UNIT - II
INNER PRODUCT SPACES
Definition
The vector space V over F is said to be an inner product space, if there is defined for any two
vectors u, v ∈ V an element (u, v) in F such that,
1.
( u , v ) = ( v, u ) ;
2.
( u , u ) ≥ 0 and ( u , u ) = 0 is and only if u = 0
3.
(α u + β v , w) = α ( u, w) + β ( u , w ) for any u, v, w ∈ V and α , β ∈ F .
Note : A function satisfying the properties 1, 2, 3 is called an inner product.
Example :
1)
In F (n ) define for u = ( u1.....un ) and v = ( v1.....vn ) . ( u , v ) = u1v1 + ...... + unvn , this defines
an inner product of F (n ) .
2)
( u , v ) = 2u1v1 + u1v2 + u2v1 + u2v2 this defines an inner product on F(2).
3)
Let V be the set of all continuous complex valued functions on the closed unit interval [ 0, 1].
1
If f ( t ) , g ( t ) ∈V define ( f ( t ) , g ( t ) ) = ∫ f (t ) g ( t ) dt .
0
Definition :
If v ∈ V then the length of v (norm of v) written as v is defined by
u =
( u, v )
26
Leema :
If u, v ∈ V and α , β ∈ F then
(α u + β v, α u + β v ) = αα ( u, u ) + αβ ( u, v )
Proof : by property 3,
(α u + β v ,α u + β v ) = α ( u, αu + β v ) + β ( v, αu + βv )
( u ,α u + β v ) = α ( u, u ) + β ( u + v ) and ( v ,α u + β v ) = α ( v, u ) + β ( v , v )
but
∴ (α u + β v, α u + β v ) = αα ( u , u ) + αβ ( u, v ) + αβ ( v ,u ) + ββ ( v , v )
Hence the result.
Corollary :
αu = α u
Proof : α u
2
∴ αu
(
= (α u, α u ) = αα ( u, u ) since by above Lemma. Qαα = α
2
=α
2
u
2
2
and ( u , u ) = u
2
)
taking positive square roots yields α u = α u .
Lemma :
If a, b, c are real numbers such that a > 0 and aλ 2 + 2bλ + c ≥ 0 for all real number λ then
b 2 ≤ ac .
1(
b2 
2 
)
Proof : Completing the squares, aλ + 2bλ + c = aλ + b +  c −  .
a

a 
2
Since it is greater than or equal to 0 for all λ , in particular this must be true for λ = −
 b2 
c −   ≥ 0 and since a > 0 we get b 2 ≤ ac .
 a 
Theorem : If u, v ∈ V then ( u, v ) ≤ u v .
Proof : If u = 0 then both ( u , v ) = 0 and u v = 0 so that the result is true there.
27
b
. Thus
a
Case - I :
Suppose (for the moment) that ( u , v ) is real and u ≠ 0 .
We know if u, v ∈ V and α , β ∈ F then
(α u + β v, α u + β v ) = αα ( u, u )+ αβ ( u, v )+ αβ ( v , u ) + ββ ( v, v )
For any real number λ ,
0 ≤ ( λu + v, λu + v ) = λ 2 ( u, u ) + 2 ( u , v ) λ + ( v, v )
Let a = ( u, u ) , b = ( u, v ) and c = ( v, v ) for these the hypothesis ⇒ b 2 ≤ ac .
That is ( u , v )2 ≤ ( u , u )( v, v ) ; from this it is immediate that ( u , v ) ≤ u v .
Case - II :
If α = ( u, v ) is not real then it certainly is not 0 so that
u
is meaningful.
α
1
u  1
Now  α , v  = α ⋅ ( u, v ) = ( u , v ) ⋅ ( u, v ) = 1 and it is certainly real.


