±
±
±
LS.3
Complex
and
Repeated
Eigenvalues
1.
Complex
eigenvalues.
In
the
previous
chapter,
we
obtained
the
solutions
to
a
homogeneous
linear
system
with
constant
coeﬃcients
A
x
=
0
under
the
assumption
that
the
roots
of
its
characteristic
equation

A
−
I

=
0
—
i.e.,
the
eigenvalues
of
A
—
were
real
and
distinct
.
In
this
section
we
consider
what
to
do
if
there
are
complex
eigenvalues.
Since
the
chan
racteristic
equation
has
real
coeﬃcients,
its
complex
roots
must
occur
in
conjugate
pairs:
¯
=
a
+
bi,
=
a
−
bi
.
Let’s
start
with
the
eigenvalue
a
+
bi
.
According
to
the
solution
method
described
in
Chapter
LS.2,
the
next
step
would
be
to
Fnd
the
corresponding
eigenvector
λ±
,
by
solving
the
equations
(LS.2,
(27))
(
a
−
)
a
1
+
ba
2
= 0
ca
1
+
(
d
−
)
a
2
= 0
for
its
components
a
1
and
a
2
.
Since
is
complex,
the
a
i
will
also
be
complex,
and
therefore
the
eigenvector
λ±
corresponding
to
will
have
complex
components.
Putting
together
the
eigenvalue
and
eigenvector
gives
us
formally
the
complex
solution
λ
(1)
x
=
±
e
(
a
+
bi
)
t
.
Naturally,
we
want
real
solutions
to
the
system,
since
it
was
real
to
start
with.
To
get
them,
the
following
theorem
tells
us
to
just
take
the
real
and
imaginary
parts
of
(1).
Theorem
3.1
Given
a
system
x
=
A
x
,
where
A
is
a
real
matrix.
If
x
=
x
1
+
i
x
2
is
a
complex
solution,
then
its
real
and
imaginary
parts
x
1
,
x
2
are
also
solutions
to
the
system.
Proof.
Since
x
1
+
i
x
2
is
a
solution,
we
have
(
x
+
i
x
2
)
=
A
(
x
+
i
x
2
)
;
equating
real
and
imaginary
parts
of
this
equation,
1
=
A
x
1
,
x
2
=
A
x
2
,
x
which
shows
that
the
real
vectors
x
1
and
x
2
are
solutions
to
x
=
A
x
.
Example
1.
±ind
the
corresponding
two
real
solutions
to
x
=
A
x
if
a
complex
eigenvalue
and
corresponding
eigenvector
are
i
λ
=
−
1
+
2
i
,
±
=
.
2
−
2
i
Solution.
±irst
write
λ±
in
terms
of
its
real
and
imaginary
parts:
0
1
λ±
=
+
i
.
2
−
2
13
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18.03
NOTES:
LS.
LINEAR
SYSTEMS
The
corresponding
complex
solution
x
=
λ
e
t
to
the
system
can
then
be
written
±
±±
1
²
³
0
x
=
+
i
e
−
t
cos
2
t
+
i
sin
2
t
,
2
−
2
so
that
we
get
respectively
for
the
real
and
imaginary
parts
of
x
±
±
±
±
0
1
−
sin
2
t
−
t
x
1
=
e
cos
2
t
−
i
sin
2
t
=
e
−
t
2
−
2
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 Fall '09
 vogan
 Linear Algebra, Complex number, real solutions

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