Linear Algebra

Section III. Nilpotence 381To produce a string basis, first pick ⃗β 2 and ⃗β 4 from R(t) ∩ N (t)⎛ ⎞ ⎛ ⎞0000⃗β 2 =1⃗β⎜ ⎟ 4 =0⎜ ⎟⎝0⎠⎝0⎠01(other choices are possible, just be sure that {⃗β 2 , ⃗β 4 } is linearly independent).For ⃗β 5 pick a vector from N (t) that is not in the span of {⃗β 2 , ⃗β 4 }.⎛ ⎞1−1⃗β 5 =0⎜ ⎟⎝ 0⎠0Finally, take ⃗β 1 and ⃗β 3 such that t(⃗β 1 ) = ⃗β 2 and t(⃗β 3 ) = ⃗β 4 .⎛ ⎞ ⎛ ⎞0010⃗β 1 =0⃗β⎜ ⎟ 3 =0⎜ ⎟⎝0⎠⎝1⎠00Now, with respect to B = 〈⃗β 1 , . . . , ⃗β 5 〉, the matrix of t is as desired.⎛⎞0 0 0 0 01 0 0 0 0Rep B,B (t) =0 0 0 0 0⎜⎟⎝0 0 1 0 0⎠0 0 0 0 02.14 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 isdetermined by t.This illustrates the argument below, which describes three kinds of basisvectors (we shown them in squares if they are in the null space and in circles ifthey are not).❦3 ↦→ ❦ 1 ↦→ · · · · · · ↦→ ❦ 1 ↦→ 1 ↦→ ⃗0❦3 ↦→ ❦ 1 ↦→ · · · · · · ↦→ ❦ 1 ↦→ 1 ↦→ ⃗0.❦3 ↦→ 1 ❦ ↦→ · · · ↦→ 1 ❦ ↦→ 1 ↦→ ⃗02 ↦→ ⃗0.2 ↦→ ⃗0