Section 9.4 Trace-Determinant Plane
Key Terms:
• Eigenvalues
• Trace
• Determinant
• Trace-Determinant pairs
• Classification of equilibria
Previously we developed a formula for the characteristic polynomial p(λ) of a general
real 2 × 2 matrix
a b 
A=

 c d
in terms of the trace and determinant of matrix A.
p(λ ) = det ( A − λI ) = λ 2 − tr ( A ) λ + det ( A )
The eigenvalues of A are the roots of the characteristic polynomial and are given by
T ± T 2 − 4D
λ1 , λ2 =
2
Since the eigenvalues λ1 and λ2 are roots of the characteristic polynomial p(λ) so we have
p(λ) = (λ - λ1)(λ – λ2).
Multiplying out the right side and collecting terms we have
p(λ) = λ2 – (λ1 + λ2 )λ + λ1 λ2 .
Comparing this expression with p(λ) = λ2 – tr(A) λ + det(A) we have that
D = det(A) = λ1 λ2
and
T = tr(A) = λ1 + λ2 .
There is a connection between pairs of eigenvalues and trace-determinant pairs. We
now develop a way to use the trace and determinant of the 2 × 2 matrix A to analyze
the possible equilibrium points.
Since A is a real matrix both the trace and the determinant are real, we can use the
trace-determinant plane to provide visual assistance to our efforts. This is simply
a coordinate plane with axes T and D.
From connection between the entries of matrix A and its eigenvalues determined
by the relationship
T ± T 2 − 4D
λ1 , λ2 =
2
we see that the discriminant T2 – 4D plays an important role. In fact we have
1. two distinct real roots (when T 2 − 4D > 0)
2. two complex conjugate roots (when T 2 − 4D < 0)
3. one real root of multiplicity 2 (when T 2 − 4D = 0)
In the TD-plane the discriminant
T2 – 4D is a parabola through the
origin T = 0, D = 0 so it divides
the TD-plane into two regions:
The graph of T2 – 4D
determines points where
T2 – 4D = 0.
Note that the point T = 0, D = 1
satisfies T2 – 4D < 0, so the
region above the graph of the
parabola is such that T2 – 4D < 0.
In a similar fashion T = 1, D = 0
satisfies T2 – 4D > 0, so the
region below the graph of the
parabola is such that T2 – 4D < 0.
(See the Figure.)
Next we use T and D to indicate information about the eigenvalues of matrix A and hence
obtain a classification regarding the behavior at equilibrium points. We also determine
the corresponding region of the TD plane for the corresponding behavior.
two complex conjugate eigenvalues λ = α ± βi (when T 2 − 4D < 0)
When α = 0 we have a center. When
α < 0 we have a spiral sink. When α <
0 we have a spiral source. These
results follow from the fact that
T ± T 2 − 4D
λ1 , λ2 =
2
So the real part of the complex
number is T/2 hence the sign of T
determines the sign of α.
Center: on D-axis above the parabola.
Spiral Sink; in the second
quadrant above the parabola.
Spiral Source; in the first quadrant
above the parabola.
two distinct real eigenvalues λ1 ≠ λ2 (when T 2 − 4D > 0)
Here we use that D = det(A) = λ1 λ2 and T = tr(A) = λ1 + λ2 .
Both eigenvalues positive
implies D and T are positive
and are below the parabola but
in the first quadrant; nodal
source.
Both eigenvalues negative
implies D > 0 and T < 0 and are
below the parabola but in the
second quadrant; nodal sink.
One eigenvalue negative and
the other is positive implies D <
0 , T nonzero and are below
the T-axis; saddles.
So given
a b 
A=

 c d
we compute the
trace and the
determinant.
Depending on the
values of the
discriminant T2 – 4D,
T and D we can
determine the
behavior around the
equilibrium point.
We need not
determine the
eigenvalues.
−10 7 

 −5 4 
Example: Determine the behavior around (0, 0) for yꞌ = Ay when A = 
T = -6, D = -40 + 35 = -5, T2 – 4D = 36 + 20 > 0  saddle
Example: Determine the behavior around (0, 0) for yꞌ = Ay when
 3 2
A=

 −4 1
T = 4, D = 3 + 8 = 11, T2 – 4D = 16 – 44 < 0  Spiral Source
Example: Do yꞌ = Ay and yꞌ = ATy have the same behavior around (0, 0)? Explain.