20
A = T^ block diag. [C, C, ..., C] T
p btocks
then for some n* e Rpr, z(t) can be generated as follows
(2-31)
z(t) = K2n (t)
n*(t) = An*(t) n*(0) = n*
(2-32)
Furthermore, the initial state nQ is unique for any given z(t).
Proof: By the conditions given in Proposition 2.2 each of the p
components of z(t) must satisfy (2-30) and hence the entire vector z(t)
can be expressed uniquely in terms of a pr dimensional initial condition
vector. By assumption, this vector can lie anywhere in pr dimensional
space. From the definition of the A matrix, each element of the vector
*
K2n (t) must also satisfy (2-30). Hence, if we show the initial
k
condition vector representing (t) can be made to lie anywhere in
k
pr dimensional space by appropriate choice of the initial state nQ,
the proof will be complete. Such an initial state can be shown to exist
k
by noting that any state nQ can be observed through the output
k
K2n (t). Consequently, a linearly independent set of initial states
k
must result in a linearly independent set of outputs K2n (t). Thus,
*
since nQ spans pr dimensional space, the initial condition vector
k
defining K2n (t) will span pr dimensional space.
To prove uniqueness, let n* be another initial state such that
z(t) = K2n (t)  (t) = An (t) n (0) = nQ
(2-33)