last modified Jul 26, 1996
whose sum is equal to a
constant
I, where I is the identity.
As we shall see, the case of
three idempotents can be completely resolved using algebra, indeed
using NCAlgebra. It is known in various cases that if k is equal
to four,
and if
is a Banach algebra, the conclusion is greatly simplified.
This, of course, can not be analyzed using NCAlgebra.
Now we turn to the case of three idempotents and derive the
following theorem.
Theorem
Let
be an algebra and A, B and C be elements
of
.
If A, B and C are idempotents and
A + B + C =
,
then
satisfies a 5th degree polynomial
whose roots are 0, 1, 2, 3 and
³/2.
Furthermore, A, B, and C can be classified, and
A, B and C commute in all cases except
= ³/2.
Now we turn to proving the theorem and deriving the classification claimed by the theorem. The first main step in the proof of the theorem is the following proposition.
Proposition 1
Let
be an algebra and A, B and C be elements
of
.
If A, B and C are idempotents and
A + B + C =
,
then
satisfies a 5th degree polynomial
-
+ 20 ³
-
²
+ 9 ,
whose roots are 0, 1, 2, 3 and
³/2.
Proof
Let I be the ideal of
K[A,B,C,
]
generated by
,
A - A
,
B - B
,
C - C
}
The
spreadsheet
shows that
is a root of a 5th degree polynomial
-
+ 20 ³
-
²
+ 9 .
The Solve command in Mathematica shows that the roots
of this polynomial are 0, 1, 2, 3 and
³/2.
= 0:
Let
be an algebra. If A, B and C are elements
of
which are idempotents and A + B + C = 0, then
A = 0, B = 0 and C = 0.
Proof See the sum of three idempotents equals zero problem.
= 1:
There exists an algebra
and idempotents a, b and c
of
such that a + b + c = 1. Moreover, one can pick
a, b and c to be nonzero.
Proof
equal the complex numbers and a = 1, b = 0 and
c = 0.
equal to the collection of 3 × 3
matrices over the complex numbers and let
| 1 | 0 | 0 |
| ||
| a | = | 0 | 0 | 0 | ||
| 0 | 0 | 0 |
|
| 0 | 0 | 0 |
| ||
| b | = | 0 | 1 | 0 | ||
| 0 | 0 | 0 |
|
| 0 | 0 | 0 |
| ||
| c | = | 0 | 0 | 0 | ||
| 0 | 0 | 1 |
= Q[A,B,C]/I and
a = A + I, b = B + I and c = C + I, then
a, b and c are idempotents which satisfy
a + b + c = 1.
The Gröbner Basis is shown
here.
Note that it is not necessary for A, B,
and C to be zero, but
A and B commute, and their product is zero.
From the given relation
C = 1 - A - B,
we can see that C also commutes with A and B.
Also, C multiplied by either A or B is zero.
= 2
There exists an algebra
and idempotents a, b and c
of
such that a + b + c = 2. Moreover, one can pick
a, b and c to be nonzero.
Proof The proof is analogous to the one above where the Gröbner Basis is here. Notice that A and B commute again, and therefore C commutes with both of them.
= 3
For every algebra
,
there exists idempotents a, b and c
of
such that a + b + c = 3. Moreover, a, b and
c must be equal to the identity element of
.
Proof The proof is analogous to the one above where the Gröbner Basis is here.
=
³/2
For every algebra
,
there exists idempotents a, b and c
of
such that a + b + c =
³/2.
Proof The proof is analogous to the one above where the Gröbner Basis is here. In this case, A and B need not commute. The equation to take note of is