u
u 
v .
Therefore by Case I, 1 =  , v  ≤
α  α
∴
u
1
=
u
α
α
we get 1 ≤
u v
.
α
Whence α ≤ u v , putting α = ( u, v ) we obtain ( u, v ) ≤ u v the desired result.
Example 1 : If V = F (n) with ( u , v ) = u1v1 + ...... + unvn where u = ( u1.......un ) , v = ( v1.......vn )
then
1
Example 2 :
(
2
2
u1v1 + .... + unvn ≤ u1 + .... + un
∫
0
2
1
1
0
0
2
)( v
1
2
+ .... + vn
2
)
2
2
f (t )g (t )dt ≤ ∫ f ( t ) dt ∫ g ( t ) dt
Definition : If u, v ∈ V then u is said to be orthogonal to v if (u, v) = 0.
Note : If u is orthogonal to v then v is orthogonal to u, for ( v, u ) = ( u , v ) = ( 0 ) = 0 .
28
Definition : If W is subspace of V, the orthogonal complement of W, W ⊥ is defined by
W ⊥ = {x ∈ V | ( x , w) = 0∀w ∈ W}
Leema : W ⊥ is a subspace of V..
If a , b ∈ W ⊥ then for all α , β ∈ w and
for w ∈ W , (α a + β b, w ) = α ( aw) + β ( bw) ) = 0
∴ aα + β b ∈W ⊥
Note : W I W ⊥ = (0) for if w ∈ W I W ⊥ it must be self orthogonal. if ( w , w) = 0
∴ w = 0 ⇒w = 0
Example : W be subspace of V, W ⊥ is orthogonal complement of W which also subspace of V, then
V = W +W ⊥
(ii) W I W ⊥ = {0}
(i) V =W +W ⊥
i)
There exists an orthogonal basis ( w1 , w2 ,..., wr ) of w which is a part of an orthogonal basis
( w1...wr wr+1...wn ) of V so that
1 if i = j
wi w j = δ ij 
, ⇒ wr +1...wn ∈ w⊥
0 if i ≠ j
Let v ∈ V be arbitrary and hence ∃ unique scalars α1....α n such that
n
v = ∑ αi wi =
i =1
∴v =u + w
r
∑αi wi +
i= 1
n
∑
i= r +1
r
n
i =1
i =r +1
αi wi taking u = ∑ αi wi , w = ∑ αi wi
u ∈W , w ∈W ⊥
also this representation is unique for the scalars α1.....α n .
Thus v = W + W ⊥
29
ii)
Let u ∈ W I W ⊥ be arbitrary than u ∈W , u ∈W ⊥ .
⇒ (u , u ) = 0 ⇒ u = 0 ⇒ u = 0
∴W IW ⊥ = {0}
Definition : The set of vectors {vi } in V is an orthonormal set if
(i) each vi is of length 1, (if ( vi , vi ) = 1 ).
(ii) for if i ≠ j , ( vi , v j ) = 0 .
Lemma : If B = {v1, v2 ,...vn } is an orthogonal set then the vectors in B are linearly independent.
n
If w = ∑ α i vi then α i = ( w, vi ) vi .
r =1
Proof : Suppose that α1v1 + .... + α nvn = 0 .
∴ (α1v1 + .... + α nvn , vi ) = 0 ⇒ α1 ( v1, vi ) + .... + αn ( vn , vi ) = 0 since ( v j , vi ) = 0
For j ≠ i while ( vi , vi ) = 1 this equation reduces to α i = 0 .
Thus the v j ' s arelinearly independent.
If w = α1v1 + .... + α n vn then
( w, vi ) = ( α1v1 + .... + αn vn , vi ) =α1 ( v1, vi ) + .... + α n ( v1 , vn )
0 i ≠ j
Q ( v j , vi ) = 
1 i = j
= αi
⇒ α i = ( w, vi ) .
Lemma : If {vi ...vn } is an orthogonal set in V and if w ∈V then
u = w − ( w, v1 ) v1 − ( w, v2 ) .... − ( w ,vi ) vi .... ( w, vn ) vn is orthogonal to each of v1,...., vn .
30
Proof : Let
( u , vi ) = ( w − ( w, v1 ) v1... ( w, vn )vn , vi )
= ( w, vi ) − ( w, v1 ) ( v1, vi ) .... ( w, vi )( vi , vi ) .... ( w, vn )( vn , vi )
= ( w, vi ) − ( w, vi )
0 i ≠ j
Q ( v j , vi ) = 
1 i = j
=0
and v is arbitrary.
⇒∴u is orthogonal to each v1....vn .
Theorem :
Let V be a finite dimensional inner product space, then V has an orthogonal set as a basis.
Proof : Let V be of finite dimension n over F and let v1, v2 ,...., vn be a basis of V. Now from this basis
we shall construct an orthogonal set of n vectors
Let
u1 = v1
u2 = v2 − ( v2 , u1 )
u3 = v 3 − ( v3 , u2 )
u1
u1
u2
u2
linear space of u2 , u1 .
2
2
− ( v3 , u1 )
u1
u1
linear space of u3, u1,u 2 .
2
M
i
ui +1 = vi +1 − ∑ ( vi + j , u j )
j =1
Now
uj
uj
( u1, u2 ) =  v1 , v2 − ( v2 , v1 )

v1 
( v2 , v1 ) v , v
=
v
,
v
−
(
)
1
2
2 
2 ( 1 1)
v1 
v1
= ( v1 , v2 ) − ( v2 , v1 )
( u2 , u3 ) =  v2 − ( v2 , u1 )

v1
v1
u1
u1
linear space of ui +1, u1....u i .
2
2
2
2
=0
, v3 − ( v3, u2 )
31
u2
u2
2
− ( v3 , u1 )
u1 
2
u1 
= ( v2 , v3 ) −
( v3 , u2 )
u2
+
2
( u 2, u2 )−
( v3 , u1 )
u1
( v2 , u1 ) ( v3, u2 )
u1
2
u2
2
2
( v2, u1 ) −
( u1, u2 ) +
( v2 , u1 )
u1
2
( u1, u3 )
( v2 , u1 ) ( v3 , u1 )
u1
2
u1
2
u1 

 v3 , v2 − ( v2 , u1 )
2
u1  
u1

= ( v2 , v3 ) −
 u2 , v2 − ( v2 , u1 )
2
2
u1
u1

( u1, u1 )
 ( v3, u1 ) ( v2 , u1 )
−
2
u1

 (u , v ) ( v , u ) (v , u )  
( v , u )( v , u )  ( v , u ) ( v , u )
= ( u2 , v3 ) −  3 22 − 3 12 2 2 1  −  ( v2 , u2 ) − 2 1 22 1  − 3 1 21 1
 u2
 

u1 u2
u1
u1

 

