Can One Hear Fastening of a Rod?

0.17 MB·

by  A. M. Akhtyamov

CAN ONE HEAR FASTENING OF A ROD?
Akhtyamov A.M., Mouftakhov A.V.,
Tei
her M., Yamilova L.S.
(Bashkir State University, Ufa;
Bar-Ilan University, Ramat-Gan,Israel)
1. Introdu
tion. Rods are parts of various devi
es (see [1℄-[5℄). If it
is impossible to observe the rod dire
tly, the only sour
e of information
8
0 about possible defe
ts of its fastening
an be the natural frequen
ies of
0
2 its (cid:29)exural vibrations. The question arises whether one would be able to

n
dete
t damage in rod fastening by the natural frequen
ies of its (cid:29)exural
a
J vibrations. This paper gives and substantiates a positive answer to this

2
1 question.

The problem in question belongs to the
lass of inverse problems
]
P
and is a
ompletely natural problem of identi(cid:28)
ation of the boundary
S
h.
onditions.
t
a Closely related formulations of the problem were proposed in [6, 7℄.
m
Contrary to this, in this paper it is not the form of the domain or size
[

of an obje
t whi
h are sought for but the nature of fastening. The
1
v problem of determining a boundary
ondition has been
onsidered in
2
7 [8℄. However, as data for (cid:28)nding the boundary
onditions, we take not
8
1 a set of natural frequen
ies, but not
ondensation and inversion (as in
.
1 [8℄).
0
8 Similarly formulated problems also o

ur in the spe
tral theory of
0
: di(cid:27)erential operators, where it is required to determine the
oe(cid:30)
ients
v
i of a di(cid:27)erential equation and the boundary
onditions using a set of
X
r eigenvalues (for more details, see [9℄(cid:21)[14℄). However, as data for (cid:28)nding
a
the boundary
onditions, we take one spe
trum but not several spe
tra
or other additional spe
tral data (for example, the spe
tral fun
tion, the
Weyl fun
tion or the so-
alled weighting numbers) that were used in [9℄(cid:21)
[13℄. Moreover, the prin
ipal aim there was to determine the
oe(cid:30)
ients
in the equation and not in boundary
onditions. The aim of this paper
is to determine the boundary
onditions of the eigenvalue problem from
its spe
trum in the
ase of a known di(cid:27)erential equation.
The problem of determining a boundary
ondition using a (cid:28)nite set
of eigenvalues has been
onsidered previously in [15℄(cid:21)[17℄. In
ontrast
to papers [15℄(cid:21)[16℄, we think it is ne
essary to determine not the type
of fastening of plates, but the type of fastening of a rod. The problem
1
of (cid:28)nding of fastening of one of the end of a rod is
onsidered in [17℄.
In
ontrast to papers [17℄ we think it is ne
essary to determine not the
type of fastening of one of the ends of a rod, but the type of fastening
of both ends of a rod.
2. Formulation of the dire
t problem. The problem of (cid:29)exural
os
illations of rod is redu
ed to following boundary problem [1, 2℄
(αu ) = ρF u¨,
′′ ′′
(1)
(αu ) c u = 0, αu c u = 0,
′′ ′ − −1 x=0 ′′ − −2 x=0 (2)
(cid:2)(αu ) + c+u(cid:3) = 0, (cid:2)αu + c+u(cid:3) = 0,
′′ ′ 1 ′ x=l ′′ 2 ′ x=l (3)
(cid:2) (cid:3) (cid:2) (cid:3)
α l ρ
where is the (cid:29)exural rigidity, is the length, is the density and
F c 0 c
±i ±i
is the
ross-se
tion area of the rod, ( ≤ ≤ ∞) are rigidity

oe(cid:30)
ients of springs.
α ρ F λ2 ρF ω2/α
Let , and be
onstant. By denote .
For vibrations, we write
u(x,t) = y(x) cosωt
y(x)
and hen
e obtain the following eigenvalue problem for :
y(4) = s2y,
(4)
U (y) = a y(0) + a y (0) = 0, U (y) = a y (0) + a y (0) = 0,
1 1 4 ′′′ 2 2 ′ 3 ′′
− −
(5)
U (y) = a y(l) + a y (l) = 0, U (y) = a y (l) + a y (l) = 0,
3 5 8 ′′′ 4 6 ′ 7 ′′
(6)
s2 = ρF ω2/α a 0 i = 1,2,...,8
i
where , ≥ ( ).
a
i
The
oe(cid:30)
ients
hara
terize
onditions for fastening the rod (rigid

