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18.06
 Spring
2005
 Problem
Set
8
Solution
to
the
Challenge
Problem
Challenge
Problem:
Consider
the
3
×
3
matrix
a
b
c
A
=
⎞
1
d
e
�
0
1
f
Determine
the
entries
a,
b,
c,
d,
e,
f
so
that:
•
the
top
left
1
×
1
block
is
a
matrix
with
eigenvalue
2;
•
the
top
left
2
×
2
block
is
a
matrix
with
eigenvalues
3
and
3;
•
the
top
left
3
×
3
block
is
a
matrix
with
eigenvalues
0,
1
and
2.
Solution.
Let
A
i
denote
the
top
left
i
×
i
block
of
A
.
The
matrix
A
1
is
the
matrix
(
a
).
Since
a
is
the
only
eigenvalue
of
this
matrix,
we
conclude
that
a
=
2.
We
now
move
on
to
determining
the
entries
of
the
matrix
A
2
,
the
top
left
2
×
2
block
of
A
:
⎠
⎛
2
b
A
2
=
1
d
Since
the
sum
of
the
eigenvalues
of
A
2
is
0
by
hypothesis,
and
it
is
also
equal
to
the
trace
of
A
2
,
we
obtain
that
2
+
d
=
0,
or
d
=
−
2.
Moreover,
the
product
of
the
eigenvalues
of
A
2
is
9
by
hypothesis,
and
it
is
equal
to
the
determinant
of
A
2
.
Thus
we
have
−
9
=
2
d
−
b
=
−
4
−
b
and
we
deduce
that
b
=
5
and
therefore
⎠
⎛
2
5
A
2
=
1
−
2
Finally,
consider
A
=
A
3
.
Again,
the
sum
of
the
eigenvalues
of
A
is
1
and
it
is
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 Fall '08
 SOYSAL
 Linear Algebra, Characteristic polynomial, c − 2e

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