The companion matrix and its phase portrait;
The matrix exponential: initial value problems.
[1]
We spent a lot of time studying the second order equation
x" + bx' + kx
=
0
and if
b
and
k
are
nonnegative we interpreted them as the damping
constant and spring constant (divided by the mass).
The companion system is obtained by setting
x'
=
y
y'
=
 kx  by
whose matrix of coefficients is the "companion matrix"
A
=
[ 0 1 ; k b ]
Note that
tr(A) = b ,
det(A) = k, and the characteristic polynomial
of
A
is
lambda^2 + b lambda + k,
i.e., it is the same as the
characteristic polynomial of
the original second order equation.
If
lambda_1
is an eigenvalue of a companion matrix, then to find
an eigenvector we look for
v
such that
[ lambda_1 , 1 ;
* , * ] v
=
0
v = [ 1 ; lambda_1 ]
does nicely. This makes sense:
x = e^{lambda_1 t}
has derivative
x' = lambda_1 e^{lambda_1 t} ,
so
[ x ; x' ]
=
e^{lambda t} [ 1 ; lambda_1 ]
is a solution to the companion system.
QUESTION 1:
What region in the
(tr,det)
plane corresponds to
c > 0,
k > 0?
Ans: the upper left quadrant.
QUESTION 2:
What region in the
(tr,det)
plane corresponds to
overdamping?
Ans: The part of the upper left quadrant which is below the critical
parabola,
where there are stable nodes.
For example
x" + (3/2)x' + (1/2)x = 0
is overdamped. We saw long ago
that
solutions to overdamped equations decay as time increases, can cross the
x = 0
slope).
axis at most once, and can have at most one critical point (zero
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 Winter '08
 Staff
 Linear Algebra, initial condition, e^

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