lamping, free support, free edge, (cid:29)oating (cid:28)xing, elasti
(cid:28)xing).
We note that the fun
tions
y (x,s) = (cos√sx + cosh√sx)/2,
1
y (x,s) = (sin√sx + sinh√sx)/(2√s),
2
y (x,s) = ( cos√sx + cosh√sx)/(2s), (7)
3
−
y (x,s) = ( sin√sx + sinh√sx)/(2√s3),
4
−
are linearly independent solutions of the equation (4) whi
h satisfy the

onditions
0 j = r,
(r 1)
y − (0,s) = when 6 j,r = 1,2,3,4.
j (cid:26) 1 j = r, (8)
when
2
ω s
i i
The natural frequen
ies ( ) are the
orresponding positive eigen-
values of problem (4)(cid:21)(6) (see [2, 5℄). The non-zero eigenvalues of prob-
lem (4)(cid:21)(6) are the roots of the determinant
U (y ) U (y ) U (y ) U (y )
1 1 1 2 1 3 1 4
(cid:12) U (y ) U (y ) U (y ) U (y ) (cid:12)
2 1 2 2 2 3 2 4
∆(s) = (cid:12) (cid:12)
(cid:12) U (y ) U (y ) U (y ) U (y ) (cid:12)
3 1 3 2 3 3 3 4
(cid:12) (cid:12)
(cid:12) U (y ) U (y ) U (y ) U (y ) (cid:12)
(cid:12) 4 1 4 2 4 3 4 4 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
We shall now formulate the dire
t eigenvalue problem (4)(cid:21)(6): it is
required to (cid:28)nd the unknown natural frequen
ies of the os
illations of
a i = 1,2,...8
i
the rod from ( ).
ω i = 1,2,...
i
Thus, (cid:28)nding the natural frequen
ies ( ) is equivalent
s i = 1,2,... ∆(s)
i
to founding of the roots ( ) of .
a i = 1,2,...8 s i = 1,2,...
i i
Thus, knowing ( ) it is possible to (cid:28)nd ( )
by standard methods [2, 5℄. The solution to this dire
t problem presents
no di(cid:30)
ulties. The question arises whether one would be able to do
a i = 1,2,...8 s i = 1,2,...
i i
the reverse and (cid:28)nd ( ) knowing ( ). In
a broader sense it may be stated as follows. Is it possible to derive
s
i
unknown boundary
onditions with a knowledge of ? The answer to
this question is given in the next se
tion.
3. Formulation of the inverse problem. We shall denote the ma-
a j = 1,2,...,8 U (y)
j i
trix,
onsisting of the
oe(cid:30)
ients ( ) of the forms
i = 1,2,3,4 A M i,j,m,k = 1,2,...,8
ijkm
( ) by and its minors by ( ):
a 0 0 a 0 0 0 0
1 4
(cid:13) 0 a a 0 0 0 0 0 (cid:13)
2 3
A = (cid:13) (cid:13), M = a a a a .
ijkm i j k m
(cid:13) 0 0 0 0 a 0 0 a (cid:13) ±
5 8
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 a a 0 (cid:13)
(cid:13) 6 7 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
In terms of eigenvalue problem (4)(cid:21)(6), the inverse problem
on-
a j =
j
stru
ted above should be formulated as follows: the
oe(cid:30)
ients (
1,2,...8 U (y) i = 1,2,3,4
i
) of the forms ( ) of problem (4)(cid:21)(6) are un-
known, the rank of the matrix A to make up these
oe(cid:30)
ients is equal to
s
k
four, the eigenvalues of problem (4)(cid:21)(6) are known and it is required
A
to (cid:28)nd the matrix (the
lass of linear equivalent matrixes).
4. The duality of the solution. Together with problem (4)(cid:21)(6),
let us
onsider the following eigenvalue problem
y(4) = s2y,
(9)
3
U (y) = a y(0) + a y (0) = 0, U (y) = a y (0) + a y (0) = 0,
1 1 4 ′′′ 2 2 ′ 3 ′′
− −
f e e f e e (10)
U (y) = a y(l) + a y (l) = 0, U (y) = a y (l) + a y (l) = 0.
3 5 8 ′′′ 4 6 ′ 7 ′′
(11)
f e e f e e ai
We denote the matrix
omposed of the
oe(cid:30)
ients of the forms
U (y) A M
j ijkm
by and its minors by : e
f a e0 0 a 0 0 0f 0
1 4
(cid:13) 0 a a 0 0 0 0 0 (cid:13)
2 3
A = (cid:13) e e (cid:13), Mijkm = aiaj ak am.
(cid:13) 0 0 0 0 a 0 0 a (cid:13) ±
5 8
(cid:13) (cid:13)
e e
e (cid:13) 0 0 0 0 0 a a 0 (cid:13) f
(cid:13) 6 7 (cid:13) e e e f
e e
(cid:13) (cid:13)
(cid:13) (cid:13)
A e e
∗
Let be matrix
e a 0 0 a 0 0 0 0
5 8
(cid:13) 0 a a 0 0 0 0 0 (cid:13)
6 7
A∗ = (cid:13) e e (cid:13).
(cid:13) 0 0 0 0 a 0 0 a (cid:13)
1 4
(cid:13) (cid:13)
e e
e (cid:13) 0 0 0 0 0 a a 0 (cid:13)
(cid:13) 2 3 (cid:13)
e e
(cid:13) (cid:13)
(cid:13) (cid:13)
e e
Theorem 1 (on the duality of the solution of the inverse problem).
Suppose the following
onditions are satis(cid:28)ed
rankA = rankA = 4.
(12)
e
s s
i i
If the eigenvalues { } of problem (4)(cid:21)(6) and the eigenvalues { } of
problem (9)(cid:21)(11) are identi
al, with a

