7—Operators and Matrices 168This provides only one eigenvector, a multiple of
(
1
0
)
. You need two for a basis.
Change this matrix in any convenient way to make the two roots of the characteristic equation
different from each other. For example,
M=
(
1 + 2
0 1
)
The eigenvalue equation is now
(1 +−λ)(1−λ) = 0
and the resulting equations for the eigenvectors are
λ= 1 : v 1 + 2v 2 = 0, 0 = 0 λ= 1 +: 0v 1 + 2v 2 = 0, v 2 = 0
Now you have two distinct eigenvectors,
λ= 1 :
(
1
−/ 2
)
, and λ= 1 +:
(
1
0
)
You see what happens to these vectors as→ 0.
Differential Equations at Critical
Problem4.11was to solve the damped harmonic oscillator for the critical case thatb^2 − 4 km= 0.
m
d^2 x
dt^2
=−kx−b
dx
dt
(7.59)
Write this as a pair of equations, using the velocity as an independent variable.
dx
dt
=vx and
dvx
dt
=−
k
m
x−
b
m
vx
In matrix form, this is a matrix differential equation.
d
dt
(
x
vx
)
=
(
0 1
−k/m −b/m
)(
x
vx
)
This is a linear, constant-coefficient differential equation, but now the constant coefficients are matrices.
Don’t let that slow you down. The reason that an exponential form of solution works is that the
derivative of an exponential is an exponential. Assume such a solution here.
(x
vx
)
=
(
A
B
)
eαt, giving α
(
A
B
)
eαt=
(
0 1
−k/m −b/m
)(
A
B
)
eαt (7.60)
When you divide the equation byeαt, you’re left with an eigenvector equation where the eigenvalue
isα. As usual, to get a non-zero solution set the determinant of the coefficients to zero and the
characteristic equation is
det