# Unitary transformations

1. Jun 14, 2010

### LAHLH

Hi,

I just have a quick question, I understand that $$(T\vec{u},T\vec{v})=(\vec{u},\vec{v})$$ defines a unitary transformation $$T$$, but how does one go from this relation to $$(T\vec{u},\vec{v})=(\vec{u},T^{-1}\vec{v})$$ as the other way to define a unitary transformation?

Thanks

2. Jun 14, 2010

### Fredrik

Staff Emeritus
You define the adjoint of T by $$(T^\dagger\vec u,\vec v)=(\vec u,T\vec v)$$ (after proving that it always exists). Then you can define "unitary" by saying that T is unitary if $$T^{-1}=T^\dagger$$. If T is unitary according to this alternative definition, we have

$$(T\vec u,T\vec v)=(T^\dagger T\vec u,\vec v)=(T^{-1}T\vec u,\vec v)=(\vec u,\vec v)$$

It's not much harder to prove that if T is unitary according to your definition, it's also unitary according to the alternative definition, but I'll let you do that one for yourself.

3. Jun 14, 2010

### Landau

It's impossible. You have to require that U is surjective. Of course, if U preserves the inner product (which is LAHLH's definition) then it is an isometry (actually, these are equivalent), hence injective. Surjectivity is needed to guarantee bijectiveness, else T^-1 does not make sense. In finite dimensions injective already implies surjective, but in general (infinite dimension) not.

Consider E.g. the unilateral shift T on $\ell(\mathbb{N})$which sends (x1,x2,x3,...) to (0,x1,x2,x3,...). It is an isometry, so it preserves the inner product. So T would be unitary according to the first definition. But T is not invertible (not surjective): its image does not contain sequences whose first coordinate is non-zero.

Last edited: Jun 14, 2010