ount taken for their multeipli
-
A A A A
∗
ities, the
lasses of linearly equivalent matrixes and or and
are also identi
al. e e
Proof. The eigenvalues of problem (4)(cid:21)(6) are the roots of the deter-
minant
∆(s) = M [f (s)/s2] + (M + M )[f (s)] + M [s2f (s)]
1256 − 2457 1368 − 3478 −
−
+(M + M M M )[f+(s)] + M [z(s)/s] M [sz(s)]
1278 3456 2457 1368 1357 2468
− − −
+(M + M )[g (s)/√s] (M + M )[√s3g (s)]+
2456 1268 − 3468 2478 −
−
+(M + M )[g+(s)/√s3] (M + M )[√sg+(s)],
1356 1257 3457 1378
−
(13)
where
f (s) = (1 cos√s cosh√s)/2, z(s) = (sin√s sinh√s)/2,
±
±
g (s) = ( sin√s cosh√s cos√s sinh√s)/2.
±
− ±
4
Note that the fun
tions in square bra
kets are linearly independent.
∆(s) 0 ∆(s) 0
Sin
e 6≡ , 6≡ are entire fun
tions in s of order 1/2, it
follows from Hadameard's fa
torization theorem (see [18℄) that determi-
∆(s) ∆(s)
nants and are
onne
ted by the relation
e ∆(s) C ∆(s),
≡
k C e
where is a
ertain integer and is a
ertain non-zero
onstant. From
this and we obtain the equalities
M = K M ,
1256 1256
(14)
M1357 = K Mf1357,
(15)
M2468 = K Mf2468,
(16)
M3478 = K Mf3478,
(17)
M1257 + M1356 = K (Mf1356 + M1257),
(18)
M1268 + M2456 = K (Mf1268 + Mf2456),
(19)
M1378 + M3457 = K (Mf1378 + Mf3457),
(20)
M2478 + M3468 = K (Mf2478 + Mf3468),
(21)
M1278 + M3456 = K (Mf1278 + Mf3456),
(22)
M1368 + M2457 = K (Mf1368 + Mf2457).
(23)
f f
Hen
e,
a = a , a = a , a = a , a = a ,
1 1 2 2 3 3 4 4
a = a , a = a , a = a , a = a ,
5 e5 6 e6 7 e7 8 e8
e e e e
or
a = a , a = a , a = a , a = a ,
1 5 2 6 3 7 4 8
a = a , a = a , a = a , a = a .
5 e1 6 e4 7 e3 8 e4
e e e e
These equations
an be proved by standard method of algebra. Let
M = 0 M = 0 a = 0
3478 3478 3
us
onsider only one
ase 6 for example. If 6 , 6 ,
a = 0 a = 0 a = 0 a = 0 a = 0 a = 0 a = 0
4 7 8 3 4 7 8
6 , 6 , 6 , 6 , 6 , 6 , 6 and so matrixes
A A
and have the formse e e e
ae 0 0 1 0 0 0 0 a 0 0 1 0 0 0 0
1 1
(cid:13) 0 a 1 0 0 0 0 0 (cid:13) (cid:13) 0 a 1 0 0 0 0 0 (cid:13)
2 2
A = (cid:13) (cid:13), A = (cid:13) e (cid:13).
(cid:13) 0 0 0 0 a 0 0 1 (cid:13) (cid:13) 0 0 0 0 a 0 0 1 (cid:13)
5 5
(cid:13) (cid:13) (cid:13) (cid:13)
e
(cid:13) 0 0 0 0 0 a 1 0 (cid:13) e (cid:13) 0 0 0 0 0 a 1 0 (cid:13)
(cid:13) 6 (cid:13) (cid:13) 6 (cid:13)
e
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
e
5
K = 1.
From this and from (17) we obtain
It follows from (15), (20) that
a a = a a , a + a = a + a .
1 5 1 5 1 5 1 5
e e e e
Therefore by Vieta's theorem, we have
a = a , a = a a = a , a = a .
1 1 5 5 1 5 5 1
or (24)
e e e e
Similarly, equations (16), (21) imply
a = a , a = a , a = a , a = a .
2 2 6 6 2 6 6 2
or (25)
e e e e
These equations are equivalent to four
ases:
a = a , a = a , a = a , a = a
1 1 5 5 2 2 6 6
Case 1: .
a = a , a = a , a = a , a = a
1 5 5 1 2 6 6 2
Case 2: e e e e.
a = a , a = a , a = a , a = a
1 1 5 5 2 6 6 2
Case 3: e e e e.
a = a , a = a , a = a , a = a
1 5 5 1 2 2 6 6
Case 4: e e e e.
Cases 3 and 4eare spe
iale
ases of 1 aend 2. Indeeed, in
ase 3 it follows
from (22) or (23) that
(a a )(a a ) = 0.
1 5 2 6
− − (26)
a a = 0 a a = 0. a a = 0
1 5 2 6 1 5
Hen
e − or − If − , then
ase 3 is spe
ial
a = a = a = a a a = 0
1 1 5 5 2 6

