Linear Algebra, Self-Composition

This subsection is optional, although it is necessary for later material in this section and in the next one.

A linear transformations t : V → V, because it has the same domain and
codomain, can be iterated.[1]
That is, compositions of t with itself such as {t}^{2} = t ∘ t
and {t}^{3} = t ∘ t ∘ t are defined.

Note that this
power notation for the linear transformation functions dovetails with
the notation that we've used earlier for their squared matrix representations
because if {R e p}_{B , B} ( t ) = T then {R e p}_{B , B} ( {t}^{j} ) = {T}^{j}.

Example 1.1

For the derivative map d / d x : {𝒫}_{3} → {𝒫}_{3}
given by

a + b x + c {x}^{2} + d {x}^{3} \overset{d / d x}{⟼} b + 2 c x + 3 d {x}^{2}

the second power is the second derivative

a + b x + c {x}^{2} + d {x}^{3} \overset{{d}^{2} / d {x}^{2}}{⟼} 2 c + 6 d x

the third power is the third derivative

a + b x + c {x}^{2} + d {x}^{3} \overset{{d}^{3} / d {x}^{3}}{⟼} 6 d

and any higher power is the zero map.

Example 1.2

This transformation of the space of {}2 \ × \ 2 matrices

( \begin{matrix}a & b \\ c & d\end{matrix} ) \overset{t}{⟼} ( \begin{matrix}b & a \\ d & 0\end{matrix} )

has this second power

( \begin{matrix}a & b \\ c & d\end{matrix} ) \overset{{t}^{2}}{⟼} ( \begin{matrix}a & b \\ 0 & 0\end{matrix} )

and this third power.

( \begin{matrix}a & b \\ c & d\end{matrix} ) \overset{{t}^{3}}{⟼} ( \begin{matrix}b & a \\ 0 & 0\end{matrix} )

After that, {t}^{4} = {t}^{2} and {t}^{5} = {t}^{3}, etc.

These examples suggest that on iteration more and more zeros appear
until there is a settling down.
The next result makes this precise.

Lemma 1.3

For any transformation t : V → V, the rangespaces of the powers
form a descending chain

V ⊇ ℛ ( t ) ⊇ ℛ ( {t}^{2} ) ⊇ ⋯

and the nullspaces form an ascending chain.

\{ \overset{→}{0} \ \} ⊆ 𝒩 ( t ) ⊆ 𝒩 ( {t}^{2} ) ⊆ ⋯

Further, there is a k such that for powers less than k the subsets are proper (if j < k then ℛ ( {t}^{j} ) ⊃ ℛ ( {t}^{j + 1} ) and 𝒩 ( {t}^{j} ) ⊂ 𝒩 ( {t}^{j + 1} )), while for powers greater than k the sets are equal (if j ≥ k then ℛ ( {t}^{j} ) = ℛ ( {t}^{j + 1} ) and 𝒩 ( {t}^{j} ) = 𝒩 ( {t}^{j + 1} )).

Proof

We will do the rangespace half and leave the rest for
Problem 6.
Recall, however, that for any map the dimension of its rangespace
plus the dimension of its nullspace equals
the dimension of its domain.
So if the rangespaces shrink then the nullspaces must grow.

That the rangespaces form chains is clear
because if \overset{→}{w} ∈ ℛ ( {t}^{j + 1} ), so that
\overset{→}{w} = {t}^{j + 1} ( \overset{→}{v} ),
then \overset{→}{w} = {t}^{j} ( \ t ( \overset{→}{v} ) \ ) and so \overset{→}{w} ∈ ℛ ( {t}^{j} ).
To verify the "further" property,
first observe that if any pair of rangespaces 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 is because if
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} )
(and induction shows that it holds for all higher powers).
So if the chain of rangespaces ever stops being strictly decreasing then it is
stable from that point onward.

But the chain must stop decreasing. Each rangespace is a subspace of the one before it. For it to be a proper subspace it must be of strictly lower dimension (see Problem 4). These spaces are finite-dimensional and so the chain can fall for only finitely-many steps, that is, the power k is at most the dimension of V.

Example 1.4

The derivative map a + b x + c {x}^{2} + d {x}^{3} \overset{d / d x}{⟼} b + 2 c x + 3 d {x}^{2}
of Example 1.1 has this chain of rangespaces

{𝒫}_{3} ⊃ {𝒫}_{2} ⊃ {𝒫}_{1} ⊃ {𝒫}_{0} ⊃ \{ \overset{→}{0} \ \} = \{ \overset{→}{0} \ \} = ⋯

and this chain of nullspaces.

