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218 PAULO TIRAO
Now is not difficult to see that one can obtain the following shape for A1,
B2 and B3,
⎡
1
⎢
A1 = ⎢
⎣
. ..
0
⎤
⎥
1⎥
⎦ , B2
⎡
⎢ 0
⎢
= ⎢
⎣ 1
. ..
1
⎤
∗ ⎥
⎦
0
, B3
⎡
⎢ ∗
⎢
= ⎢
⎣ 1
. ..
1
⎤
∗ ⎥
⎦
0
.
To continue, we first reduce the ∗ block of B2. If in this block the rows
are linearly dependent, then we get a new partition of the upper part of B3
and, after operating on B3, we have
⎡ ⎤ ⎡
1
⎢
A1 = ⎢
⎣
. .. ⎥
1⎥
,
⎥
⎦
⎢
B2 = ⎢
⎣
0
0 I
0 0
I 0
⎤
⎡
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ , B3 = ⎢
⎥ ⎢
⎦ ⎣
I 0
0 I
I 0
By writing down the corresponding matrices A and B it is clear that the
representation decomposes.
Hence, it should be
⎡ ⎤
1
⎡
⎢ .
⎢ .. ⎥ ⎢
⎢ ⎥ ⎢ 0
A1 = ⎢
1⎥
, B2 = ⎢
⎥ ⎣
⎣ ⎦
1
⎤
. .. ⎥
1 ⎥
⎦ , B3
⎡
⎢ 0
= ⎢
⎣
1
⎤
. .. ⎥
1 ⎥
⎦ ,
0
I 0
I 0
which again gives a decomposable representation.
Therefore, A1 must be square and in that case we obtain
1 1 1
A1 = . .. , B2 = . .. , B3 = . .. .
1
1
1
The corresponding representation, given by A and B, is clearly decomposable
if any of A1, B2 or B3 is of rank m ≥ 2. So, it remains possibly
indecomposable the representation defined by
A =
−1 1 0 0 0
1 −1 0 1
1 0
However, the unimodular matrix
P =
, B =
1 −1 0 0 1
−1 0 1 1
1 0 0
1 0
1
−1 0 1 1 −1
−1 1 0 −1
−1 0
.
⎤
⎥ .
⎥
⎦