Nonexistence of Continuous Linear Transformations over Finite Dimensional Vector Space whose Commutator equals Identity

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Let $\map {CL} X := \map {CL} {X, X}$ be the space of continuous linear transformations.

Let $A, B \in \map {CL} {X, X}$ be continuous linear transformations.

Let $I$ be the identity operator.


Then there is no $A, B$ such that $A \circ B - B \circ A = I$

Proof

Aiming for a contradiction, suppose there is $A, B$ such that $A \circ B - B \circ A = I$.

Lemma

$\forall n \in \N_{> 0} : A \circ B^n - B^n \circ A = n B^{n - 1}$

where

$B^n = \underbrace{B \circ B \circ \ldots \circ B}_{n \text{ times} }$

and $B^0 = I$.

Proof

This will be a proof by mathematical induction.


Basis for the Induction

Let $n = 1$

Then:

\(\ds A \circ B^1 - B^1 \circ A\) \(=\) \(\ds 1 \cdot B^{1 - 1}\)
\(\ds \leadsto \ \ \) \(\ds A \circ B - B \circ A\) \(=\) \(\ds I\)


Induction Hypothesis

This is the induction hypothesis:

$A \circ B^n - B^n \circ A = n B^{n - 1}$

It is to be demonstrated that it follows that:

$A \circ B^{n + 1} - B^{n + 1} \circ A = \paren {n + 1} B^n$


Induction Step

\(\ds A \circ B^{n + 1} - B^{n + 1} \circ A\) \(=\) \(\ds A \circ B^n \circ B - B^n \circ B \circ A\)
\(\ds \) \(=\) \(\ds \paren {n B^{n - 1} + B^n \circ A} \circ B - B^n \circ B \circ A\)
\(\ds \) \(=\) \(\ds n B^n + B^n \circ A \circ B - B^n \circ B \circ A\)
\(\ds \) \(=\) \(\ds n B^n + B^n \circ \paren {A \circ B - B \circ A}\)
\(\ds \) \(=\) \(\ds n B^n + B^n \circ I\)
\(\ds \) \(=\) \(\ds n B^n + B^n\)
\(\ds \) \(=\) \(\ds \paren {n + 1} B^{\paren {n + 1} - 1}\)

$\Box$


Take the supremum operator norm of $A \circ B^n - B^n \circ A = n B^{n - 1}$:

\(\ds \norm {n \paren {B^{n - 1} } }\) \(=\) \(\ds n \norm {B^{n - 1} }\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \norm {A \circ B^{n - 1} \circ B - B \circ B^{n - 1} \circ A}\)
\(\ds \) \(\le\) \(\ds \norm {A \circ B^{n - 1} \circ B} + \norm {B \circ B^{n - 1} \circ A}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(\le\) \(\ds 2 \norm A \norm B \norm {B^{n - 1} }\)

Lemma

$\forall n \in \N_{> 0} : B^{n - 1} \ne \mathbf 0$

Proof

Aiming for a contradiction, suppose $\exists n \in \N_{> 0} : B^{n - 1} = \mathbf 0$

For $n = 1$ we have $B^0 = I \ne \mathbf 0$.

Hence, $n \ne 1$.

Suppose:

$\exists n \in \N_{> 0} : B^{n - 1} \ne \mathbf 0$

If $B^n = \mathbf 0$ then:

\(\ds \mathbf 0\) \(=\) \(\ds A \circ \mathbf 0 - \mathbf 0 \circ A\)
\(\ds \) \(=\) \(\ds A \circ B^n - B^n \circ A\)
\(\ds \) \(=\) \(\ds n B^{n - 1}\)
\(\ds \) \(\ne\) \(\ds \mathbf 0\)

Altogether:

$\forall n \in \N_{> 0} : \paren {B^{n - 1} \ne \mathbf 0} \implies \paren {B^n \ne \mathbf 0}$

This is a contradiction.


$\Box$

Hence:

$\forall n \in \N_{> 0} : \norm {B^{n - 1} } > 0$.

Thus:

$\forall n \in \N_{> 0} : n \le 2 \norm A \norm B$

This holds only if $\norm A \norm B = \infty$.


This page needs the help of a knowledgeable authority.
In particular: "Finite Dimensional Vector Space" in the title added to dodge questions about derivatives as continuous linear transformations. In this proof finiteness is not used explicitely, but is probably hidden behind the finite operator norm. The connection between finite/infinite and bounded/unbounded aspects needs clarification to give a proper name for this entire theorem
If you are knowledgeable in this area, then you can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Help}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


However, the image of $\norm {\, \cdot \,}$ is a subset of $\R$:

$\Img {\norm {\, \cdot \,}} \subseteq \R$

and does not contain the infinity.

This is a contradiction.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Nonexistence_of_Continuous_Linear_Transformations_over_Finite_Dimensional_Vector_Space_whose_Commutator_equals_Identity&oldid=626127"