Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-III
69
Vector space over F. Let x ∈ W, x = α
1v1,
α
1
∈F
and T( x) = α
1T( v1 ) = α
1λ
1v1 ∈W.
Hence W is T-invariant.
Let V
= V W = dim, dim ( V ) = dim V W
e j = din V-din W = n-1.
T induces a linear transformation
on
=
defined by
Also minimal polynomial over F of T , divides the minimal polynomial of T over F. Hence all the roots of the
minimal polynomial of are roots of minimal polynomial of T. Therefore all eigen values, of lie in F. Now
satisfies the hypothesis of the theorem. Since dim
so by induction hypothesis, ∃ a basis
of V
=e j over F such that
V W
is triangular.
i.e.
T ( v ) = α v + α v
3 32 2 33 3
Μ Μ Μ Μ
T ( v ) = α v + α v + ........ + α v
n n 2 n 3
nn n
2 3
Let v 2
, v 3
,........ v n
be elements of V mapping to v 2
, v 3
,..... v n
respectively, then it is easy to prove that
v , v ,........ form a basis of V. Now T ( v )[ − α v =
1 2
v n
].
Ve T ∃v αλ
+ nW W j
) = 1, Tv ( ) + ∀ v+ W∈
V 12 ( 21 , Tv =
3 ( v+ W)
− α
22 n 22 ( v ) W
+ ∈W
=
2
) ∈Fα ,........ α
− =
211
α22 31v α n
222 Now ∈ = +
2W α 32 21
= v1
2v,
+
1,
α
33v3
W
Hence
Μ Μ Μ Μ
T( vn ) = α n v1 + α n v2
+ ............ + α nnv
1 2
n
⇒ T( v ) = α v + α v
2 21 1 22 2
2 22 2 0
= W i.e.,
Similarly,
Also T( v ) = λ v = α v (Taking
1 1 1 11 1
).
Hence a basis
of V over F, such that T( v i ) = linear combination of v i
and its predecessors
in the basis. Therefore, matrix of T in this basis :
L
NM
α 11
Mα
is triangular.
α
21 22
α α α
31 32 33
α
41
α
42
α
43
α
44
Ο
− − − − − − − − − Ο
Ο
α α α _ _ _ _ _ α
O
P
Q
n1 n2 n3
nnP