Course 111: Algebra, 27th April 2007
To be handed in at tutorials on April 30th and May 1st.
1. Suppose A is a 4 × 4 matrix with eigenvalues λ1 = 5 and λ2 = 2 with
multiplicity nλ1 = 3 and nλ2 = 1 respectively.
Determine eAt .
First recall we have
eλi t = R(λi ).
(1)
Since A is a 4 × 4 matrix its characteristic equation is of degree 4 and
so R(λ) is of degree 3 or less. Then write
R(λ) = α0 + α1 λ + α2 λ2 + α3 λ3 .
and we need the αi .
To make the algebra a little easier observe that if λ is an eigenvalue of
the matrix A then λ is also an eigenvalue of the matrix At where t is a
scalar.
Then for an eigenvalue with multiplicity 3 we can write the equations
e λ = α 0 + α 1 λ + α 2 λ2 + α 3 λ3 ,
deλ
= eλ = α1 + 2α2 λ + 3α3 λ2
dλ
d 2 eλ
= eλ = 2α2 + 6α3 λ
dλ2
So we have four equations to solve:
e5
e5
e5
e2
=
=
=
=
α0 + 5α1 + 25α2 + 125α3 ,
α1 + 10α2 + 75α3
2α2 + 30α3
α0 + 2α1 + 4α2 + 8α3
Giving, 27α3 = (21e5 − 9e2 ), α0 = 9e2 − 98α3 , α1 = −4e5 + 75α3 and
α2 = 1/2(e5 − 30α3 ).
So all the αi can be determined and then
eAt = α0 I + α1 A + α2 A2 + α3 A3 .
(2)
2. Consider the matrix
1 2
3 4
A=
!
(3)
Use the Cayley-Hamilton theorem to determine A4 + 2A3 − A2 .
Write P (s) = s4 +2s3 −s2 and reduce the order of the polynomial using
long division. Using the Cayley-Hamilton theorem the remainder, R(s)
is equivalent to P (s).
The characteristic equation of A is
∆(s) = s2 − 5s − 2
Then
s4 + 2s3 − s2
= (s2 + 7s + 36) + (194s + 72)
2
s − 5s − 2
The remainder is 194s + 72 and so
P (A) = 194A + 72 =
266 388
582 848
!
3. Prove that the conjugate transpose of a the matrix product of two
Hermitian matrices is (AB)H = B H AH .
Proof
(AB)Tij = (B T AT )ij
= (bTik aTkj )
= (bTik )(aTkj )
H
= bH
ik akj
= B H AH
as required. Note the subscripts (ij) etc denote entries of the matrix
A or B.