= ( u2 , v3 ) − ( v3 , v2 ) +
( v3 , u1 ) ( v2 , u1 ) − ( v3 , u1 ) ( v2 , u1 )
u1
2
u1
2
=0
∴ u1, u2 ,..., un are orthogonal.
w1 =
u1
u
u
, w2 = 2 ,..., wn = n
u1
u2
un
∴{w1 , w2 ,...., wn} is orthogonal set which is linearly independent and dim (V) = no. of elements
in this set. Therefore, it forms basis.
Linear Transformations
1)
HOM (V, V) : the set of all vector space homomorphisms of V into itself.
2)
HOM (V, V) forms a vector space over F under the operations addition and scalar multiplication
defined as
T1 , T2 ∈ HOM (V ,V ) , then ( T1 + T2 ) ( v) = T1 ( v ) + T2 ( v ) , ∀u ∈ V and for α ∈ F ,
(α T1 ) ( v ) = α ( T1 ( v ) )
Example :
1)
For T1 , T2 ∈ HOM (V ,V ) and T1 ( v ) ∈ V for any v ∈ V then show that
T1T2 ∈ HOM (V , V ) .
32
Solution : Let T1 , T2 ∈ HOM (V ,V ) we define T1T2 ( v ) = T1 ( T2 ( v ) ) for any v ∈ V .
Let α , β ∈ F and u, v ∈ V to show that ( T1T2 ) (α u + β v ) = α (T1T2 ( u ) ) + β ( T1T2 ( v ) ) .
Consider T1T2 ( α u + β v ) = T1 T2 (α u + β v )  =T 1 α T2 ( u ) + β T2 ( v ) 
= T1 (α T2 ( u ) ) + T1 ( β T2 ( v ) ) = α ( T1 ( T2 ( u ) ) ) + β (T1 ( T2 ( v) ) )
= α ( T1T2 ) ( u )  + β ( T1T2 ) ( v ) 
Thus T1T2 ∈ HOM (V , V ) .
2)
( T1 + T2 ) T3 = T1T3 + T2T3
Solution : Let α , β ∈ F and u, v ∈ V .
( T1 +T 2 ) T3 ( α u + β v) = (T1 + T2 ) T3 ( α u + β v )  = ( T1 + T2 ) α T3 ( u ) + βT3 ( v )
= T1 α T3 ( u ) + β T3 ( v )  +T 2 αT3 ( u ) + β T3 ( v )
= αT1T3 ( u ) + βT1T3 ( v ) + αT2T3 ( u ) + βT2T3 ( v )
= α T1T3 ( u ) + T2T3 ( u )  + β T1T3 ( v ) +T2T3 ( v ) 
= α ( T1T3 + T2T3 ) ( u )  + β  (T1T3 + T2 T3 ) ( v )
= α ( T1 +T 2 ) T3  ( u ) + β ( T1 +T 2 ) T3  ( v )
∴ (T1 + T2 ) T3 = T1T3 + T2T3
3)
T3 ( T1 + T2 ) = T3T1 + T3T2 same as above.
4)
T1 (T2 , T3 ) = ( T1 , T2 ) T3
Solution : T1 (T2 , T3 ) (α u + βv ) = T1  (T2 , T3 ) ( α u + β v)  = T1 α ( T2 , T3 ) ( u ) + β ( T2 , T3 ) ( v ) 
= α ( T1 , T2 , T3 ) ( u ) + β (T1, T2 , T3 ) ( v )
= α ( T1 , T2 ) T3 ( u ) + β (T1 , T 2 ) T3 (v )
33
= ( T1, T2 ) α T3 ( u )  + (T1, T2 )  β T3 ( v )
= ( T1 , T2 ) T3 ( α u) + ( T1 , T2 ) T3 ( β v )
= ( T1 , T2 ) T3 (α u + β v)
5)
α (T1 , T2 ) = ( αT1 ) T2 = T1 ( αT 2 )
Solution : α (T1, T2 ) ( u ) = ( αT1 ) (T 2 ( u ) ) = ( T1 ) ( αT2 ( u ) ) = T1 (αT2 ) (u )
∴ From properties 1, gives clouser property w.r. to multplication 2, 3, 4 give HOM (V, V) an
associative ring.
and I ∈ HOM (V , V ) defined as Iv = v , ∀v ∈V and
TI = IT = T for every T ∈ HOM (V ,V ) .
∴ HOM (V ,V ) is ring with unity..
Definition : An associative ring A is called an algebra over F if A is a vector space over F such that
∀a, b ∈ A and α ∈ F
α ( ab ) = (α a ) b = a (α b )
Note : HOM (V, V) is an algebra over F. We denote if A (V) and whenever we want to emphasize the
role of the field F. We shall denote it by AF (V).
Definition : A linear transformation on V over F is an element of AF (V). A (V) is the ring or algebra
of linear transformations on V.
Lemma : If A is an algebra with unit element over F, then A is isomorphic to a subalgebra of A (V) for
some vector sapce V over F.
Proof : Since A is an algebra over F it must be a vector space over F. We shall use V = A to prove the
lemma.
If a ∈ A let Ta : A → A be defined by Ta ( v ) = va for every v ∈ A .
34
We assert that Ta is a linear transformation on V (= A).
By the right distribution law.
Ta ( u1 + v2 ) = ( v1 +u 2 ) a = v1a + u 2 a
= Ta ( v1 ) + Ta ( v2 )
Therefore, A is an algebra Ta (α v ) = ( αv ) a = α ( va ) = α (Ta ( v ) ) , for v ∈ A , α ∈ F .
Thus Ta is indeed a linear transformation on A. Consider the mapping ψ : A → A (V ) defined
as ψ ( a ) = Ta for every a ∈ A .
Claim : ψ is an isomorphism of A into A (V).
a=b
Ta = Tb ⇒ ψ ( a ) = ψ ( b )
ψ is well deined.
If a , b ∈ A and α , β ∈ F then ∀v ∈ A .
Tα a+ β b ( v ) = v (α a + β b ) = α ( va ) + β ( vb )
Q by left distribution law and AB algebra
= α ( Ta ( v ) ) + β (Tb ( v ) ) = (α Ta + βTb ) ( v )
Ta and Tb are L.T..
∴Tα a + β b = α Ta + β Tb ⇒ ψ is a vector space homomorphous of A into A (V).
i.e.
ψ ( α a + β b ) = αψ (a ) + βψ (b )
Now a , b ∈ A .
Tab ( v ) = u ( ab ) = ( va ) b = Tb (Ta ( v ) ) = ( TaTb ) ( v )
Q A is algebra associative law in A.
⇒ Tab = TaTb ⇒ ψ ( ab ) = ψ ( a )ψ (b)
∴ψ is also a ring homomorphism of A.
∴ψ is a homomorphism of A as an algebra into A (V).
Now Kev (ψ ) = {a ∈ A | ψ ( a) = 0} . i.e. ψ ( a) = 0 i.e. Ta = 0 and Ta ( v ) = 0 , ∀v ∈ V .
Now V = A and A has a unit element e hence Ta ( e ) = 0 . However 0 = Ta ( e ) = ea = a .
Providing that a = 0. The Kernel of ψ must consist of O.
ψ is one-one and dim( A) = dim ( A(V ) ) gives ψ is onto
∴ψ is an isomorphism of A into A (V). This completes the proof.
35
Lemma : Let A be an algebra with unit element over F, and suppose that A is of dimension n over F.
Then every element in A satisfies some nontrial polynomial in F [x] of degree at most m.
Proof : Let e be the unit element of A, if a ∈ A consider the elements e, a, a2, ... am in A. Since A is
no dimensional over F, we know “If v1....vn is basis of V over F and if w1....wm in V are Linear
independent over F then m ≤ n ” .
Let
dim A = m < m + 1
∴ e, a, a2, ... am be linearly dependent over F. In other words there are elements
α 0 , α1,..., α m ∈ F not all zero. Such that α 0e + α1a + .... + α ma m = 0 . But then a satisfies the nontrivial polynomial q ( x ) = α 0 +α1x + .... + α m x m of degree at most m in F [x].
Theorem : If V is an n-dimensional vector space over F, then given any element T in A (V), there
exists a non-trivial polynomial q ( x ) ∈ f ( x ) of degree at most n2 such that q ( T ) = 0 .
Proof : As above.
Definition : V is finite dimensional, T ∈ A (V ) some polynomial q ( x ) exist for which q ( T ) = 0 a
non-trivial polynomial of lowest degree with this property p ( x ) exists in f ( x ) we call p ( x ) a
minimal polynomial for T over F. If T satisfies then p ( x ) | h ( x ) .
Definition : An element T ∈ A (V ) is called right invertible if there exists an δ ∈ A (V ) such that
TS = 1 (1 is unit element in A (V)). Similarly we can define left in verify if U ∈ A (V ) such that
UT = 1.
If TS = UT = 1 then T is invertible.
Example : If TS = UT = I then S = U
Solution : S = IS = (UT) S = U (TS) = UI = U.
Definition : An element T in A (V) is invertible or regular if it is both right and left invertible i.e. if there
is an element S ∈ A( V ) such that ST = TS = 1 we write S as T–1.
Note : An element in A (V) which is not regular is called singular.
36
Example : Let F be the field of real numbers and let V be f ( x ) the set of all polynomials on x over
F.
x
d
Solution : Let S be defined by S ( q ( x ) ) = ( q ( x ) ) and T by T ( q ( x ) ) = ∫ q ( x ) dx
dx
1
Then where as TS = 1.
Note : An element in A (V) is right invertible but is not invertible.
Theorem : If V is finite dimensional over F, then T ∈ A (V ) is invertible if and only if the constant term
of the minimal polynomial for T is not 0.
Let p ( x ) = α 0 +α1x + .... + α k xk , α k ≠ 0 be the minimal polynomial for T over F..
If α k ≠ 0 since 0 = p ( T ) = α kT k + α k −1T k −1 + .... + α1T + α 0 we obtain
 1

