Page 210 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes that is,
df ct d  ct
t
 cf ( )  e (e f ), etc.
dt dt
r
r
r
r
–ct
r
Thus (D – cI) f = 0 if and only if D (e f) = 0. A function g such that D g = 0, i.e., d g/dt = 0, must be
a polynomial function of degree (r – 1) or less:
...
g(t) = b + b t + + b t r–1
0 1 r–1
Thus f satisfies (2) if and only if f has the form
f(t) = e (b + b t + ... + b t )
ct
r–1
0 1 r–1
ct
r–1
ct
Accordingly, ‘the functions’ e , te , ..., t e span the space of solutions of (2). Since 1, t,..., t are
r–1 ct
j ct
linearly independent functions and the exponential function has no zeros, these r functions t e ,
0  j  r – 1, form a basis for the space of solutions.
Returning to the differential equation (1), which is
p(D)f = 0
r
p = (x – c ) 1 ... (x – c ) k r
1 k
we see that the n functions t e j t , 0  m  r – 1, 1  j  k, form a basis for the space of solutions
m c
j
to (1). In particular, the space of solutions is finite-dimensional and has dimension equal to the
degree of the polynomial p.
Example 2: Prove that the matrix A
 1 1 1 
 
A   1  1  1
 
 1 1 0 
 
is nilpotent. Find its index of nilpotency.
Proof:
 1 1 1   1 1 1   1 1 0



2
A   1  1  1    1  1  1   1  1 0 
     
  1 1 0   1 1 0     0 0 0 
 
 1 1 0  1 1 1   0 0 0


3
A   1  1 0    1  1  1   0 0 0 
     
 0 0 0  1 1 0   0 0 0
     
2
3
So A = 0. Hence A is nilpotent of the index of nilpotence 3. Notice that A  0. (matrix)
3
Also the characteristic polynomial of A is p(x) = x .
Self Assessment
1. If V is the space of all polynomials of degree less than or equal to n over a field F, prove
that the differentiation operator on V is nilpotent. Show that its characteristic polynomial
n
is x .

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