Page 129 - Matrices theory and applications

6. Matrices with Entries in a Principal Ideal Domain; Jordan Reduction
112
Let K be a field and A ∈ M n (K). Let X, Y ∈ K
T
T
that X AY = 0. We normalize by X AY =1 and define
T
Show that in the factorization

0
I r
PAQ = B := A − (AY )(X A). n be vectors such
P, Q ∈ GL n (K),
,
0 0 n−r
one can choose Y as the first column of Q and X T as the first row of
P. Deduce that rk B =rk A − 1.
T
More generally, show that if X, Y ∈ M n×m (K), X AY ∈ GL m (K),
and if
T
T
B := A − (AY )(X AY ) −1 (X A),
then rk B =rk A − m.
If A ∈ Sym (IR)and if A is positive semidefinite, and if X = Y ,
n
show that B is also positive semidefinite.
6. For A ∈ M n (CC), consider the linear differential equation in CC n
dx
= Ax. (6.1)
dt
(a) Let P ∈ GL n (CC)and let t → x(t) be a solution of (6.1). What
is the differential equation satisfied by t → Px(t)?
(b) Let (X −a) m be an elementary divisor of A. Show that for every
at
k =0,... ,m − 1, (6.1) possesses solutions of the form e Q k (t),
where Q k is a complex-valued polynomial map of degree k.
7. Consider the following differential equation of order n in CC:
x (n) (t)+ a 1 x (n−1) (t)+ ··· + a n x(t)=0. (6.2)
n
(a) Define P(X)= X + a 1 X n−1 + ··· + a n and let M be the
companion matrix of P.Let

P(X)= (X − a) n a
a∈A
be the factorization of P into irreducible factors. Compute the
Jordan form of M.
(b) Using either the previous exercise or arguing directly, show that
the set of solutions of (6.2) is spanned by the solutions of the
form
at
t → e R(t), R ∈ CC[X], deg R< n a .
8. Consider a linear recursion of order n in a field K
u m+n + a 1 u m+n−1 + ··· + a n u m =0, m ∈ IN. (6.3)