Linear Algebra

Section III. Nilpotence 365Proof. We will verify the second sentence, which is equivalent to the first. Thefirst clause, that the dimension n of the domain of t n equals the rank of t n plusthe nullity of t n , holds for any transformation and so we need only verify thesecond clause.Assume that ⃗v ∈ R ∞ (t) ∩ N ∞ (t) = R(t n ) ∩ N (t n ), to prove that ⃗v is ⃗0.Because ⃗v is in the nullspace, t n (⃗v) = ⃗0. On the other hand, because R(t n ) =R(t n+1 ), the map t: R ∞ (t) → R ∞ (t) is a dimension-preserving homomorphismand therefore is one-to-one. A composition of one-to-one maps is one-to-one,and so t n : R ∞ (t) → R ∞ (t) is one-to-one. But now — because only ⃗0 is sent bya one-to-one linear map to ⃗0 — the fact that t n (⃗v) = ⃗0 implies that ⃗v = ⃗0. QED2.2 Note Technically we should distinguish the map t: V → V from the mapt: R ∞ (t) → R ∞ (t) because the domains or codomains might differ. The secondone is said to be the restriction ∗ of t to R(t k ). We shall use later a point fromthat proof about the restriction map, namely that it is nonsingular.In contrast to the j = 0 and j = n cases, for intermediate powers the spaceV might not be the direct sum of R(t j ) and N (t j ). The next example showsthat the two can have a nontrivial intersection.2.3 Example Consider the transformation of C 2 defined by this action on theelements of the standard basis.( ( ) ( ( ( )1 n 0 0 n 0 0 0↦−→↦−→ N = Rep0)1 1)0)E2,E 2(n) =1 0The vector(0⃗e 2 =1)is in both the rangespace and nullspace. Another way to depict this map’saction is with a string.⃗e 1 ↦→ ⃗e 2 ↦→ ⃗02.4 Example A map ˆn: C 4 → C 4 whose action on E 4 is given by the string⃗e 1 ↦→ ⃗e 2 ↦→ ⃗e 3 ↦→ ⃗e 4 ↦→ ⃗0has R(ˆn) ∩ N (ˆn) equal to the span [{⃗e 4 }], has R(ˆn 2 ) ∩ N (ˆn 2 ) = [{⃗e 3 , ⃗e 4 }], andhas R(ˆn 3 ) ∩ N (ˆn 3 ) = [{⃗e 4 }]. The matrix representation is all zeros except forsome subdiagonal ones.⎛ ⎞0 0 0 0ˆN = Rep E4 ,E 4(ˆn) = ⎜1 0 0 0⎟⎝0 1 0 0⎠0 0 1 0∗ More information on map restrictions is in the appendix.