1MA01: Mathematical Methods Tutorial Sheet
1
Remember, to find the eigenvalues (λ) you use the characteristic equation of a matrix A:
det(A − λI) = 0
and then for each λ solve Av = λv for the eigenvector v.
1. Given the matrix
3 6
1 4
A=
!
.
(1)
Find its eigenvalues and corresponding eigenvectors.
To find the eigenvalues, solve the characteristic equation det(A − λI) =
0. ie
A − λI =
=
3 6
1 4
!
λ 0
0 λ
−
3−λ
6
1
4−λ
!
!
Then
!
0 =
=
=
=
3−λ
6
det
1
4−λ
(3 − λ)(4 − λ) − 6(1)
λ2 − 7λ + 6
(λ − 6)(λ − 1)
and in the usual way λ = 6 and λ = 1, are the eigenvalues.
Calculate first the eigenvector for λ = 6. Solve
0 = (A − λI)v
!
3 6
=
−
1 4
=
1
−3 6
1 −2
!
6 0
0 6
a
b
!
a
b
!
!
Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/˜ryan/123.html
yielding −3a+6b = 0 and a−2b = 0 and these equations mean that any
vector with components satisfying a = 2b is an eigenvector, eg (2,1).
For λ = 1, solve
0 = (A − λI)v
!
3 6
=
−
1 4
=
2 6
1 3
!
1 0
0 1
a
b
!
a
b
!
!
yielding 2a + 6b = 0 and a + 3b = 0 and these equations mean that any
vector with components satisfying a = −3b is an eigenvector, eg (-3,1).
2. Find the eigenvalues and eigenvectors of the matrix
A=
0
1
−2 −3
!
.
(2)
3. A square triangular matrix has all zeros either above or below the
main diagonal. The matrices below are all examples of this. Compute
their determinants in the usual way and notice that for these special
matrices det(A) = a11 a22 . . . ann ie the determinant is just the product
of the entries on the diagonal!


5 0 0

A=
 0 −3 0  ; B =
0 0 4
det(A) =
=
det(B) =
=
det(C) =
=
6 0
2 −1
!


10 5 1

; C=
 0 0 −4  (3)
0 0 6
5[−3(4) − 0] − 0[0(4) − 0(0)] + 0[0(0) − (−3)(0)]
−60
6(−1) − 0(2)
−6
10[0(6) − (−4)(0)] − 5[0(6) − (−4)(0)] + 1[0(0) − 0(0)]
0
4. Consider the matrix
A=
6 0
9 −4
!
(4)
Determine its eigenvalues. What do you notice about the eigenvalues
when compared with the matrix entries?
Compute the eigenvalues by solving the characteristic equation.
0 =
=
=
=
6−λ
0
det
9
−4 − λ
(6 − λ)(−4 − λ) − 0
λ2 − 2λ − 24
(λ − 6)(λ + 4)
!
and so the eigenvalues are λ = 6 and λ = −4. Note that these are
the same numbers that appear on the main diagonal of the matrix in
question.
In general for triangular matrices - square matrices which have all zeroes
either above or below the main diagonal - the eigenvalues are simply
the entries on the main diagonal.
5. Use the Cayley Hamilton theorem to determine A5 given,
3 1
2 4
A=
!
(5)
Cayley-Hamilton says a matrix satisfies its own characteristic equation.
Here the characteristic equation is
det(A − λI) = (3 − λ)(4 − λ) − 2 = λ2 − 7λ + 10 = 0
So, A2 − 7A + 10 = 0.
Now, A3 = AA2 = A(7A − 10) = 7A2 − 10A = 7(7A − 10) − 10A =
39A − 70; A4 = AA3 = A(39A − 70) = 39A2 − 70A = 39(7A − 10) −
70A = 203A − 390; A5 = AA4 = A(203A − 390) = 203A2 − 390A =
203(7A − 10) − 390A = 1031A − 2030.
Then
A
5
= 1031
3 1
2 4
!
−
2030
0
0
2030
!
=
3(1031) − 2030
1(1031)
2(1031)
4(1031) − 2030
=
1063 1031
2062 2094
!
!
6. A Leslie matrix is given by
L=
0.728 1.302
0.52 0.62
!
(6)
Find the long-term (percentage) growth rate and the stable age population.
The growth rate is given by the largest eigenvalue and the stable age
population by its corresponding eigenvector.
The eigenvalues are found by solving
!
0.728 − λ 1.302
0 = det
0.52
0.62 − λ
= (0.728 − λ)(0.62 − λ) − 1.302(0.52)
= λ2 − 1.348λ − 0.22568
Solved by λ1,2 = 0.674 ± 0.8246. The largest eigenvalue is λ1 = 0.674 +
0.8246 = 1.498 ∼ 1.5
The growth rate is then 1 − λ as a percentage ie ∼50%.
The corresponding eigenvalue is found from
0 = (L − λI)v
!
0.728 − 1.498
1.302
=
0.52
0.62 − 1.498
v1
v2
!
Giving in both cases, v1 = 1.69v2 or v1 ≈ 1.7v2 . So the stable age
population is one in this ratio of 1.7 (young) to 1 (adults).