ase 2: . If − , then
ase 3 is spe
ial
ase 1:
a = a = a = a
6 2 6 6
e . e
e In the same weay, we
an show that the
ase 4 is spe
ial
ase of
ases
M = 0
3478
1 and 2. We have proved that in the
ase 6 it is possible only
two
ases: 1 and 2. In a similar manner we
an prove the analogous
results for other
ases. This
ontradi
tion proves the theorem.
5. Exa
t solution. It has been shown above that the problem
A
of (cid:28)nding the matrix (in the sense that
lass of linearly equivalent
A
matrixes of ) from the natural frequen
ies of (cid:29)exural os
illations of a
rod has a duality solution. The next question is how this solution
an
be
onstru
ted.
This se
tion deals with solving this problem and
onstru
ting exa
t
ω
i
solution by the (cid:28)rst 9 natural frequen
ies .
s s s
1 2 9
Suppose , , ..., are the values
orresponding to the (cid:28)rst nine
ω s s s
i 1 2 9
natural frequen
ies . We substitute the values , , ..., into (13)
6
and obtain a system of nine homogeneous algebrai
equations
x [f (s )/s2] + x [f (s )] + x [s2f (s )]
1 − i i 2 − i 3 i − i
+x [f+(s )] + x [z(s )/s ] + x [s z(s )]
4 i 5 i i 6 i i
+x [g (s )/√s ] + x [ s3g (s )]+ (27)
7 − i i 8 i − i
+x g+(s )/ s3] + x p[√s g+(s )] = 0, i = 1,2,...,9
9 i i 10 i i
p
in the ten unknowns
x = M , x = M + M , x = M ,
1 1256 2 2457 1368 3 3478
−
x = M + M M M , x = M ,
4 1278 3456 2457 1368 5 1357
− −
x = M , x = M + M , x = (M + M ),
6 2468 7 2456 1268 8 3468 2478
− −
x = M + M , x = (M + M ).
9 1356 1257 10 3457 1378
−
(28)
The resulting set of equations (27) has an in(cid:28)nite number of solutions.
x
i
If the resulting set has a rank of 9, the unknown
an be found in
A
a

