canonical form of T will contain one Jordan block of length 1 corresponding to
λ
2
= 6. To find a basis
for
K
6
=
E
6
consisting of cycles, it suffices to find any 6eigenvector of T; working in
β
0
coordinates
and solving the matrix equation (
B

6
I
)
x
= 0, we see that we may take
β
2
=
1

3
3
0
7
as our basis for
K
6
.
Thus, the Jordan canonical form of T is
J
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
6
,
and a Jordan canonical basis is
β
=
β
1
∪
β
2
=
1
0
0
0
,
0
1
1
0
,
0
0
0
1
,
1

3
3
0
.
(f)
V
is the vector space of polynomial functions in two real variables
x
and
y
of degree at most
2
, as
defined in Example 4, and
T
is the linear operator on
V
defined by
T
(
f
(
x, y
)
)
=
∂
∂x
f
(
x, y
) +
∂
∂y
f
(
x, y
)
.
8
dots. Thus, the dot diagram for
λ
= 0 must be
•
(T

0 I)
2
v
•
(T

0 I)
w
•
u
•
(T

0 I)
v
•
w
•
v
Proceeding as in part (b), we first find
v
such that T
2
v
6
= 0; we may take
v
=
x
2
, which gives the
cycle
β
1
=
(T

0 I)
2
v ,
(T

0 I)
v , v
=
2
,
2
x, x
2
.
Next, to find a cycle
{
(T

0 I)
w , w
}
of length two, we find
w
such that (T

0 I)
2
w
= 0 but
(T

0 I)
w
6
= 0, taking care to ensure that the resulting cycle is linearly independent of the previously
constructed cycle. One might be tempted to guess
w
=
y
, which will generate a cycle of length 2;
however, the initial vector of the resulting cycle would be 1, which is not linearly independent of the
previous cycle. However, taking
w
=
x
2

y
2
yields the cycle
β
2
=
(T

0 I)
w , w
=
2
x

2
y , x
2

y
2
,
which does the job. Finally, to get our last cycle of length 1 we just need a 0eigenvector of T that
is linearly independent of the previously collected vectors. Examining the matrix
A
(or using Math
54 techniques to find its nullspace), we see that we can take
β
3
=
{
u
}
=
{
2
xy

x
2

y
2
}
.
Putting all of this together, we obtain the Jordan canonical form
J
=
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
and a Jordan canonical basis
β
=
β
1
∪
β
2
∪
β
3
=
{
2
,
2
x, x
2
,
2
x

2
y, x
2

y
2
,
2
xy

x
2

y
2
}
.
9