50 2.
OPERATOR
ALGEBRA
2.1.2. Box. Two endomorphisms T, U E L(V) are equal
ifT
lai} = Ulai}
for aUla,}
E B, where B is a basis
ofV.
Therefore, an endomorphism is
uniquely determined by its action on the vectors
of
a basis.
The equality of operators
can
also be established by other, more convenient,
methods when an inner product is defined on the vector space. The following two
theorems contain the essence of these alternatives.
2.1.3.
Theorem.
An endomorphism T
of
an inner product space is 0
if
and only
if!
{blT la} es (bl Ta) = Ofor aUla) and
Ib).
Proof Clearly,
ifT
= 0 then {blT la} =
O.
Conversely,
if
(blT la) = 0 for all ]a)
and Ib), then, choosing Ib)
=T la) =ITa), we obtain
(TaITa)=O
Via) *
Tla)=O
Via) *
T=O
by positive definiteness of the inner product.
o
2.1.4.
Theorem.
A linear operator T on an inner product space is 0 ifand only if
(alT la) =
Of
or aUla}.
Proof
Obviously,
ifT
= 0, then (alT la) =
O.
Conversely, choose a vectora la)+
fJ Ib), sandwich T between this vector and its dual, and rearrange terms to obtain
polarization
identity
what is known as the
polarization
identity:
a*fJ (alT
Ib)
+afJ*(blT la) = (aa +
fJbl
T laa +
fJb)
-lal
2
(alTla) -
IfJI
2
(blT Ib).
According to the assumption
of
the theorem, the RHS is zero. Thus, if we let
a = fJ = 1 we obtain (alT
Ib)
+ (hiT la) =
O.
Similarly, with a
=1
and fJ = i
we get i (alT
Ib)
- i (blT la) =
O.
These two equations give (al T Ib) = 0 for all
la) , Ib). By Theorem 2.1.3, T =
O.
0
To show that two operators U and T are equal, one can either have them act
on an arbitrary vectorand show that they give the same result, or one verifies that
U- Tis the zero operatorby means
of
one
of
the theorems above. Equivalently, one
shows that
(alT
Ib)
= (al U
Ib)
or (alT la) = (al U la) for all le},
Ib}.
In addition
to the zero element, which is presentin
all algebras, L(V) has an identity element,
1, which satisfies the relation1
la) = la) for all ]«) E V. With 1 in our possession,
we can ask whether it is possible to find an operator
T-
1
with the property that
T-
I
T =
n-
I
= 1. Generally speaking, ouly bijective mappings have inverses.
Therefore,
only automorphisms
of
a vector space are invertible.
1It is convenient here to use the notation ITa) for T la). This would then allow us to write the dual
of
the vector as (Tal,
emphasizing that
it is indeed the bra associated with T la).