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

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
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:

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

Lemma

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

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

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:

Altogether:


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

This is a contradiction.

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 $\norm A \norm B = \infty$

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.