Linear Algebra

Section IV. Jordan Form 395
2.15 Example A bit of computing with
⎛
−1
⎜ 0
T = ⎜ 0
⎝ 3
4
3
−4
−9
0
0
−1
−4
0
0
0
2
⎞
0
0 ⎟
0 ⎟
−1⎠
1 5 4 1 4
shows that its characteristic polynomial is (x − 3) 3 (x +1) 2 . This table
power p (T − 3I) p N ((t − 3) p 1
⎛
−4
⎜ 0
⎜ 0
⎜
⎝ 3
4 0 0
0 0 0
−4 −4 0
−9 −4 −1
) nullity
⎞ ⎛ ⎞
0 −(u + v)/2
⎟ ⎜ ⎟
0 ⎟ ⎜−(u
+ v)/2⎟
⎟ ⎜ ⎟ �
0 ⎟ { ⎜ (u + v)/2 ⎟ �
⎟ u, v ∈ C} 2
⎟ ⎜ ⎟
−1⎠
⎝ u ⎠
2
1
⎛
16
⎜ 0
⎜ 0
⎜
⎝−16
5 4
−16
0
16
32
1
0
0
16
16
0
0
0
0
1
⎞
0
⎟
0 ⎟
0 ⎟
0⎠
v
⎛ ⎞
−z
⎜ ⎟
⎜−z⎟
⎜ ⎟ �
{ ⎜ z ⎟ �
⎟ z,u,v ∈ C}
⎜ ⎟
⎝ u ⎠
3
3
0
⎛
−64
⎜ 0
⎜ 0
⎜
⎝ 64
−16
64
0
−64
−128
−16
0
0
−64
−64
0
0
0
0
0
0
⎞
0
⎟
0 ⎟
0 ⎟
0⎠
v
–same– —
0 64 64 0 0
shows that the restriction of t − 3toN∞(t − 3) acts on a string basis via the
two strings � β1 ↦→ � β2 ↦→ �0 and � β3 ↦→ �0.
A similar calculation for the other eigenvalue
power p (T +1I) p N ((t +1) p 1
⎛
0
⎜
⎜0
⎜
⎜0
⎜
⎝3
4 0 0
4 0 0
−4 0 0
−9 −4 3
⎞
0
⎟
0 ⎟
0 ⎟
−1⎠
)
⎛ ⎞
−(u + v)
⎜ ⎟
⎜ 0 ⎟
⎜ ⎟ �
{ ⎜ −v ⎟ �
⎟ u, v ∈ C}
⎜ ⎟
⎝ u ⎠
nullity
2
2
1
⎛
0
⎜
⎜0
⎜
⎜0
⎜
⎝8
5
16
16
−16
−40
4 1
0 0
0 0
0 0
−16 8
5
⎞
0
⎟
0 ⎟
0 ⎟
−8⎠
v
–same– —
8 24 16 8 24