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P. 320
Linear Algebra
Notes An example of a self-adjoint algebra is L(V, V) itself. Since the intersection of any collection of
self-adjoint algebras is again a self-adjoint algebra, the following terminology is meaningful.
Definition: If is a family of linear operators on a finite-dimensional inner product space, the
self-adjoint algebra generated by is the smallest self-adjoint algebra which contains .
Theorem 8: Let be a commuting family of diagonalizable normal operators on a finite-
dimensional inner product space V, and let be the self-adjoint algebra generated by and the
identity operator. Let {P ,..., P } be the resolution of the identity defined by . Then is the set
1 k
of all operators on V of the form
k
c P
j j
T = … (15)
j 1
where c , ..., c are arbitrary scalars.
1 k
Proof: Let denote the set of all operators on V of the form (15). Then contains the identity
operator and the adjoint
T* = c P
j j
j
of each of its members. If T = c P d P , then for every scalar a
j j and U =
j j
j j
aT + U = (ac d j )P j
j
and
TU = c d P P
i j i j
, i j
= c d P
i j j
j
= UT.
Thus is a self-adjoint commutative algebra containing and the identity operator. Therefore
contains .
Now let r , ..., r be all the roots of . Then for each pair of indices (i, n) with i n, there is an
1 k
operator T in such that r (T ) r (T ). Let a = r (T ) – r (T ) and b = r (T ). Then the linear
in i in n in in i in n in in n in
operator
Q = II a 1 (T – b I)
i in in in
n i
is an element of the algebra . We will show that Q = P (1 i k). For this, suppose j i and that
i i
is an arbitrary vector in V(r ). Then
j
T = r (T )
ij j ij
= b
ij
so that (T – b I) = 0. Since the factors in Q all commute, it follows that Q = 0. Hence Q agrees
ij ij i 1 i
with P on V(r ) whenever j i. Now suppose is a vector in V(r ). Then T = r (T ) and
i j i in i in j
a –1 (T – b I) = a –1 [r (T ) – r (T )] = .
in in in in i in n in
Thus Q = and Q agrees with P on V(r ); therefore, Q = P for -i = 1, ... , k. From this it follows
i i i i i i
that = .
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