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b. Express each vector of
(coming as before from ( 8.14 )) as an element of N ( S ),
with coordinates coming from the basis
B
P
.
,
for the nullspace N ( S ) of the stoichiometric matrix S as in Exercises 8.5 and 8.8 .As
per Exercises 8.13 and 8.15 ,dim
Exercises 8.13 and 8.15 explore two bases, namely the sets of vectors
B
and
P
6, so any vector in N ( S ) can be expressed
uniquely in six coordinates taken either with respect to
(
N
(
S
)) =
B
, or with respect to
P
.For
11 , we previously checked
example, viewing N ( S ) as a subspace of Euclidean space
R
that
T
).
(As per usual conventions, the coordinates here are in terms of the standard basis
{
v
:= (
1
,
1
,
0
,
0
,
0
,
1
,
0
,
1
,
0
,
0
,
1
)
N
(
S
11 .) However, in terms of coordinates with respect to the ordered
e 1 ,...,
e 11 }
of
R
basis
B ={
z 1 ,...,
z 6 }
of N ( S ) as a six-dimensional vector space,
T
v
=
z 1 = (
1
,
0
,
0
,
0
,
0
,
0
)
B .
(8.15)
Likewise, since y 1
=
z 1 , in terms of coordinates with respect to the ordered set
P ={
y 1 ,...,
y 6 }
, one has
T
v
= (
1
,
0
,
0
,
0
,
0
,
0
)
P .
(8.16)
On the other hand,
T
11
w
:= (
0
,
1
,
1
,
0
,
0
,
1
,
1
,
0
,
0
,
0
,
0
)
N
(
S
) R
,
satisfies
T
w
=
z 4 = (
0
,
0
,
0
,
1
,
0
,
0
)
B ,
(8.17)
but
T
= (
,
,
,
,
,
)
P =
y 4
y 1 .
w
1
0
0
1
0
0
(8.18)
A change-of-basis matrix 9 from the basis
B
to the basis
P
is a square 6
×
6matrix
A
with the following property:
B,P
If
T
u
:= (
u 1 ,...,
u 6 )
B ,
:=
then the matrix product q
A B,P u gives the coordinate expression for u in the basis
P
, i.e.,
T
q
= (
q 1 ,...,
q 6 )
P .
The matrix A
can be created by setting each of its columns
[
A
] j ,
1
j
6, to
B , P
be the basis vector z j
B
written in terms of coordinates of
P
. Thus, in this case, by
Eqs. ( 8.15 ) and ( 8.16 ),
T
[
A
] 1 = (
1
,
0
,
0
,
0
,
0
,
0
)
,
while by Eqs. ( 8.17 ) and ( 8.18 ),
T
[
A
] 4 = (
1
,
0
,
0
,
1
,
0
,
0
)
.
9 Also called a “transition matrix” in some texts, while in others, this term is reserved for Markov chain
processes.
 
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