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