Linear Algebra

384 Chapter Five. Similarity2.17 Example The matrix⎛⎞0 0 0 0 01 0 0 0 0−1 1 1 −1 1⎜⎟⎝ 0 1 0 0 0⎠1 0 −1 1 −1is nilpotent.power p N p N (N p )⎛⎞ ⎛ ⎞0 0 0 0 0 01 0 0 0 001−1 1 1 −1 1{∣ u − vu, v ∈ C}⎜⎟ ⎜ ⎟⎝ 0 1 0 0 0⎠⎝ u ⎠1 0 −1 1 −1 v⎛⎞ ⎛ ⎞0 0 0 0 0 00 0 0 0 0y21 0 0 0 0{∣ zy, z, u, v ∈ C}⎜⎟ ⎜ ⎟⎝1 0 0 0 0⎠⎝u⎠0 0 0 0 0 v3 –zero matrix– C 5That table shows that any string basis must satisfy: the null space after onemap application has dimension two so two basis vectors map directly to zero,the null space after the second application has dimension four so two additionalbasis vectors map to zero by the second iteration, and the null space after threeapplications is of dimension five so the final basis vector maps to zero in threehops.⃗β 1 ↦→ ⃗β 2 ↦→ ⃗β 3 ↦→ ⃗0⃗β 4 ↦→ ⃗β 5 ↦→ ⃗0To produce such a basis, first pick two independent vectors from N (n)⎛ ⎞ ⎛ ⎞0000⃗β 3 =1⃗β⎜ ⎟ 5 =0⎜ ⎟⎝1⎠⎝1⎠01then add ⃗β 2 , ⃗β 4 ∈ N (n 2 ) such that n(⃗β 2 ) = ⃗β 3 and n(⃗β 4 ) = ⃗β 5⎛ ⎞ ⎛ ⎞0011⃗β 2 =0⃗β⎜ ⎟ 4 =0⎜ ⎟⎝0⎠⎝1⎠00