urate to a
oe(cid:30)
ient. The unknown matrix is found from (31) by
dire
t
al
ulations as in proof of theorem 1.
These reasons prove
Theorem 2 (on the duality of the solution of the inverse problem). If
the matrix of system (27) has a rank of 9, the solution of the inverse
problem of the re
onstru
tion boundary
onditions (10), (11) is duality.
Remark. Theorem 2 is stronger than theorem 1. Theorem 2 use only
9 natural frequen
ies for the re
onstru
tion of boundary
onditions and
not all natural frequen
ies as in theorem 1. But theorem 1 is proves
duality of the solution in the
ommon
ase (if the set (27) has not a
rank of 9).
s
i
Continuity of the solution of the inverse problem with respe
t to is
s
i
proved as in [16℄. This shows that small perturbations of eigenvalues
i = 1,2,3
( ) lead to small perturbations of the boundary
onditions. It
follows from this and theorem 1 that the inverse problem is well posed,
s
i
sin
e its solution exists, is unique and
ontinuous with respe
t to
(i = 1,2,3)
.
Computer
al
ulations
on(cid:28)rm the stability of the solution of the
inverse problem. The order of error often hardly di(cid:27)erent from the error
s s
k i
in the
loseness of values of and and only in some
ases it
an be
deteriorated by four orders ofemagnitude. So the measurement a

ura
y
of instruments to measure natural frequen
ies must ex
eed a

ura
y to
measure boundary
onditions by four orders of magnitude.
7
3. Numeri
al results. We use dimensionless variables in the nu-
meri
al examples.
Example 1 (rigid
lamping (cid:22) free support). Suppose
s = 15.4182057169801, s = 49.9648620318002,
1 2
s = 104.247696458861, s = 178.269729494609,
3 4
s = 272.030971305025, s = 385.531421917553,
5 6
s = 518.771081332259, s = 671.749949549144,
7 8
s = 844.468026568208
9
ω
i

orrespond to the (cid:28)rst 9 natural frequen
ies determined using instru-
ments for measuring the natural frequen
ies, then the solution of set
(27) has the form
x = M 0 C, x = M + M 0 C,
1 1256 2 2457 1368
− ≈ · ≈ ·
x = M 0 C, x = M + M M M 0 C,
3 3478 4 1278 3456 2457 1368
≈ · − − ≈ ·
x = M 0 C, x = M 0 C,
5 1357 6 2468
≈ · − ≈ ·
x = M + M 0 C, x = (M + M ) 0 C,
7 2456 1268 8 3468 2478
− ≈ · − ≈ ·
x = M + M C, x = (M + M ) 0 C
9 1356 1257 10 3457 1378
≈ − ≈ ·
(29)
10 9
−
with an a

ura
y of .
1 C M + M = 1
1356 1257
Substituting for in (29), we get . This means
M = 0 M = 0
1356 1257
that 6 or 6 .
M = 0 a a a a = 0
1356 1 3 5 6
Suppose 6 . Then 6 . So
1 0 0 a 0 0 0 0
4
(cid:13) 0 a 1 0 0 0 0 0 (cid:13)
2
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 a (cid:13)
8
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 1 a 0 (cid:13)
(cid:13) 7 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
A
(in the sense that
lass of linearly equivalent matrixes of ).
M = 0 M = 0 M +M = 0 a = 0
1256 1357 2457 1368 2
Using , and , we have ,
a = 0 a = 0 M + M M M = 0
7 8 1278 3456 2457 1368
and . From this and − − ,
a = 0
4
we get .
Thus we have
1 0 0 0 0 0 0 0
(cid:13) 0 0 1 0 0 0 0 0 (cid:13)
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 0 (cid:13)
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 1 0 0 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
8
M = 0 a a a a = 0
1257 1 2 5 7
Suppose now 6 . Then 6 . So
1 0 0 a 0 0 0 0
4
(cid:13) 0 1 a 0 0 0 0 0 (cid:13)
3
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 a (cid:13)
8
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 a 1 0 (cid:13)
(cid:13) 6 (cid:13)
(cid:13) (cid:13)
M = 0 (cid:13)M + M = 0 (cid:13)a = 0 a = 0
1256 2457 1368 6 4
Using and , we get and .
M = 0 M + M
1357 1278 3456
If we
ombine this with equations and −
M M = 0 a = 0 a = 0
2457 1368 3 8
− , we get and .
Thus we have
1 0 0 0 0 0 0 0
(cid:13) 0 1 0 0 0 0 0 0 (cid:13)
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 0 (cid:13)
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 0 1 0 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
So boundary
onditions are or
U (y) = y(0) = 0, U (y) = y (0) = 0,
1 2 ′′
U (y) = y(l) = 0, U (y) = y (l) = 0,
3 4 ′
either
U (y) = y(0) = 0, U (y) = y (0) = 0,
1 2 ′
U (y) = y(l) = 0, U (y) = y (l) = 0.
3 4 ′′
s
i
Note that the numbers presented above are the same as the (cid:28)rst
nine exa
t values
orresponding to fastening of rigid
lamping and free
support. This means that the unknown rod fastening ina