1 = T  − ( α kT k −1 + α k −1T k − 2 + .... + α1 ) 
α
 0

∴S = −
1
(α k T k −1 + .... + α1 ) acts as an inverse for T..
α0
Whence T is invertible.
Suppose on the other hand that T is invertible, let α 0 = 0
Then
0 = α1T + α 2 T 2 + .... + α kT k = (α1 + α 2 + .... + α kT k −1 ) T
Multiplying this relation from the right by T–1 yields
α1 + α 2T + .... + α k T k −1 = 0
Whereby T satisfy the polynomial q ( x ) = α1 + α 2 x + .... + α k x k −1 in f ( x ) .
and deg ( q ( x ) ) is less than that of f ( x ) this is impossible.
∴ p ( x ) is minimal polynomial consequently α 0 ≠ 0 .
Hence the Theorem.
37
Corollary : If V is finite dimensional over F and if T ∈ A (V ) is invertible then T–1 is a polynomial
expression in T over T.
T is invertible
∴α 0 + α1T + .... + α kT k = 0 with α 0 ≠ 0
Then
T −1 = −
1
( α1 + α 2T + .... + α kT k −1 )
α0
Corollary : If V is finite-dimensional over F and if T ∈ A (V ) is singular then there exists an S ≠ 0 in
A (V) such that
ST = TS = 0
Proof : Because T is not regular the constant term of its minimal must be 0.
i.e. p ( x ) = α1x + .... + α k x k where 0 = α1T + .... + α kT k
If S = α1 + .... + α kT k −1 then S ≠ 0
∴α1 + .... + α k x k −1 is of lower degree then p ( x ) .
∴ ST = TS = 0
Corollary : If V is finite-dimensional over F and if T ∈ A (V ) is right invertible, then it is invertible.
Proof : Let TU = I, if T were singular there would be on S ≠ 0 such that ST = 0 .
However 0 = ( ST ) U = S ( TU ) = SI = S ≠ 0 a contradiction. Thus T is regular..
i.e. T is invertible.
Theorem : If V is finite-dimensional over F then T ∈ A (V ) is singular if and only if ∃ a v ≠ 0 in V
such that T ( v ) = 0 .
Proof : We know T is singular if and only if there is an S ≠ 0 in A (V) such that ST = TS = 0 .
Since S ≠ 0 there is an element w ∈V such that S ( w) ≠ 0 .
Let v = S ( w) then T ( v ) = T ( S ( w) ) = ( TS ) ( w )
= 0 ( w) = 0
We produced anon-zero vector v in V which is annihilated by T.
38
Conversely, if T ( v ) = 0 with v ≠ 0 .
Let ∃ of S ∈ A( V ) , v = S ( w) for some w ∈V .
∴ 0 = T ( S ( w ) ) = TS ( w) ⇒ TS = 0
∴T is singular, i.e. T is not invertible.
Definition : If T ∈ A (V ) , then the range of T, T (V) is define by
T (V ) = {v | v ∈ V } .
Theorem : If V is finite dimensional over F then T ∈ A (V ) is regular if and only if T maps V onto V..
V is finite dimensional vector space over F.
Proof : Let T ∈ A (V ) is regular. For v ∈ V we have T ( T −1 ( v ) ) = v .
∴T (V ) = V and hence T is onto.
Conversly suppose T is onto.
Suppose that T is not regular.
∴T is singular then there exists a vector v1 ≠ 0 in V such that T ( v1 ) = 0 , Q v1 ≠ 0 .
We can extend to form a basis for V as v1, v2 ,..., vn . Then every element in T(V) is a linear
combination of the elements w1 = T ( v2 ) w1 = T ( v2 ) ....wn = T ( vn ) .
Therefore dimT (V ) ≤ n − 1 < n = dimV . But then T (V) must be different from V. i.e. T is
not onto a contradiction hence T must be regular.
Definition : If V is finite dimensional over F, then the rank of T is the dimension of T (V), the range of
T over F.
We denote rank of T by r (T).
Note :
1)
If r ( T ) = dim V , then T is regular..
2)
If r ( T ) = 0 then T = 0 and so T is singular..
39
Lemma : If V is finite dimensional over F then for S , T ∈ A (V ) .
1)
r ( ST ) ≤ r ( T )
2)
r ( TS ) ≤ r ( T )
3)
r ( ST ) = r (TS ) = r (T ) for S regular in A (V).
(so r ( ST ) ≤ min {r ( T ) , r ( S )} )
Proof :
1)
Since S (V ) < V
∴ (TS )(V ) = T ( S (V ) ) ≤ T (V )
∴ dim (TS (V ) ) ≤ dim T (V ) i.e. r ( TS ) < r ( T )
2)
Suppose that r ( T ) = m , ∴T (V ) has a basis of m elements w1.....wm .
But the S ( T (V ) ) is spanned by S ( w1 ) , S ( w2 ) ,...., S ( wn ) .
Hence has dimension at most m.
Since r ( ST ) = dim ( ( ST ) (V ) ) = dim ( S (T (V ) ) ) ≤ m = dim T (V ) = r (T )
3)
If S is invertible then S (V ) = V .
∴TS (V ) = T (S (V ) ) = T (V )
∴ r ( ST ) = dim ( ( TS ) (V ) ) = dim (T (V ) ) = r ( T )
On the other hand if T (V) has w1....wm as a basis the regularity of S implies that
S ( w1 ) ,....., S ( wm ) are linearly independent.
Therefore for α1....α m ∈ F , α1S ( w1 ) + .... + α m S ( wm ) = 0
S (α1w1 ) + .... + S (α m wm ) = 0
Q S is linear..
⇒ S (α1w1 + .... + α m wm ) = 0
multiply by S' on left.
α1w1 + .... + α m wm = 0
Q S is regular..
⇒ α1 =α 2 =..... =α m = 0
∴ w1.....wm basis of T (V)
and it spans ( ST ) (V ) they form a basis of ST (V ) .
But then r ( ST ) = dim ( ST (V ) ) = dim ( T (V ) ) = r (T )
40
Corollary : If T ∈ A (V ) and if S ∈ A( V ) is regular then
r ( T ) = r ( STS −1 )
Proof : By 3 above r ( S −1T ) = r ( TS − 1 ) = r (T )
∴ r ( STS −1 ) = r ( S (TS −1 ) ) = r ( (TS −1 ) S ) = r ( T )
Example : Let V and w be vector space over the field F and let T be a linear transformation from V
into w. If T is invertible then the inverse function T–1 is a linear transformation from w onto V.
Solution : When T is one-one and onto, there is a uniquely determined inverse function T–1 which
maps w and V. such that T–1T identity on V and TT–1 identity on W.
Claim : T–1 is linear i.e. to show for α , β ∈ F , w1, w2 ∈ W .
T −1 (αw1 + β w2 ) = αT −1 ( w1 ) + β T −1 ( w2 )
Now let w1, w2 ∈ W , ∴∃v1 , v2 ∈ V such that T ( v1 ) = w1 , T ( v2 ) = w2 .
i.e. v1 = T− 1 ( w1 ) and v2 = T −1 ( w2 )
T (α v1 + β v2 ) = α T ( v1 ) + βT ( v2 ) = α w1 + β w2
∴T −1 (αw1 + β w2 ) = T −1 T ( α v1 + β v2 )  = α v1 + β v2 = α T −1 ( w1 ) + β T −1 ( w2 )
Characteristics Roots
V will always denote a finite dimensional vector space over a field F.
Definition : If T ∈ A (V ) then λ ∈ F is called a characteristic root (or eigen value) of T if λ I − T is
