Linear Algebra, 2020a

426 Chapter Five. Similarity
Proof First recall that for any map the dimension of its range space plus
the dimension of its null space equals the dimension of its domain. So if the
dimensions of the range spaces shrink then the dimensions of the null spaces
must rise. We will do the range space half here and leave the rest for Exercise 14.
We start by showing that the range spaces form a chain. If ⃗w ∈ R(t j+1 ),so
that ⃗w = t j+1 (⃗v) for some ⃗v, then ⃗w = t j ( t(⃗v)).Thus⃗w ∈ R(t j ).
Next we verify the “further” property: in the chain the subsets containments
are proper initially, and then from some power k onward the range spaces
are equal. We first show that if any pair of adjacent range spaces in the
chain are equal R(t k ) = R(t k+1 ) then all subsequent ones are also equal
R(t k+1 )=R(t k+2 ), etc. This holds because t: R(t k+1 ) → R(t k+2 ) is the
same map, with the same domain, as t: R(t k ) → R(t k+1 ) and it therefore has
the same range R(t k+1 )=R(t k+2 ) (it holds for all higher powers by induction).
So if the chain of range spaces ever stops strictly decreasing then from that point
onward it is stable.
We end by showing that the chain must eventually stop decreasing. Each
range space is a subspace of the one before it. For it to be a proper subspace it
must be of strictly lower dimension (see Exercise 12). These spaces are finitedimensional
and so the chain can fall for only finitely many steps. That is, the
power k is at most the dimension of V.
QED
1.5 Example The derivative map a + bx + cx 2 + dx 3 d/dx
↦−→ b + 2cx + 3dx 2 on P 3
has this chain of range spaces.
R(t 0 )=P 3 ⊃ R(t 1 )=P 2 ⊃ R(t 2 )=P 1 ⊃ R(t 3 )=P 0 ⊃ R(t 4 )={⃗0}
All later elements of the chain are the trivial space. It has this chain of null
spaces.
N (t 0 )={⃗0} ⊂ N (t 1 )=P 0 ⊂ N (t 2 )=P 1 ⊂ N (t 3 )=P 2 ⊂ N (t 4 )=P 3
Later elements are the entire space.
1.6 Example Let t: P 2 → P 2 be the map d 0 + d 1 x + d 2 x 2 ↦→ 2d 0 + d 2 x. As the
lemma describes, on iteration the range space shrinks
R(t 0 )=P 2 R(t) ={a 0 + a 1 x | a 0 ,a 1 ∈ C} R(t 2 )={a 0 | a 0 ∈ C}
and then stabilizes, so that R(t 2 )=R(t 3 )=···. The null space grows
N (t 0 )={0} N (t) ={b 1 x | b 1 ∈ C} N (t 2 )={b 1 x + b 2 x 2 | b 1 ,b 2 ∈ C}
and then stabilizes N (t 2 )=N (t 3 )=···.