The trace-determinant plane
Math 33B lecture 20
Ryan Reich
http://www.math.ucla.edu/~ryanr/33b.2.12s
ryanr@math.ucla.edu
MS 5338
23 May 2012
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Outline
1
Eigenvalues and equilibria
2
On the border
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Eigenvalues and equilibria
Trace, determinant, and the characteristic polynomial
The characteristic polynomial of a 2 × 2 matrix A = ac
straightforward to compute, but there is an easier way:
a−λ
b
pA (λ) = det(A − λ) = det
c
d −λ
b
d
may be
!
= (λ − a)(λ − d) − bc = λ2 − (a + d)λ + (ad − bc)
= λ2 − tr(A)λ + det(A) = λ2 − T λ + D.
Therefore, its roots, the eigenvalues of A, can be expressed in
terms of just T and D:
√
T ± T 2 − 4D
λ1 , λ2 =
.
2
Going the other way, T and D can be expressed in terms of the
eigenvalues:
T = λ1 + λ2
Ryan Reich (MS 5338)
D = λ1 λ2 .
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Eigenvalues and equilibria
Phase portrait types
The eigenvalues are real if the discriminant is positive and
complex if it is negative:
λ’s real: T 2 > 4D
λ’s complex: T 2 < 4D.
If they are real, then since D = λ1 λ2 , they have the same sign if
D > 0 and opposite sign if D < 0.
nodal portrait: real, D > 0
saddle portrait: D < 0.
If they are complex, then Re(λ) = T /2 (either λ), so:
spiral source: complex, T > 0
spiral sink: complex, T < 0
Finally, if they are real and nodal, both λ’s have the same sign,
so T has that sign also:
nodal source: real, T > 0
nodal sink: real, T < 0.
Over all, the portrait is a source if T > 0 and a sink if T < 0.
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Eigenvalues and equilibria
The trace-determinant plane
sink
spiral
source
nodal
T
saddle
D
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Eigenvalues and equilibria
Example
Problem
Quickly classify the linear systems ~x 0 = A~x with A equal to:
(1)
(3)
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4 −3
15 −8
!
−2
0
1 −1
!
(2)
(4)
8
5
−10 −7
!
1
4
−1 −3
!
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Eigenvalues and equilibria
Solution
1
4 −3
15 −8
2
8
5
−10 −7
3
−2
0
1 −1
: T = −2 − 1 = −3,
4
1
4
−1 −3
: T = 1 − 3 = −2,
!
: T = 4 − 8 = −4,
!
!
!
D = −32 + 45 = 13,
T 2 − 4D = −36 =⇒ spiral sink.
: T = 8 − 7 = 1,
D = −56 + 50 = −6
=⇒ saddle.
D = 2,
T 2 − 4D = 1 =⇒ nodal sink.
D = −3 + 4 = 1,
T 2 − 4D = 0 =⇒ ???
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Eigenvalues and equilibria
Solution, visualized
sink
spiral
1
source
nodal
3
4
T
saddle
2
D
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On the border
Generic and nongeneric portraits
Almost all systems fall into one of the categories listed above.
These are called generic portraits, and we have described them
last time.
When one of the inequalities is actually an equality, then the
portrait is nongeneric.
There are four kinds of nongeneric portraits:
1
2
3
4
When
When
When
When
T = 0. These are the centers that we saw last time.
D = 0. One of the eigenvalues is zero.
T 2 = 4D. Both eigenvalues are equal.
T = 0 and D = 0. Both eigenvalues are equal to zero.
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On the border
A zero eigenvalue
Problem
Sketch a phase portrait of the system
~x =
0
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1 2
~x .
1 2
!
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On the border
Solution
The determinant is clearly zero, and the trace is 3, so the
eigenvalues are 0 and 3 (since their sum is the trace and their
product is the determinant).
The eigenvectors are:
1 2
~v0 :
1 2
!
−2
2
~v3 :
1 −1
=⇒ ~v0 =
!
=⇒ ~v3 =
2
−1
!
1
1
!
The general solution is therefore
3t
~x = C1~v0 + C2 e ~v3 = C1
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2
1
+ C2 e 3t
.
−1
1
!
!
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On the border
Portrait
4
~v3
2
0
−2
−4
−4
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~v0
−3
−2
−1
0
1
2
3
4
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On the border
Repeated eigenvalues
Problem
Draw a phase portrait of the system
~x =
0
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1
4
~x .
−1 −3
!
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On the border
Solution
We already found that T = −2, D = 1, and T 2 − 4D = 0, so there
is only one eigenvalue, which is half of T : λ = −1.
We therefore need both an eigenvector and a generalized
eigenvector:
~v−1
w
~ −1
2
4
:
~v = 0 =⇒ ~v−1 =
−1 −2 −1
!
2
4
:
−1 −2
!
a
b
!
= ~v−1
=⇒ a + 2b = 1,
w
~ −1 =
2
−1
!
!
−1
1
The solution is:
~x = (C1 + C2 t)e
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2
−1
+ C2 e −t
.
−1
1
!
−t
!
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On the border
Portrait
4
2
0
w
~ −1
−2
−4
−4
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−3
−2
−1
0
1
2
3
4
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On the border
Zero eigenvalues
Problem
Draw a phase portrait for the system
~x =
0
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2
1
~x .
−4 −2
!
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On the border
Solution
The trace is zero and the determinant is zero, so the characteristic
polynomial is λ2 and both eigenvalues are also zero.
We again need an eigenvector and a generalized eigenvector.
2
1
~v0 :
−4 −2
!
2
1
w
~0 :
−4 −2
!
=⇒ ~v0 =
a
b
!
−1
2
!
= ~v0
=⇒ 2a + b = −1,
w
~0 =
!
−1
.
1
The solution is
!
!
−1
−1
~x = (C1 + C2 t)
+ C2
.
2
1
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On the border
Portrait
4
w
~ −1
2
0
−2
−4
−4
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−3
−2
−1
0
1
2
~v−1
3
4
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