singular.
Theorem : The element λ ∈ F is a characteristic root of T ∈ A (V ) if and only if for some v ≠ 0 in
V, T ( v ) = λ ( v ) .
Proof : If λ is a characteristic root of T then λ − T is singular..
We know “If V–F.D.V.S. over F then T ∈ A (V ) is singular if and only if there exists a v ≠ 0
in V such that T ( v ) = 0 .”
....
41
∴ There is a vector v ≠ 0 in V such that ( λ − T ) ( v ) = 0 .
⇒ λ ( v ) − T ( v ) = 0 ⇒ T ( v ) = λv
Conversely, let T ( v ) = λ v for some v ≠ 0 in V.
∴ λv −T (v ) = 0 i.e. ( λ − T ) ( v ) = 0 by (*) must be singular and so λ is a characteristics
root of T.
Lemma : If λ ∈ F is a characteristic root of T ∈ A (V ) , then for any polynomial q ( x ) ∈ F [ x ] ,
q ( λ ) is a characteristic root of q ( T ) .
Proof : Suppose that λ ∈ F is a characteristics root of T, by above theorem there is a non-zero
vector v in V such that Tv = λ v .
Now apply T on both side, we have
T 2 ( v) = T ( λ ( v ) ) = T ( λ v ) = λT ( v ) = λ 2 v
Continuing this way, we obtain T k ( v ) = λ k v ∀ positive integers k.
If q ( x ) = α 0 x m + .... + α m , α i ∈ F then q ( T ) = α 0T m + .... + α m apply on v.
q ( T ) ( v ) = (α 0 T m + α1T m −1 + .... + α m ) ( v ) = α 0 T m ( v ) + .... + α m v
= ( α 0 λ m + α1λ m −1 + .... + α m ) v = q ( λ ) v
Thus ( q ( λ ) − q (T ) ) ( v ) = 0 hence by above theorem q ( λ ) is a characteristic root of q ( T ) .
Theorem : If λ ∈ F is a characteristic root of T ∈ A (V ) then λ is a root of the minimal polynomial
of T. In particular T only has a finite number of characteristics roots in F.
Proof : Let p ( x ) be the minimal polynomial over F or T, thus p ( T ) = 0 .
If λ ∈ F is a characteristic root of T, there is a v ≠ 0 in V with T ( v ) = λ v .
As we know “If λ ∈ F is characteristic root of T ∈ A (V ) then for any polynomial
q ( x ) ∈ F [ x ] , q ( λ ) is a characteristic root of q ( T ) .”
Therefore, we have p ( T ) ( v ) = p ( λ ) v but p ( T ) = 0 , which implies that p ( λ ) v = 0 ,
Q v ≠ 0 by property of vector space we must have p ( λ ) = 0 . Therefore λ is a root of p ( x ) . Since
p ( x ) has only a finite number of roots (in fact∴ log p ( x ) ≤ n2 where dim V = n 2 , p ( x ) has at most
n2 roots) in F, there can only be a finite number of characteristic rots of T in F.
42
Lemma : If T , S ∈ A (V ) and if S is regular, then T and STS −1 have the same minimal polynomial.
Let T , S ∈ A (V ) and S is reguar then we have
( STS −1 ) 2 = STS −1STS −1 = STITS −1 = ST 2S −1
( STS −1 )3 = STS −1STS −1STS −1 = STITITS −1 = ST 3S −1....
( STS −1 ) k
= ST k S − 1
Now for any q ( x ) ∈ F [ x ] , q ( STS −1 ) = Sq ( T ) S −1
∴ if q ( x ) = α 0 +α1x + ... + α m x m
q ( STS −1 ) = α 0 + α1STS −1 + ... + α m ( STS −1 )
m
= α 0 + α1STS −1 + ... + α m ST m S −1
= Sα 0 S −1 + Sα1TS −1 + ... + Sα mT m S −1
= S (α 0 S −1 + α1TS −1 + ... + α mT m S − 1 )
Q S is linear
= S (α 0 + α1T + ... + α mT m ) S −1
Q S − 1 is linear
= Sq ( T ) S −1
In particular if q ( T ) = 0 then q ( STS −1 ) = 0 .
Thus if p ( x ) is the minimal polynomial for T then it follows easily that p ( x ) is also the
minimal polynomial for STS −1 .
Hence the proof.
Definition : The element 0 ≠ v ∈V is called a characteristic vector of T. Belonging to the characteristic
root λ ∈ F if T ( v ) = λ v .
Theorem : If λ1....λk in F are distinct characteristic roots of T ∈ A (V ) and if v1....vk are characteristic
vectors of T belonging to λ1....λk respectively, then v1....vk are linearly independent over F..
Proof : If k =1 the result trivialy true.
Therefore one assume that k > 1
43
Suppose v1....vk are linearly dependent over F then there is a relation of the form
α1v1 + ... + α k vk = 0 where α1....α k ∈ F and not all of them are o. In all such relations, there is one
having as few non-zero coefficients as possible.
By suitable renumbering the vectors we can assume this shortest relation to be
β1v1 + ... + β j v j = 0
β1 ≠ 0....β j ≠ 0
.... (1)
We know that T ( vi ) = λi vi so applying T to equation (1) we obtain
( λ2 − λ1 ) β2 v2 + .... + ( λ j − λ1 ) β j v j = 0
Now λi − λ1 ≠ 0 for i > 1 and β1 ≠ 0 whence ( λi −λ 1 ) β i ≠ 0 .
But then we have produced a shorter relation than that in (1) between v1....vk . This contradiction
proves the theorem.
Corollary : If T ∈ A (V ) and if dimV = n then T can have at most n distinct characteristic roots in F..
Proof : Any set of linearly independent vectors in V can have at most n elements. Since any set of
distinct characteristic roots of T by above theorem gives rise to a corresponding set of linearly
independent characteristic vectors which is at most n.
Corollary : If T ∈ A (V ) and dimV = n if and if T has n distinct characteristic roots in F then there
is a basis of V over F which consists of characteristic vectors of T.
Matrices
Let V be a n-dimensional vector space over F and let v1....vn be basis of V over F..
If T ∈ A (V ) then T is determined on any vector as soon as we know its action on a
basis of V.
Tv1, Tv2 ,..., Tvn are the elements of V..
Each of these can be written as a linear combination of v1....vn unique way..
Thus
T ( v1 ) = α1 1v1 + α12 v2 + .... + α n vn , T ( v2 ) = α 21v1 +α 22 v2 + .... + α 2 n vn
...T ( vn ) = α n1v1 +α n 2v2 + .... + α nnvn
44
This system can be written more compactly as
n
T ( vi ) = ∑ α ij v j
for i = 1,..., n
j =1
The set of n2 numbers α ij ∈ F completely discribes T..
Definition : Let V be an n-dimensional vector space over F and let v1....vn be a basis of V over F..
If T ∈ A (V ) then the matrix of T in the basis v1....vn witten as m (T) is
 α11 α12 K α1n 
α

