C3 THEORY OF COMPUTATIONAL DYNAMICS
EXAMPLE SHEET 5
Stars indicate level of difficulty.
1.* Let f : [0,1] → [0,1] be the tent map given by
f(x)
2x

2(1 − x )
=
0≤x ≤
1
2
1
2
≤ x ≤1
(recall C2 Example Sheet 2, Q3). Suppose that multiple shooting is used to find period 2 orbits of
f, starting with initial conditions y 1 and y 2. Show that the method converges in one iteration.
Which period-2 points does it find in each of the following situations:
2.
1
i)
0 < y1 < y2 < 2.
ii)
0 < y1 < 2 < y2 < 1.
iii)
1
2
1
< y1 < y2 < 1.
C3 May 1995, Q5.
3.** Let A be a n×n matrix and denote its matrix of co-factors by B (as in C3 Example Sheet 4, Q4).
Recall that AB† = det A. This provides an efficient formula for the computation of B† when A is
invertible, ie. B † = (det A) A-1 and both det A and A-1 can be easily computed from the same LR
decomposition of A. However, when A is singular this approach fails. Assuming that A has a
simple 0 eigenvalue it is however still possible to give a formula for B†. This question leads you
through its derivation.
Let v and u† be respectively the right and left eigenvectors of A with eigenvalue 0, so that Av = 0
and u†A = 0. Deduce that if bi† is the ith row of B, then b i = αiv for some αi∈R. Similarly deduce
that if b (j) is the jth column of B then (b(j))† = βj u† for some βj∈R (hint: B†A = 0). By using bi = αiv for
all i, show that also b(j) = ( α1vj, …, αnvj)†, where v = (v1, …, vn)† and hence that αivj = βjui for all i,j.
Hence show that there is a constant γ∈R such that αi = γui and so bij = γuivj, where b ij is the ijth
entry of B. Thus B = γu †v. The constant γ depends on the normalization of u and v (recall that
these are only determined up to a scalar multiple). It can be evaluated by picking an i and j such
that ui ≠ 0, vj ≠ 0 and computing the relevant co-factor by LR decomposition.
4.
Let A be a n×n matrix and as in lectures define
M
=
 A b
 †

 c 0
where c is chosen such that Av = 0, v ≠ 0 implies that c.v ≠ 0. Let V = { v : c.v = 0 } and choose any b
≠ 0 such that b∉AV, (ie b ≠ Av for any v∈V). Deduce that if c.v = 0, v ≠ 0 then αb ≠ Av for any α∈R.
Hence show that if
 A b  v 
 †
  
 c 0  α 
=
 0
 
 0
then v = 0, α = 0, and hence M is not singular.
C3 Exercise Sheet 5
2
5.* Let A be a n×n matrix. Recall that in lectures we described a method for finding zeros of det A by
constructing a function ψ(A) which has the same zeros as det A. This is given as the solution of
 A b  v 
 †
  
 c 0 ψ 
=
 0
 
1 
In lectures we showed that ψ(A) = 0 implies det A = 0. Here we prove the converse. So suppose
that A is singular and choose some u ≠ 0 such that Au = 0. Show that we can choose an α∈R, α ≠
0, such that c†(v + αu) ≠ 0, and use this to construct a vector v' of the form v' = γ(v + αu) such that
c†v' = 1. Show that Av' = γAv, and hence
 A b  v ′ 
 †
  
 c 0  γψ 
=
 0
 
1 
Deduce that v = v' and ψ = γψ and hence show that ψ (A) ≠ 0 leads to the contradiction α u = 0.
Hence we must have ψ(A) = 0, as required.
6.
C3 May 1994, Q6.
7.
C3 May 1995, Q6.
Do not attempt C3 May 1994 Q5 or C3 May 1996 Q7 as the necessary material has not been covered
in lectures this year.