Is Unitarity Preserved by Different Inner Products?

[e(ho0n3]

Let V be a finite dimensional complex inner product space with inner product < , >. Let U be unitary with respect to this inner product. If ( , ) is another inner product, is U also unitary with respect to ( , )?

The definition of unitary I'm working with is the one that says: U is unitary if <Uv, Uw> = <v w>, i.e. it preserves inner products.

Now it is easy to show that U is unitary with respect to < , > if and only if U'U = 1, where U' is the adjoint and 1 is the identity transformations. But by replacing < , > with ( , ), the prior statement says that U is also unitary with respect to ( , ).

Am I missing something?

 

[quasar987]

I don't know if the result is true of not, I haven'T thinking about it but you seem to be missing that if the inner product changes, then the adjoint changes also! So mayeb we lose U'U=1 when we change the inner product.

 

[e(ho0n3]

You know what, you're right! That hadn't occur to me.

I think the result should be as follows: If U is unitary with respect < , >, then it is similar to a unitary with respect to ( , ).

 

[morphism]

The answer to the original question is "no." Any inner product <.,.> on a finite-dimensional vector space is of the form

$$\langle x,y\rangle = \overline{x}^t H y$$

for some hermitian, positive-definite matrix H. For U to be unitary with respect to <.,.>, one needs that $$\overline{U}^t H U = H$$. With this in mind, it's easy to cook up examples of unitary matrices with respect to one inner product failing to be unitary matrices with respect to another.

On the other hand, the answer to the second question is "yes." It's enough to show this for when <.,.> is arbitrary and (.,.) is the standard dot product (so: let's fix a basis for our space and think of it as $\mathbb{C}^n$). Let H be the matrix associated to <.,.> as above. Since H is positive definite and hermitian, one can find an invertible matrix S such that $$H = \overline{S}^t S$$ (exercise!). Now if U is unitary with respect to (.,.), i.e. $$U^{-1} = \overline{U}^t$$, then taking $V = S^{-1}US$, we find that

[tex]\overline{V}^t H V = \overline{S}^t \overline{U}^t \overline{S}^{-t} H S^{-1}US = \overline{S}^t \overline{U}^t \overline{S}^{-t} \overline{S}^t S S^{-1}US = \overline{S}^t \overline{U}^t US = \overline{S}^t S = H,[/itex]<br /> <br /> that is, V is unitary with respect to <.,.>.<br /> <br /> Conversely, if U is unitary with respect to <.,.>, i.e. [tex]\overline{U}^t H U = H[/itex], then one can check that $SUS^{-1}$ is unitary with respect to (.,.).<br /> <br /> Hopefully I haven't made any mistakes.[/tex][/tex]

 

[e(ho0n3]

The answer to the original question is "no."

I'm convinced of that now. Thanks.

Since H is positive definite and hermitian, one can find an invertible matrix S such that $$H = \overline{S}^t S$$ (exercise!).

Since H is Hermitian, you can use the spectral theorem to get an S such that H = S², where S is positive definite and hence Hermitian.

Hopefully I haven't made any mistakes.

I didn't recognize any. Thanks again.