Linear Algebra, 2020a

434 Chapter Five. Similarity
Finally, take ⃗β 1 and ⃗β 3 such that t(⃗β 1 )=⃗β 2 and t(⃗β 3 )=⃗β 4 .
⎛ ⎞ ⎛ ⎞
0
0
1
0
⃗β 1 =
0
⃗β
⎜ ⎟ 3 =
0
⎜ ⎟
⎝0⎠
⎝1⎠
0
0
Therefore, we have a string basis B = 〈⃗β 1 ,...,⃗β 5 〉 and with respect to that basis
the matrix of t has blocks of subdiagonal 1’s.
⎛
⎞
0 0 0 0 0
1 0 0 0 0
Rep B,B (t) =
0 0 0 0 0
⎜
⎟
⎝0 0 1 0 0⎠
0 0 0 0 0
2.16 Theorem Any nilpotent transformation t is associated with a t-string basis.
While the basis is not unique, the number and the length of the strings is
determined by t.
This illustrates the proof, which describes three kinds of basis vectors (shown
in squares if they are in the null space and in circles if they are not).
❦3 ↦→ 1 ❦ ↦→ ··· ··· ↦→ 1 ❦ ↦→ 1 ↦→ ⃗0
❦3 ↦→ 1 ❦ ↦→ ··· ··· ↦→ 1 ❦ ↦→ 1 ↦→ ⃗0
.
❦3 ↦→ 1 ❦ ↦→ ··· ↦→ 1 ❦ ↦→ 1 ↦→ ⃗0
2 ↦→ ⃗0
.
2 ↦→ ⃗0
Proof Fix a vector space V. We will argue by induction on the index of
nilpotency. If the map t: V → V has index of nilpotency 1 then it is the zero
map and any basis is a string basis ⃗β 1 ↦→ ⃗0, ..., ⃗β n ↦→ ⃗0.
For the inductive step, assume that the theorem holds for any transformation
t: V → V with an index of nilpotency between 1 and k − 1 (with k>1) and
consider the index k case.
Observe that the restriction of t to the range space t: R(t) → R(t) is also
nilpotent, of index k − 1. Apply the inductive hypothesis to get a string basis
for R(t), where the number and length of the strings is determined by t.
B = 〈⃗β 1 ,t(⃗β 1 ),...,t h 1
(⃗β 1 )〉 ⌢ 〈⃗β 2 ,...,t h 2
(⃗β 2 )〉 ⌢···⌢〈⃗β i ,...,t h i
(⃗β i )〉