Hence the assertion is true and the result follows.
PROOF.
Let T : U V 7! U + V where U and V are subspaces of some W , such
that T (u, v) = u + v.
Thus Im T = U + V , and
Ker T
= cfw_(u, v) | u 2 U, v 2 V, and u + v = 0
= cfw_(t, t) | t 2 U \ V
Clearly
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