21 α 22 K α 2 n 

(
)
m T =
 M

M
 α

 n1 α n2 L α nn 
n
where T ( vi ) = ∑ α ij v j
j =1
Example : Let F be a field and V be the set of all polynomials in x of degree n–1 or less over F. On
V let D be defined by
D ( β 0 + β1 x + ... + β n −1 x n −1 ) = β1 + 2 β2 x + ... +( n −1) β n −1x n − 2
(it is called differentiation operator)
1)
Show that D is L.T. on V.
2)
Find m (D) w.r.t. basis 1, x, x 2 ,..., x n −1
Solution :
1)
α , β ∈ F , p ( x) ,q ( x ) ∈ V
α p ( x ) + β q ( x ) ∈V
D (α p ( x) + β q ( x)) = D ( α p ( x) ) + D ( β q ( x))
= D (α ) p ( x )+ α D ( p ( x ) ) + D ( β ) q ( x ) + β D ( q ( x ) )
= α D ( p ( x )) + β D ( q ( x ))
∴ D is linear transformation.
45
Q D (α ) = 0 = D ( β )
2)
The basis for V is 1, x, x 2 ,..., x n −1
∴ D ( v1 ) = D (1) = 0 , D ( v2 ) = D ( x ) = 1 = 0v1 + 0v2 + .... + 0vn
n
= ∑ ovi
i =1
D ( v3 ) = D ( x 2 ) = 2 x = 0 v1 + 2v2 + 0v3 + .... + ovn
M
D ( vi ) = D ( xi −1 ) = ( i − 1) x i −2 = 0v1 + .... + 2vi −2 + ( i −1) vi−1 + 0vi + ... + 0v1
M
D ( vn ) = D ( x n−1 ) = ( n −1) x n−2 = 0v1 + .... + 2vn− 2 + ( n − 1) vn− 1 + 0vn
(
)
gives basis
0
1

