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Unformatted text preview: 2 MA 36600 LECTURE NOTES: FRIDAY, APRIL 23 Example. Again, consider the 2 × 2 matrix
1
A=
4
1
1 =⇒ tr A = −2 det A = −3 =⇒ disc A = 16 pA (r) = r2 − 2 r − 3 = (r − 3) (r + 1) . Hence the eigenvalues are r1 = 3 and r2 = −1, with corresponding eigenvectors
a12
1
a12
1
(1)
(2)
ξ=
=
and
ξ=
=
.
r1 − a11
2
r2 − a11
−2
This gives the matrices D=
r1
0
0
3
=
r2
0 0
−1 and T=
ξ11
ξ21
ξ12
1
=
ξ22
2
1
.
−2 Repeated Eigenvalues
SetUp. So far we have considered the case of 2 × 2 matrices with distinct eigenvalues. We now consider
the case where disc A = 0.
To gain some intuition, consider the following second order diﬀerential equation with constant coeﬃcients:
a y + b y + c y = 0.
Say that the characteristic equation
a r2 + b r + c = 0
has repeated roots i.e., b − 4 a c = 0. The roots are r1 = r2 = −b/(2 a) so that the general solution of the
diﬀerential equation is
2 y (t) = c1 y1 (t) + c2 y2 (t) in terms of y1 (t) = er1 t ,
y2 (t) = t er2 t . Recall that we can express this diﬀerential equation as a system of ﬁrst order diﬀerential equations:
d
0
1
y
x=
=⇒
x = Ax
in terms of
A=
c
b.
y
−a −a
dt
We see that disc A = b2 − 4 a c /a2 = 0. The general solution to this system is
1 r1 t
t
(1)
(2)
(1)
(2)
x(t) = c1 x (t) + c2 x (t)
in terms of
x (t) =
e,
x (t) =
er1 t .
r1
r1 t + 1
We see that the general solution cannot be found by writing
x(t) = ξ ert
for some constant vector ξ and some constant scalar r.
Example. Consider the following 2 × 2 matrix: We have the characteristic polynomial
r−1
pA (r) = det (r I − A) = det
−1
1
A=
1
−1
.
3
1
2
= (r − 1) (r − 3) + 1 = r2 − 4 r + 4 = (r − 2) .
r−3 Hence we have equal eigenvalues r = 2. To compute the eigenvectors, we set
ξ
0
1
1 ξ1
ξ1 + ξ2
= (r I − A) ξ =
=
ξ= 1
=⇒
ξ2
0
−1 −1 ξ2
−1 (ξ1 + ξ2 )
so that ξ2 = −ξ1 . Up to scalar multiple,
r=2 has the eigenvector
1
ξ=
.
−1 ...
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 Spring '09
 EdrayGoins
 Eigenvectors, Vectors

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