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Solution:
The characteristic polynomial
det
±
3
−
λ
2
−
15
−
λ
²
=
λ
2
−
8
λ
+17
has the roots
λ
=4
±
i
,and
±
3
−
(4 +
i
)2
−
11
−
(4 +
i
)
²
~v
=
~
0
has the solution
~v
=
±
2
1+
i
²
. [It actually has the solution
~v
=
±
2
z
(1 +
i
)
z
²
,
but
z
= 1 is the simplest choice.] Writing
=
~u
+
i~w
where
~u
and
~w
have real
components, we have
T
(
~u
)+
iT
(
)=
T
(
~v
)=(4+
i
)
=(4+
i
)(
~u
+
)=4
~u
−
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This note was uploaded on 04/12/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee

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