\{ \overset{→}{0} \ \} ⊂ {𝒫}_{0} ⊂ {𝒫}_{1} ⊂ {𝒫}_{2} ⊂ {𝒫}_{3} = {𝒫}_{3} = ⋯

Example 1.5

The transformation π : {ℂ}^{3} → {ℂ}^{3} projecting onto the
first two coordinates

( \begin{matrix}{c}_{1} \\ {c}_{2} \\ {c}_{3}\end{matrix} ) \overset{π}{⟼} ( \begin{matrix}{c}_{1} \\ {c}_{2} \\ 0\end{matrix} )

has {ℂ}^{3} ⊃ ℛ ( π ) = ℛ ( {π}^{2} ) = ⋯ and ${ \overset{→}{0} \ } ⊂ 𝒩 ( π ) = 𝒩 ( {π}^{2} ) = ⋯ $.

Example 1.6

Let t : {𝒫}_{2} → {𝒫}_{2} be the map
{c}_{0} + {c}_{1} x + {c}_{2} {x}^{2} ↦ 2 {c}_{0} + {c}_{2} x .
As the lemma describes, on
iteration the rangespace shrinks

ℛ ( {t}^{0} ) = {𝒫}_{2} \ ℛ ( t ) = \{ a + b x \ | \ a , b ∈ ℂ \} \ ℛ ( {t}^{2} ) = \{ a \ | \ a ∈ ℂ \}

and then stabilizes ℛ ( {t}^{2} ) = ℛ ( {t}^{3} ) = ⋯,
while the nullspace grows

𝒩 ( {t}^{0} ) = \{ 0 \} \ 𝒩 ( t ) = \{ c x \ | \ c ∈ ℂ \} \ 𝒩 ( {t}^{2} ) = \{ c x + d \ | \ c , d ∈ ℂ \}

and then stabilizes 𝒩 ( {t}^{2} ) = 𝒩 ( {t}^{3} ) = ⋯.

This graph illustrates Lemma 1.3.
The horizontal axis gives the power j of a transformation.
The vertical axis gives
the dimension of the rangespace of {t}^{j}
as the distance above zero— and thus also shows the dimension of the
nullspace as the distance below the gray horizontal line,
because the two add to the dimension n of the domain.

As sketched, on iteration
the rank falls and with it the nullity grows
until the two reach a steady state.
This state must be reached by the n-th iterate.
The steady state's distance above zero is the dimension of the
generalized rangespace
and its distance below n is the dimension of
the generalized nullspace.

Definition 1.7

Let t be a transformation on an n-dimensional space. The generalized rangespace (or the closure of the rangespace) is {ℛ}_{∞} ( t ) = ℛ ( {t}^{n} ) The generalized nullspace (or the closure of the nullspace) is {𝒩}_{∞} ( t ) = 𝒩 ( {t}^{n} ).

Exercises

Problem 1

Give the chains of rangespaces and nullspaces for the zero and
identity transformations.

Problem 2

For each map,
give the chain of rangespaces and the chain of nullspaces,
and the generalized rangespace and the
generalized nullspace.

• {t}_{0} : {𝒫}_{2} → {𝒫}_{2},
  a + b x + c {x}^{2} ↦ b + c {x}^{2}
• {t}_{1} : {ℝ}^{2} → {ℝ}^{2},

( \begin{matrix}a \\ b\end{matrix} ) ↦ ( \begin{matrix}0 \\ a\end{matrix} )

• {t}_{2} : {𝒫}_{2} → {𝒫}_{2},
  a + b x + c {x}^{2} ↦ b + c x + a {x}^{2}
• {t}_{3} : {ℝ}^{3} → {ℝ}^{3},

( \begin{matrix}a \\ b \\ c\end{matrix} ) ↦ ( \begin{matrix}a \\ a \\ b\end{matrix} )

Problem 3

Prove that function composition is associative
( t ∘ t ) ∘ t = t ∘ ( t ∘ t )
and so we can write {t}^{3} without specifying a grouping.

Problem 4

Check that a subspace must be of dimension less than or equal to the
dimension of its superspace.
Check that if the subspace is proper (the subspace does not equal the
superspace) then the dimension is strictly less.
(This is used in the proof of Lemma 1.3.)

Problem 5

Prove that the generalized rangespace {ℛ}_{∞} ( t ) is the
entire space, and the generalized nullspace {𝒩}_{∞} ( t ) is trivial,
if the transformation t is nonsingular.
Is this "only if" also?

Problem 6

Verify the nullspace half of Lemma 1.3.

Problem 7

Give an example of a transformation on a three
dimensional space whose range has dimension two.
What is its nullspace?
Iterate your example until the rangespace and nullspace stabilize.

Problem 8

Show that the rangespace and nullspace of a linear transformation
need not be disjoint.
Are they ever disjoint?

Solutions

• ↑ More information on function interation is in the appendix.

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