16.4. Counterexamples 427
After making this reduction, we consider the operator T E Mn(A) de-
fined as follows:
0 bi 0 0 0
(^0 0) b2 0 0
0 0 0 0 0
T=
0 0 0 0 bn-1
bn^0 0 0 0
Step 2 is to convince yourself that it suffices to show C* (T) contains the
matrices
[
~ ~ ~I , [~ b~ ~I , · · ·, [H : : : ~I ,
0 0 0 0 0 0 0 0 · · · bn
together with all of the matrix units of Mn(C) C Mn(A). If this is the case,
then matrix unit manipulations show we can generate all of Mn(A).
Step 3 is to use functional calculus on TT* (remember we separated the
spectra of the bJ's) to show that C*(T) contains all the matrices
II
0
!I·[!
0
ol lo
0
b~I
(^0) b~ 0 0 0
0 0
b ' ' b
0
It follows that C* (T) contains all the matrices
1~
0
f I · If
0
ol lo
0
btl
0 b2^0 0 0
b ' ' b
o· 0 0 0
and
[bf
0
~r I~
0
0 b-1 ol lo 0... 0 I
~ ' ... ' ~ ~ ·.·.· ~ '
2
.....
0 0 0 0 0 b-n^1
since bi is the unique positive square root of bJ and C* -algebras are inverse
closed. · ·
. The final step is to play around with all of the matrices we now know
belong to C* (T) and to show that all of the matrix units must also be in