# Wielandt theorem for unital normed algebras

Theorem. (Wielandt (1949)) Let $A$ be a normed unital algebra (with unit $e$). If $x,y\in A$ then $xy-yx\ne e$.

###### Proof.

Assume there are $x,y\in A$ such that $xy-yx=e$. Then for all $n\in \mathbb{N}$ we have

${x}^{n}y-y{x}^{n}$ | $=n{x}^{n-1}\ne 0$ |

We prove this by induction^{} over $n\in \mathbb{N}$. It holds for $n=1$ by assumption^{}. Assume it is valid for $n\in \mathbb{N}$. Then ${x}^{n}\ne 0$ and

${x}^{n+1}y-y{x}^{n+1}$ | $={x}^{n}(xy-yx)+({x}^{n}y-y{x}^{n})x$ | ||

$={x}^{n}e+n{x}^{n-1}x={x}^{n}e+n{x}^{n}=(n+e){x}^{n}$ |

From this identity it follows that

$n\parallel {x}^{n-1}\parallel $ | $=\parallel {x}^{n}y-y{x}^{n}\parallel \le 2\parallel {x}^{n}\parallel \parallel y\parallel \le 2\parallel {x}^{n-1}\parallel \parallel x\parallel \parallel y\parallel $ |

It follows that $n\le 2||x||||y||$ for all $n\in \mathbb{N}$ which is impossible. ∎

Corollary. The identity operator on a Hilbert space^{} $\mathscr{H}$ cannot be expressed as a commutator of two bounded linear operators in $\mathcal{L}(\mathscr{H})$.

Remark. The above can be understood as a version of the uncertainty principle in one dimension. Let $H={L}^{2}(\mathbb{R})$. Let $q:H\to H$ be $q(f)(x):=xf(x)$ with $D(q)=\{f\in {L}^{2}(\mathbb{R}):x\mapsto xf(x)\in {L}^{2}(\mathbb{R})\}$, the coordinate operator and $p:H\to H,p(f)(x)=-i{f}^{\prime}(x)$ the momentum operator with $D(p):=\{f\in {L}^{2}(\mathbb{R}):f\text{absolutely continuous},{f}^{\prime}\in {L}^{2}(\mathbb{R})\}$. It follows that

$pq-qp$ | $=-i{\mathrm{id}}_{D}\text{on}D=D(q)\cap D(p)$ |

According to the corollary $D={L}^{2}(\mathbb{R})$ can never be the case.

Title | Wielandt theorem for unital normed algebras |
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Canonical name | WielandtTheoremForUnitalNormedAlgebras |

Date of creation | 2013-03-22 19:01:22 |

Last modified on | 2013-03-22 19:01:22 |

Owner | karstenb (16623) |

Last modified by | karstenb (16623) |

Numerical id | 6 |

Author | karstenb (16623) |

Entry type | Theorem |

Classification | msc 46H99 |