Linear Algebra

Section III. Nilpotence 375
We can exhibit such a m-string basis and the change of basis matrices witnessing
the matrix similarity. For the basis, take M to represent m with respect
to the standard bases, pick a � β2 ∈ N (m) and also pick a � β1 so that m( � β1) = � β2.
�β2 =
� �
1
1
�β1 =
� �
1
0
(If we take M to be a representative with respect to some nonstandard bases
then this picking step is just more messy.) Recall the similarity diagram.
C2 w.r.t. E2
⏐
id
m
−−−−→
M
C2 w.r.t. E2
⏐
�P id�P
C 2 w.r.t. B
m
−−−−→ C 2 w.r.t. B
The canonical form equals Rep B,B(m) =PMP −1 , where
P −1 =Rep B,E2 (id) =
� �
1 1
0 1
P =(P −1 ) −1 =
and the verification of the matrix calculation is routine.
�
1
0
��
−1 1
1 1
��
−1 1
−1 0
� �
1 0
=
1 1
�
0
0
2.16 Example The matrix
⎛
0
⎜ 1
⎜
⎜−1
⎝ 0
0
0
1
1
0
0
1
0
0
0
−1
0
⎞
0
0 ⎟
1 ⎟
0 ⎠
1 0 −1 1 −1
is nilpotent. These calculations show the nullspaces growing.
� �
1 −1
0 1
p N p N (N p 1
⎛
0
⎜ 1
⎜
⎜−1
⎝ 0
0
0
1
1
0
0
1
0
0
0
−1
0
⎞
0
0 ⎟
1 ⎟
0 ⎠
)
1 0 −1 1 −1
{
⎛ ⎞
0
⎜ 0 ⎟ �
⎜
⎜u
− v⎟
�
⎟ u, v ∈ C}
⎝ u ⎠
2
⎛
0 0 0
⎜
⎜0
0 0
⎜
⎜1
0 0
⎝1
0 0
0
0
0
0
⎞
0
0 ⎟
0 ⎟
0⎠
⎛ ⎞v
0
⎜
⎜y
⎟ �
{ ⎜
⎜z⎟
�
⎟ y, z, u, v ∈ C}
⎝u⎠
0 0 0 0 0
v
3 –zero matrix– C5