Linear Algebra

Section III. Nilpotence 3832.16 Example The matrix( )1 −1M =1 −1has an index of nilpotency of two, as this calculation shows.power p M p N (M p )( ) ( )1 M =1 −1{x ∣∣x ∈ C}1 −1 x( )2 M 2 =0 0C 20 0The calculation also describes how a map m represented by M must act on anystring basis. With one map application the null space has dimension one and soone vector of the basis maps to zero. On a second application, the null space hasdimension two and so the other basis vector maps to zero. Thus, the action ofthe linear transformation is ⃗β 1 ↦→ ⃗β 2 ↦→ ⃗0 and the canonical form of the matrixis this. ( )0 01 0We can exhibit such a m-string basis and the change of basis matriceswitnessing the matrix similarity. For the basis, take M to represent m withrespect to the standard bases. (We could take M to be a representative withrespect to some other basis but the standard basis is convenient.) Pick a⃗β 2 ∈ N (m) and also pick a ⃗β 1 so that m(⃗β 1 ) = ⃗β 2 .Recall the similarity diagram.( )1⃗β 2 =1C 2 wrt E 2⏐id↓PC 2 wrt Bm−−−−→( )1⃗β 1 =0MC 2 wrt E 2⏐id↓Pm−−−−→ C 2 wrt BThe canonical form equals Rep B,B (m) = PMP −1 , where( )( )P −1 1 1= Rep B,E2 (id) =P = (P −1 ) −1 1 −1=0 10 1and the verification of the matrix calculation is routine.( ) ( ) ( ) ( )1 −1 1 −1 1 1 0 0=0 1 1 −1 0 1 1 0