0
∴ m ( D) = 
0
M

0
2)
−−−
−−−
−−−
−−−
0
0 
0

0


0 0 − − − n − 1 0 
0
0
2
0
0
0
0
3
0
0
0
0
Find m ( D ) for a basis w1 = xn−1, w2 = x n− 2 ,..., wn = 1
Solution : Now D ( w1 ) = D ( x n−1 ) = ( n − 1) x n−2 = 0 w1 + ( n − 2 ) w2 + 0 w3 + ... + 0wn
D ( w2 ) = D ( x n−2 ) = ( n − 2 ) x n−3 = 0w1 + 0 w2 + ( n − 2 ) w3 + ... + 0wn
M
D ( wi ) = D ( xn − i ) = ( n − i ) x n − −i −1 = 0 w1 + .... + 0 wi + ( n − i ) wi+1 + 0 wi + 2 + ... + 0wn
M
D ( wn ) = D ( x ) = 0 = 0 w1 + .... + 0wn
46
0
 0 ( n − 1)
0
0
( n − 2)

∴m( D) =  M

0
0
0
0
0
0

3)
0 0 0
0 0 0 


0 0 1
0 0 0 
u1 = 1, u2 = 1 + x, u3 = 1 + x2 ,...., un = 1 + xn−1
is it basis for V over F and what is matrix for D.
Solution : α1u1 + α 2u 2 + .... + α nun = 0
α1 (1) + α 2 (1 + x ) + .... + α n (1 + x n−1 ) = 0
⇒ (α1 + α 2 + ... + α n ) + α 2 x + ... + α n x n−1 = 0
This is a linear combination of 1, x, x 2 ,..., x n −1 and it is a basis for V..
Therefore all α i = 0 .
∴ u1....un are L.I. and it forms a basis of V..
∴ D ( u1 ) = D (1) = 0 = 0u1 + 0u2 + .... + 0un
D ( u2 ) = D (1 + x ) = 1 = 1u1 + 0u2 + .... + 0un
D ( u3 ) = D (1+ x 2 ) = 2 x = 2 x + 2 − 2 = 2 (1 + x − 1) = 2 ( u 2 − u1 )
= −2u1 + 2u2 + 0u3 + .... + 0un
M
D ( un ) = D (1 + x n −1 ) = ( n − 1) x n −2 = ( n − 1) ( un − u1 )
= ( −n + 1) u1 + 0u 2 + .... + 0un − 2 + ( n − 1) un − 1 + 0un
 0
 1

 −2
∴ m ( D) = 
 −3
 M

 −( n − 1)
−−−
−−−
−−−
−−−
0
0 
0

0


0 0 − − − ( n − 1) 0 
0
0
2
0
0
0
0
3
0
0
0
0
47
4)
Let T is linear transformation of V of n-dimensional vector space V and if T has n distinct
characteristic roots then what is the matrix for T.
Solution : Let T is linear transformation on V and λ1...λn be n distinct characteristic roots of T..
We know “If T ∈ A (V ) and dimV = n and if T has n distinct and have ---- roots in F, then
there is a basis of V over F which consists of characteristic vectors of T.”
Therefore, we can find a basis v1, v2 ,....vn of V over F such that T ( vi ) = λi vi .
In this basis T has a matrix
 λ1 0 0 − − − 0 
 0 λ 0 −−− 0 
2

(
)
m T =
M



λx 
0 0 0
Note :
 β11 L β1n 


 , βi , j ∈ F gives rise to
If we have a basis v1....vn of V over F a given matrix  M
β

L
β
 n1
nn 
n
a linear transformation T defined on V by T ( vi ) = ∑ βij v j on this basis.
i =1
Thus every possible square away serves as the matrix of some linear transformation in the
basis v1....vn .
Let V is an n-dimensional vector space over F and v1....vn be basis suppose that S , T ∈ A (V ) ,
having matrices m ( S ) = (α ij ) , m (T ) = ( βij ) in the given basis.
Show that the collection of such matrices is an algebric structure.
S = T iff S ( v ) = T ( v ) for any v ∈ V .
Hence iff S ( vi ) = T ( vi ) for any v1....vn forming a basis of V.
V.
Equivalents S = T if and only if α ij = βij for each i and j.
If S = T if and only if m ( S ) = m ( T ) .
48
n
n
j =1
j =1
Now m ( S ) = (α ij ) and S ( vi ) = ∑ α ij v j and T ( vi ) = ∑ βij v j
∴ ( S + T ) ( vi ) = S ( vi ) + T ( vi ) = ∑ α ij v j + ∑ β ij v j = ∑ ( αij + β ij ) v j
∴ We can explicitly write down m ( S + T ) for m ( S ) = (α ij ) , m (T ) = ( βij )
This is meant by the matrix of linear transformation in a given basis, m ( S + T ) = ( λij ) where
λij = αij + βij for every i and j.
Now for γ ∈ F show that m ( γ S ) = ( µij ) when µij = γα ij for every i and j.
m ( γ S ) = ( γα ij ) = ( µij )
n
n
j =1
j =1
γ S ( vi ) = γ ∑ α ij v j = ∑ ( γαij ) v j = S ( vi )
 n
 n
ST
v
=
S
T
v
=
S
β
v
(
)
(
)
(
)

i
i
ij j  = ∑ βij S ( v j )
For m ( ST ) let
∑
 j =1
j
=
1


n
But
S ( v j ) = ∑ α jk vk
k =1
n
n n
 n n

∴ ST ( vi ) = ∑ β jk vk = ∑∑ ( βijα jk ) vk =  ∑ ∑ ( β ijα jk ) vk 
j =1
j =1 k =1
 k =1 S =1

