1.3 Linear Finite Automata 19
Corollary 1.3.5. Let M and M
be two linear finite automata over GF(q).
If the state s
i
of M and the state s
i
of M
are equivalent, i =1,...,k, then
for any c
i
in GF (q),i=1,...,k, the state c
1
s
1
+ ···+ c
k
s
k
of M and the
state c
1
s
1
+ ···+ c
k
s
k
of M
are equivalent.
Let M be a linear finite automaton over GF (q) with structure matrices
A, B, C, D and structure parameters l, m, n.Thematrix
K
n
=
⎡
⎢
⎢
⎢
⎣
C
CA
.
.
.
CA
n−1
⎤
⎥
⎥
⎥
⎦
is called the diagnostic matrix of M .
For any states s
1
and s
2
of M ,sinces
1
∼ s
2
if and only if their free
responses are the same, s
1
∼ s
2
if and only if CA
i
s
1
= CA
i
s
2
, i =0, 1,...
Since the degree of the minimal polynomial of A is at most n, s
1
∼ s
2
if
and only if CA
i
s
1
= CA
i
s
2
, i =0, 1,...,n− 1. Thus s
1
∼ s
2
if and only if
K
n
s
1
= K
n
s
2
.
Let T be a matrix consisting of some maximal independent rows of K
n
.
Then s
1
∼ s
2
if and only if Ts
1
= Ts
2
.
Theorem 1.3.4. Let M be a linear finite automaton over GF (q) with
structure matrices A, B, C, D. Assume that a matrix T over GF (q) satisfies
conditions: rows of T are linear independent and for any states s
1
and s
2
of M
s
1
∼ s
2
if and only if Ts
1
= Ts
2
.LetM
be a linear finite automaton over
GF (q) with structure matrices A
,B
,C
,D
, where A
= TAR, B
= TB,
C
= CR, D
= D, R is a right inverse matrix of T .ThenM
is minimal
and equivalent to M.
Proof. Clearly, for any state s of M , Ts = T (RT s); therefore, s ∼ RT s.
We prove TART = TA. For any state s of M, the input 0 carries states
s and RT s to states As and ART s, respectively. From RT s ∼ s,wehave
ART s ∼ As. It follows that TARTs = TAs. From arbitrariness of s,wehave
TART = TA.
We prove CRT = C. For any state s of M,sinces ∼ RT s, the output of
the input 0 on the state s and the output of the input 0 on the state RT s are
the same, namely, Cs = CRTs. From arbitrariness of s,wehaveC = CRT.
For any state s of M, Ts is a state of M
.Weproves ∼ Ts. For any input
sequence x
0
x
1
..., let the output sequences on s and on Ts be y
0
y
1
... and
y
0
y
1
..., respectively. From Theorem 1.3.1, we have
y
i
= CA
i
s + Dx
0
+
i
j=1
CA
i−j−1
Bx
j
,