# Course 111: Algebra, 20th April 2007 1. Consider the matrix ```Course 111: Algebra, 20th April 2007
To be handed in at tutorials on April 23rd and 24th.
1. Consider the matrix


1 0
1


A =  2 −2 −1 
0 0
1
Write A in its Jordan form, J by determining the matrix V which
transforms A according to J = V −1 AV .
A has characteristic equation
(1 − λ)2 (−2 − λ) = 0
and so λ(A) = {1, 1, −2} are the eigenvalues.
When λ = −2 the corresponding eigenvector is v1 = (0, 1, 0)
When λ = 1 there is one corresponding eigenvector, v2 = (3/2, 1, 0).
Solving Av3 = v2 for the generalised eigenvector v3 gives (0, 0, 0) (not
useful). Instead consider the related expression (A − λI)2 v3 = 0 (as
observed in the notes). For λ = 1 this gives



0 0 0
x



 −6 9 5   y  = (0, 0, 0)
0 0 0
z
yielding −6x + 9y + 5z = 0 and any (x, y, z) satisfying this expression
is a generalised eigenvector. Thus, v3 = (−11/6, 2/3, 1) is appropriate.
Then


0 3/2 −11/6
V =
1
2/3 
 1

0
0
1
and
⇒ V −1
− 23 1 − 17
9

11 
=  23 0
9 
0 0
1


and therefore,
J = V −1 AV
− 23 1 − 17
1 0
1
0 3/2 −11/6
9


11  
2
−2
−1
1
2/3 
=  23 0
 1

9 
0 0
1
0 0
1
0
0
1






−2 0 0


=  0 1 1 
0 0 1
Use the formula in your notes to determine e2J .
Writing J in its block diagonal form
J=
and so
e
2J
=
J1 0
0 J2
!
e2J1
0
2J2
0 e
!
where e2J1 = (e2(−2) ) = (e−4 ) and using the formula in the notes
e
2J2
=e
2(1)
1 t=2
0
1
!
=e
2(1)
1 2
0 1
!
and so
e2J




e−4 0 0
0.0183
0
0


2
2 
0
7.389 14.778 
=  0 e 2e  = 

0 0 e2
0
0
7.389
2. Consider the matrix equation, Ax = b with
A=
1
1
1 1+
!
, 0 &lt; &lt;&lt; 1
(1)
Determine the condition number of the matrix A with respect to the
∞ norm.
||A||∞ = max{2, 2 + } = 2 + and ||A−1 ||∞ = max{(2 + ), 2/} =
(2 + )/. Then
κ∞ = ||A||∞||A−1 ||∞ = (2 + )(2 + )/.
Is the matrix well or poorly conditioned?
The matrix is poorly conditioned since for 0 &lt; &lt;&lt; 1 the condition
number is large.
Given
2
2
b=
!
,
determine a solution, x to the matrix equation Ax = b.
The solution is x = (2, 0)T .
Determine what happens to the solution when
A=
1 1
1 1
!
, 0 &lt; &lt;&lt; 1
and b is defined as above.
The solution is now any x = (x1 , x2 ) with x1 + x2 = 2.
Also, determine the effect of a small change in b on the solution by
considering solutions when b = (2, 2)T and b = (2 + , 2)T and A is as
given in Eqn 1.
For b = (2, 2)T the solution is x = (2, 0)T .
For b = (2 + , 2)T the solution is x = (1, 1)T .
```