Linear Algebra, 2020a

Section III. Nilpotence 429
2.2 Corollary Where t: V → V is a linear transformation, the space is the direct
sum V = R ∞ (t)⊕N ∞ (t). That is, both (1) dim(V) =dim(R ∞ (t))+dim(N ∞ (t))
and (2) R ∞ (t) ∩ N ∞ (t) ={⃗0}.
Proof Let the dimension of V be n. We will verify the second sentence, which
is equivalent to the first. Clause (1) is true because any transformation satisfies
that its rank plus its nullity equals the dimension of the space, and in particular
this holds for the transformation t n .
For clause (2), assume that ⃗v ∈ R ∞ (t) ∩ N ∞ (t) to prove that ⃗v = ⃗0. Because
⃗v is in the generalized null space, t n (⃗v) =⃗0. On the other hand, by the lemma
t: R ∞ (t) → R ∞ (t) is one-to-one and a composition of one-to-one maps is oneto-one,
so t n : R ∞ (t) → R ∞ (t) is one-to-one. Only ⃗0 is sent by a one-to-one
linear map to ⃗0 so the fact that t n (⃗v) =⃗0 implies that ⃗v = ⃗0.
QED
2.3 Remark Technically there is a difference between the map t: V → V and
the map on the subspace t: R ∞ (t) → R ∞ (t) if the generalized range space is
not equal to V, because the domains are different. But the difference is small
because the second is the restriction of the first to R ∞ (t).
For powers between j = 0 and j = n, the space V might not be the direct
sum of R(t j ) and N (t j ). The next example shows that the two can have a
nontrivial intersection.
2.4 Example Consider the transformation of C 2 defined by this action on the
elements of the standard basis.
(
1
0
)
n
↦−→
(
0
1
) (
0
1
)
n
↦−→
(
0
0
)
(
N = Rep E2 ,E 2
(n) =
0 0
1 0
This is a shift map because it shifts the entries down, with the bottom entry
shifting entirely out of the vector.
( ) ( )
x
y
↦→
0
x
)
On the basis, this map’s action gives a string.
( ) ( ) ( )
1 0 0
↦→ ↦→ that is
0 1 0
⃗e 1 ↦→ ⃗e 2 ↦→ ⃗0
This map is a natural way to have a vector in both the range space and null
space; the string depiction shows that this is one such vector.
( )
0
⃗e 2 =
1