essible to
dire
t observation has been
orre
tly determined.
Example 2 (rigid
lamping (cid:22) elasti
(cid:28)xing). If
s = 5.60163863016235, s = 22.4984332740862,
1 2
s = 61.8604321649037, s = 120.984868139371,
3 4
s = 199.909638169628, s = 298.589053349029,
5 6
s = 417.014779762035, s = 555.183266366176,
7 8
s = 713.092945010199
9
ω
i

orrespond to the (cid:28)rst 9 natural frequen
ies determined using instru-
ments for measuring the natural frequen
ies, then the solution of set
9
(27) has the form
x = M 0, x = M + M 0,
1 1256 2 2457 1368
− ≈ ≈
x = M 0, x = M + M M M C,
3 3478 4 1278 3456 2457 1368
≈ − − ≈ −
x = M 0, x = M 0,
5 1357 6 2468
≈ − ≈
x = M + M 0, x = (M + M ) 0,
7 2456 1268 8 3468 2478
− ≈ − ≈
x = M + M 5C, x = (M + M ) 0.
9 1356 1257 10 3457 1378
≈ − ≈
(30)
C = 1 M + M = 0 M = 0
1356 1257 1356
Suppose ; then 6 . This means that 6
M = 0
1257
or 6 .
M = 0 a a a a = 0
1356 1 3 5 6
Suppose 6 . Then 6 . So
1 0 0 a 0 0 0 0
4
(cid:13) 0 a 1 0 0 0 0 0 (cid:13)
2
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 a (cid:13)
8
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 1 a 0 (cid:13)
(cid:13) 7 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
A
(in the sense that
lass of linearly equivalent matrixes of ).
M = 0 M +M = 5C
1256 1356 1257
If we
ombine this with equations , we
a = 0 1 = 5C M = 0 M + M = 0
2 1357 2457 1368
get , . Using and , we have
a = 0 a = 0 M +M M M = C
7 8 1278 3456 2457 1368
and . From this and − − − ,
a = 1/5
4
we get − − .
Thus we have
1 0 0 1/5 0 0 0 0
(cid:13) 0 0 1 0 0 0 0 0 (cid:13)
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 0 (cid:13)
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 1 0 0 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
M =(cid:13)0 a a a a = 0 (cid:13)
1257 1 2 5 7
Suppose now 6 . Then 6 . So
1 0 0 a 0 0 0 0
4
(cid:13) 0 1 a 0 0 0 0 0 (cid:13)
3
A = (cid:13) (cid:13)
(cid:13) 0 0 0 0 1 0 0 a (cid:13)
8
(cid:13) (cid:13)
(cid:13) 0 0 0 0 0 a 1 0 (cid:13)
(cid:13) 6 (cid:13)
(cid:13) (cid:13)
M = 0 (cid:13)M + M = 0 (cid:13)a = 0 a = 0
1256 2457 1368 6 4
Using and , we get and .
M = 0 M + M = 5C
1357 1356 1257
If we
ombine this with equations , ,
M + M M M = C M + M = 0
1278 3456 2457 1368 2457 1368
− − − and we get
a = 0 1 = 5C a = 1/5
3 8
, , and − − .
10