27.03.2014 Views

SEKE 2012 Proceedings - Knowledge Systems Institute

SEKE 2012 Proceedings - Knowledge Systems Institute

SEKE 2012 Proceedings - Knowledge Systems Institute

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

A S 1<br />

1 1 0 0 0 1 1 0 0 0 <br />

<br />

1 1 0 0 0 0 0 0 0 0<br />

<br />

0 0 0 0 0 0 0 0 0 0 <br />

<br />

r2<br />

r1<br />

0 0 0 0 0 0 0 0 0 0<br />

r8<br />

r7<br />

<br />

<br />

<br />

0 1 1 1 0 0 1 1 1 0 <br />

<br />

0 0 1 1 0 0 0 1 1 0 <br />

<br />

0 0 0 1 1 0 0 0 1 1<br />

<br />

<br />

0 0 0 1 1 <br />

<br />

<br />

0 0 0 0 0 <br />

<br />

After two elementary-row-transformations listed above, it’s<br />

easy to see th at the rank of satisfies R( A ) 4 S 1<br />

AS 1<br />

S1 1<br />

and Y 2 2 1 1 1<br />

T<br />

is a positiv e integer solution of<br />

S1<br />

<br />

A Y 0<br />

S1 S1<br />

. According to Theorem 1, S is a minimal support<br />

1<br />

of S-invariants. This is consistent with the facts.<br />

A S 3<br />

1 1 0 0 0 1 1 0 0 0 <br />

<br />

1 1 0 0 0 0 1 0 1 1<br />

<br />

0 0 1 1 0 r2<br />

r1<br />

0 0 1 1 0 <br />

<br />

r2<br />

r5<br />

<br />

r4<br />

r3<br />

0 0 1 1 0 0 0 0 0 1<br />

r4<br />

r6<br />

<br />

<br />

<br />

0 1 0 1 1 0 0 0 0 0 <br />

<br />

0 0 0 0 1 0 0 0 0 0 <br />

<br />

0 0 0 0 0 0 0 0 0 0<br />

<br />

<br />

0 0 0 0 0 <br />

<br />

<br />

0 0 0 0 0 <br />

<br />

However, for S s , s , s , s , s , its corresponding<br />

3 1 2 3 4 5<br />

generated sub-matrix A<br />

S 3<br />

is given above. After four<br />

elementary-raw-transformations, it’s easy to see that the last<br />

<br />

element in any solution of A Y 0<br />

S3 S3<br />

must be 0 acco rding to<br />

the 4 th <br />

line in the ladder-type matrix. So A Y 0 has no<br />

S3 S3<br />

positive integer solutions. According to Theorem 1, S isn’t a<br />

3<br />

minimal support of S-invariants. This is correct again.<br />

Theorem 2 If S is a minimal support of S-invariants, then the<br />

1<br />

<br />

rank of AS 1<br />

satisfies R( A ) S 1 and A Y 0<br />

S1 1<br />

S1 S1<br />

has positive<br />

integer solutions.<br />

Proof: Let Y be the minimal S-invariant supported by S<br />

0<br />

1<br />

,<br />

be the sub-vector of Y corresponding to S . Obviously,<br />

0<br />

1<br />

<br />

is a positiv e solution of A Y 0 according to the<br />

S1 S1<br />

definition of minimal S-invariants. Next, we only need to<br />

prove R( A ) S 1<br />

S1 1<br />

.<br />

Assuming that R( A ) S<br />

S1 1<br />

, then A<br />

S 1<br />

is a full-column-rank<br />

<br />

matrix. According to [10], the equation A Y 0 has only<br />

S1 S1<br />

zero solutions when A<br />

S 1<br />

is full-column-rank. It is<br />

contradictory with the above conclusion that is a positive<br />

<br />

solution of A Y 0<br />

S1 S1<br />

.<br />

Assuming that <br />

R AS 1<br />

S 1<br />

1<br />

, according to Lemma 2, the<br />

<br />

equation A Y 0 has a fundamental system of integer<br />

S1 S1<br />

solution, denoted by = , ,..., <br />

1 2 k ,where k 2 .<br />

1) On one hand, we will prove that any base solution in <br />

isn't constant times of . Otherwise, assuming that <br />

i<br />

satisfies c ( c R c 0 ). Since the linear<br />

i<br />

independence among base s olutions, (1 j k j i)<br />

j<br />

isn’t constant times of . So for any positive integers and<br />

1<br />

, ' <br />

isn’t constant times of .<br />

1 i 2 j<br />

2<br />

However, because and are both positive integer<br />

i<br />

j<br />

<br />

solutions of A Y 0<br />

S1 S1<br />

, so ' <br />

is also a<br />

1 i 2 j<br />

<br />

positive integer solution vector of A Y 0 . So '<br />

S1 S1<br />

corresponds to an S-invariant supported by S . By now, an S-<br />

1<br />

invariant corresponding to ' is obtained, which isn’t constant<br />

times of the minimal S-invariant corresponding to . Both of<br />

them are supported by the same minimal support S . However,<br />

1<br />

it has been proved in [4] that a ny S-invariant is a constant<br />

multiple of the minimal S-invariant when they are supported<br />

by one minimal support of S-invariants. This contradiction<br />

indicates that any fundamental system of integer solution of<br />

<br />

A Y 0 should contain no base solutions which are<br />

S1 S1<br />

constant times of ;<br />

<br />

2) On the other hand, as a solution of A Y 0<br />

S1 S1<br />

, should<br />

have the form that<br />

... <br />

------(1)<br />

1 1 2 2 k k<br />

where , ,..., are constants.<br />

1 2 k<br />

As has been proved that each vector in = , ,..., <br />

1 2 k <br />

can’t be constant times of , so there are at least two nonzero<br />

constants in , ,..., <br />

1 2 k . Assuming that 0 , then the<br />

i<br />

following equation holds:<br />

k<br />

( ) / ------(2)<br />

i j j i<br />

j 1<br />

j<br />

i<br />

According to (1) an d (2), ' , ,... , , ,..., <br />

1 2 i-1 i1<br />

t<br />

and = , ,..., <br />

1 2 t are equivalent. So ' is also a<br />

<br />

fundamental system of integer solution of A Y 0<br />

S1 S1<br />

. However,<br />

' contains vector , which is contradictory to t he<br />

conclusion obtained in the last sentence (shown in bold) of<br />

R A S is not true.<br />

part 1) in this proof. So <br />

S 1 1<br />

1<br />

To sum up, neither R AS 1<br />

S1<br />

nor <br />

which indicates R( A ) S 1<br />

S1 1<br />

.■<br />

R A S holds<br />

S 1 1<br />

1<br />

In Fig.2, S <br />

1 s , s , s , s , s<br />

1 2 5 6 7<br />

is a minimal support of S-<br />

<br />

invariants, so R( A ) S 1 holds and A Y 0 has a<br />

S1 1<br />

S1 S1<br />

positive integer solution Y 2 S1<br />

2 1 1 1<br />

T<br />

, which is<br />

consistent with Theorem 2.<br />

Now we can get a sufficient and necessary condition for a<br />

place subset to be a minimal support of S-invariants as follows.<br />

342

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!