What is "cyclic shift unitary" on $M_{n}(\mathbb{C})$?

Question

From page 264 of Brown & Ozawa's $C^*$-algebras and finite-dimensional approximations:

Let $M_{n}(\mathbb{C})$ be the $n\times n$ complex matrices, and what is the "cyclic shift unitary of order $n$" on $M_{n}(\mathbb{C})$?

Maybe it is a very basic concept in functional analysis or matrix theory?

score 6 · Accepted Answer · answered Nov 29 '14 at 18:40

6

Cyclic shift operator is the matrix $$A=\begin{bmatrix} 0&1 \\ &0&1 \\ & & \ddots \\ &&&&1\\ 1&&&&0 \end{bmatrix}$$ It has order $n$, that is to say, $A^n=I$, and it is a unitary matrix.

answered Nov 29 '14 at 18:40

Sal

2,231

Thanks for your answer, Sal. – Stephen Curry Nov 30 '14 at 06:08

What is "cyclic shift unitary" on $M_{n}(\mathbb{C})$?

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