PHZ 3113 Fall 2010
Homework #8, Due Friday, November 19
1. Let
A=
1 α2
α α3
.
What are the eigenvalues of A? What are the eigenvectors of A? For what α are the
eigenvectors orthogonal?
The eigenvalues are the roots of
det(A − λI) = det
these are
1−λ
α2
3
α
α −λ
λ = α3 ,
= λ2 − (1 + α3 )λ = 0;
λ = 0.
They also can be found from the determinant and trace of A,
det A = 0 = λ1 λ2 ,
tr A = 1 + α3 = λ1 + λ2 .
The determinant says one eigenvalue vanishes, and the trace then says the other is 1 + α3 .
The eigenvectors follow from Av = λv,
v1
v1
1 α2
=λ
.
v2
v2
α α3
For λ = 0, this says v1 + α2 v2 = 0 and αv1 + α3 v2 = 0; both lead to v1 = −α2 v2 , or
v0 = (−α2 , 1).
For λ = 1 + α3, v1 + α2 v2 = (1 + α3)v1 and αv1 + α3 v2 = (1 + α3)v2 ; both lead to v2 = αv1 ,
v1+α3 = (1, α).
Other normalizations are possible.
The vectors are orthogonal when v0 · v1+α3 = −α2 + α = 0, so there are two possibilities,
α = 0,
α = 1.
These values also make the original matrix A real symmetric, “self-adjoint”, and so by
theorem it should have real eigenvalues!
2. Let
0
0
1
P =
0 −1
1
0
0
0
!
0 1 0
1 0 1
0 1 0
1
Q= √
2
,
!
.
(a) Which of P , Q has real eigenvalues? (Cite a reason that does not involve computing the
eigenvalues.) Which is invertible? Which is a rotation? For the one that is invertible, find
the inverse.
The matrix Q is real symmetric, P is not; and so if one has real eigenvalues it is Q.
The determinants det P = 1 and det Q = 0 say P is invertible and Q is not.
The product P T P = P P T = I (the identity) says P is orthogonal: P is a rotation. The
product QT Q = QQT = Q2 is not the identity [see Problem 2(a) of Homework 7]; Q is not
a rotation.
The inverse of P is
P −1 = P T =
0
0
−1
0
1
0
1
0
0
!
.
(b) Find the eigenvalues and eigenvectors for the matrix that has real eigenvalues.
From

√
−λ
1/
2
0
√
√
det(Q − λI) = det  1/ 2 −λ 1/ 2  = λ − λ3 = λ(1 − λ)(1 + λ) = 0,
√
0
1/ 2 −λ

we obtain the eigenvalues λ = 0, ±1.
The eigenvectors are
v0 = (1, 0, −1),
Other normalizations are possible.
√
v1 = (1, 2, 1),
√
v−1 = (1, − 2, 1).
3. Three equal masses m are connected to each other along one line by springs with spring
constant k and to walls by springs with spring constant K, such that the departures of the
three masses from their equilibrium points obey
m ẍ1 = −Kx1 − k(x1 − x2 ),
m ẍ2 = −k[(x2 − x1 ) + (x2 − x3 )],
m ẍ3 = −k(x3 − x2 ) − Kx3 .
(a) Assume there are oscillatory solutions, xi = ai cos ωt. Show that this leads to a set of
algebraic (not differential) equations. Write the left- and right-hand sides of the algebraic
equations in matrix form.
Since the cos ωt time dependence gives ẍi = −ω 2 xi , we have
−mω 2 x1 = −Kx1 − k(x1 − x2 ),
−mω 2 x2 = −k[(x2 − x1 ) + (x2 − x3 )],
−mω 2 x3 = −k(x3 − x2 ) − Kx3 .
In matrix form this is
−mω 2
x1
x2
x3
!
=
−K − k
k
0
k
−2k
k
0
k
−K − k
!
x1
x2
x3
!
,
or with everything in a single matrix,
mω 2 − K − k
k
0
k
mω 2 − 2k
k
0
k
mω 2 − K − k
!
x1
x2
x3
!
= 0.
(b) What are the characteristic frequencies of oscillations? What are the oscillation mode
patterns?
As usual, if the matrix has an inverse, multiplying by the inverse on the left gives x = 0 as
the only solution; for a nontrivial solution we need
det = m3 ω 6 − 2(2k + K)m2 ω 4 + (3k 2 + 6kK + K 2 )mω 2 − 2kK(k + K)
= (mω 2 − k − K)[m2 ω 4 − (3k + K)mω 2 + 2kK] = 0.
The roots are the characteristic oscillation frequencies,
√
k+K
3k + K ± 9k 2 − 2kK + K 2
2
2
,
ω =
.
ω =
m
2m
For ω 2 = (k + K)/m, the spring displacements are in the pattern (x1 , x2 , x3 ) = (1, 0, −1):
the center mass is fixed in place, and the outer two masses oscillate with each feeling a force
from one spring on each side.
2 , the pattern is
For ω 2 = ω±
(x1 , x2 , x3 ) = 1,
−k + K ±
√
9k 2 − 2kK + K 2 ,1 .
2k
(c) Examine your frequencies and modes for K → 0. Comment. Examine your frequencies
and modes for k → 0. Comment.
For K → 0, the frequencies are
ω 2 = 0,
ω2 =
k
,
m
ω2 =
3k
,
m
with modes
(1, 0, −1),
(1, 1, 1),
(1, −2, 1).
With no connection to the walls, when the three masses move together there are no springs
compressed or extended; this is a translation mode that has zero frequency. When the central
mass is fixed, the end masses move under the force of a single spring, the Physics 1 case with
ω 2 = k/m. When the outer masses move one way and the central mass moves twice as far
the other way (leaving the center of mass fixed), the net restoring force on an outer mass is
3k times its displacement.
For k → 0, the frequencies are
ω 2 = 0,
ω2 =
K
,
m
ω2 =
K
,
m
with what might appear to be modes
(0, 1, 0),
(1, 0, −1),
(1, 0, 1).
However, the last two are degenerate, and the basis might as well be
(1, 0, 0),
(0, 0, 1).
The three masses each do their own thing, independent of the other two; two are connected to
springs K and oscillate at the Physics 1 frequency ω 2 = K/m, and the third isn’t connected
to anything at all (ω 2 = 0).