n
= ∑ βij ( α j1v1 + α j 2v2 + ... + α jn vn )
j =1
n
n
n
j =1
j =1
j =1
= ∑ β ijα j1v1 + ∑ βijα j 2v2 + .... + ∑ βijα jnvn
= ( βi1α j1 + βi 2α 21 + .... + βinα n1 ) v1 + .... + ( βi1α1n + .... + βinα nn ) vn
n
n
= ∑∑ ( βijα jk ) vk
k =1 j =1
∴ m ( ST ) = (σ ik ) when for i and j
n
σ ik = ∑ β ijα jk
j =1
49
Fn : set of all n x n matrices entry from F..
It is an algebra.
i)
(α ij ) = ( βij ) two matrix in Fn iff α ij = βij , ∀ i and j
ii)
(α ij ) + ( βij ) = ( λij ) where λij = αij + βij , ∀ i and j
iii)
γ ∈ F , γ (α ij ) = ( µij ) where µij = γα ij , ∀ i and j
iv)
(α ij )( βij ) = (σ ij ) where every i and j σ ij =
n
∑ α ik βkj
K =1
Theorem : The set of all n x n matrices over F form an associative algebra, Fn over F. If V is an ndimensional vector space over F, then A (V) and Fn are isomorphic as algebra over F.
Proof : Let v1, v2 ,..., vn be a basis of V over F, T ∈ A (V ) , T :V → V , m (T ) is the matrix of TT
w.r.t. the basis v1, v2 ,..., vn .
We deine mapping φ : A ( V ) → Fn as T → m ( T ) .
φ ( T ) = m ( T ) , claim φ is well defined, 1 - 1, onto
Let S , T ∈ A (V )
if S = T then S ( v ) = T ( v ) for every v ∈ V .
i.e. S ( vi ) = T ( vi ) , vi in the basis of V.
V.
n
n
j =1
j =1
iff ∑ α ij v j = ∑ βij v j
iff, α ij = βij , ∀ i and j
i.e. (α ij ) = ( βij )
one-one ⇒ m ( S ) = m ( T ) , φ is well defined.
A ∈ Fn , ∃T ∈ A ( V ) , φ ( T ) = Am×n .
φ is onto.
α , β ∈ F , T , S ∈ A (V ) , α T + β S ∈ A ( V ) .
50
φ ( α T + β S ) = m (α T + β S ) = α m (T ) + β m ( S )
(α T + β S ) ( vi ) = ( α T ) ( vi ) + ( β S ) ( vi ) = α T ( vi ) + β ( S ( vi ) )
n
n
n
j =1
j =1
j =1
= α ∑ α ij v j + β ∑ β ij v j = ∑ (αα ij + ββ ij ) v j
φ ( ST ) = φ ( S ) + φ (T ) = m ( ST ) = m ( S )m (T )
n
n
j =1
j =1
∴ ST ( v j ) = ∑ γ ij v j where γ ij = ∑ α ik β kj
n
n
γ ij = ∑∑ α ik β kj v j
j =1 k =1
∴φ is homomorphic.
Hence φ is isomorphic.
Theorem : If V is n-deimensional over F and if T ∈ A (V ) has the matrix m1 ( T ) in the basis v1,..., vn
and the matrix m 2 ( T ) in the basis w1 ,..., wn of V over F. Then there is an element C ∈ Fn such that
m2 ( T ) = C −1m1 (T )C .
(In fact, if S is the linear transformation of V defined by S ( vi ) = wi , ∀ i = 1, ..., n then C can
be closer to be m1 ( S ) )
Proof : Let m1 ( T ) = ( α ij ) and m2 ( T ) = ( βij )
n
n
j =1
j =1
Thus T ( vi ) = ∑ α ij v j and T ( wi ) = ∑ βij w j
Let S be the linear transformation on V defined by S ( vi ) = wi , Q v1,..., vn are basis of V over
F. S maps V onto V.
We know “ If V is finite dimensional over F then T ∈ A (V ) is regular iff T maps V onto V..”
∴ S is regular if S is invertible in A (V).
n
Now T ( wi ) = ∑ βij w j , ∴ wi = S ( vi ) on substituting this in the expression for T ( wi ) we
j =1
obtain
51
n
T ( S ( vi ) ) = ∑ βij ( S ( vi ) )
j =1
 n

⇒ TS ( vi ) = S  ∑ β ij v j 
 j=1



∴ S is invertible this further simplifies to
S −1 (TS ( vi ) ) = S − 1S ( ∑ βij v j )
⇒ ( S −1TS ) ( vi ) = ∑ βij v j
∴ by the definition of the matrix of linear transformation in the given basis,
m1 ( S −1TS ) = ( βij ) = m2 ( T )
However the mapping T → m1 ( T ) is an isomorphic of A (V) onto Fn .
(
)
∴ m1 ( S −1TS ) = m1 ( S −1 ) m1 (T ) m1 ( S ) = m1 ( S − 1 ) m1 (T ) m1 ( S )
∴ m2 ( T ) = m1 ( S −1 ) m1 (T ) m1 ( S ) which is exactly what is claimed in the theorem.
Example 1 :
A
B
1 1  1 1 1 2 
1 −1 0 1 = 1 0 


 

−1 2
( AB − BA) = 

 0 1
1 1  1 1   2 0 
0 1 1 −1 = 1 −1


 

B
A
−1 2  −1 2  1 0
( AB − BA) 2 = 

 =

 0 1  0 1  0 1
A
B
1 1  2 3  3 5
1 −1 1 2  = 1 1


 

−2 6
( AB − BA ) = 

 −2 2
 2 3  1 1   5 −1
1 2  1 −1 =  3 −1


 

B
A
−2 6   −2 6   −8 0 
( AB − BA) = 

=

 −2 2   −2 2   0 −8
1 0
= ( −8) 

0 1
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PROBLEMS :
1.
Prove that S ∈ A( V ) is regular if and only if whenever v1,..., vn ∈ V are
linearly
independent, then S ( v1 ) , S ( v2 ) ,..., S ( vn ) are also linearly independent.
2.
Prove that T ∈ A (V ) is completely determined by its values on a basis of V..
3.
Prove that the minimal polynomial of R over F divides all polynomials satisfied by T over F.
4.
If V is two-dimensional over a field F prove that every element in A (V) satisfies a polynomial
of degree 2 over F.
5.
 0 1 0


Prove that give the matrix A =  0 0 1  ∈ F3 (where the characteristic of F is not 2),
 6 −11 6 


then
(a)
A3 − 6 A2 + 11A − 6 = 0
(b)
There exists a matrix C ∈ F3 such that
CAC
6.
−1
 1 0 0


=  0 2 0
 0 0 3


If F is of characteristic 2, prove that if F2 it is possible to find matrices A, B such that
AB – BA = 1.
❏❏❏
53