Course 111: Algebra, 13th April 2007
To be handed in at tutorials on April 16th and 17th.
1. Consider the matrix
0
1
−2 −3
A=
!
.
Find the matrix eigenvalues and eigenvectors.
Solution The characteristic equation is λ2 + 3λ + 2 = 0 with solution
λ1 = −1 and λ2 = −2.
For λ = −1, the corresponding eigenvector is v1 = k1 (1, −1)T for any
scalar k1 . For λ = −2, the corresponding eigenvector is v2 = k2 (1, −2)T
for arbitrary scalar k2 .
2. Consider the matrix
4 1
−1 6
A=
!
.
Show by determining the transformation matrix V such that AV = V J
that the Jordan form of the matrix A is
J=
5 1
0 5
!
The characteristic equation is (λ−5)2 = 0 so the eigenvalues are λ(A) =
{5, 5}. The corresponding eigenvector is v1 = (1, 1)T and 1 generalised
eigenvector, v2 , is needed.
To get this solve (A − λI)v2 = v1 to get (0, 1).
The transformation matrix is then
V =
1 0
1 1
!
⇒ V −1 =
and
J =V
as required.
−1
AV =
5 1
0 5
1 0
−1 1
!
!