Linear Algebra, 2020a

Section III. Nilpotence 437
2.19 Example This matrix
⎛
⎞
0 0 0 0 0
1 0 0 0 0
−1 1 1 −1 1
⎜
⎟
⎝ 0 1 0 0 0 ⎠
1 0 −1 1 −1
is nilpotent, of index 3.
power p N p N (N p )
⎛
⎞ ⎛ ⎞
0 0 0 0 0 0
1 0 0 0 0
0
1
−1 1 1 −1 1
{
u − v
⎜
⎟ ⎜ ⎟
⎝ 0 1 0 0 0 ⎠ ⎝ u ⎠
1 0 −1 1 −1 v
⎛
⎞ ⎛ ⎞
0 0 0 0 0 0
0 0 0 0 0
y
2
1 0 0 0 0
{
z
⎜
⎟ ⎜ ⎟
⎝
⎠ ⎝ ⎠
1 0 0 0 0
0 0 0 0 0
3 –zero matrix– C 5
u
v
| u, v ∈ C}
| y, z, u, v ∈ C}
The table tells us this about any string basis: the null space after one map
application has dimension two so two basis vectors map directly to zero, the
null space after the second application has dimension four so two additional
basis vectors map to zero by the second iteration, and the null space after three
applications is of dimension five so the remaining one basis vector maps to zero
in three hops.
⃗β 1 ↦→ ⃗β 2 ↦→ ⃗β 3 ↦→ ⃗0
⃗β 4 ↦→ ⃗β 5 ↦→ ⃗0
To produce such a basis, first pick two vectors from N (n) that form a linearly
independent set.
⎛ ⎞ ⎛ ⎞
0
0
0
0
⃗β 3 =
1
⃗β
⎜ ⎟ 5 =
0
⎜ ⎟
⎝1⎠
⎝1⎠
0
1