Why is the cube of a unitary operator = identity matrix?

km707

Hi there,
If A is unitary I understand that it obeys AA⁺=1 because A^-1=A⁺.

Why does A³=1?
The explanation simply says that "A just permutes the basis vectors"..

It then goes on to say that since A³=1, then eigenvalue a³=1 also, which are 1, e^{i.2pi.theta/3}, and e^{i.4pi.theta/3}. This would make sense to me if I knew why A³=1..

Many thanks in advance!

Matterwave

What is this operator A we are talking about? Just any random Unitary operator? It doesn't seem to be true for any unitary operator...

Pengwuino

You must have a specific operator in mind because that's not true in general.

Bill_K

Because A takes each of three basis vectors into the next one, so if you apply A three times you'll get back to the original situation. For example, A³ maps i → j → k → i.