3
$\begingroup$

Question: Suppose there is a complex number $\mu$ and a unit vector $u \in V$ (where V is a complex inner product space) such that for all $v \in V$, $\phi(v)=v+(\mu -1)\langle u,v \rangle u.$ Prove that $\phi$ is unitary if and only if $\mu \bar{\mu}=1$.

Solution: $\phi$ is unitary iff $\langle \phi(v),\phi(w) \rangle=\langle v,w \rangle$ $\forall v,w \in V$.

Now,

$$\langle v+(\mu-1)\langle u,v\rangle u,w+(\mu-1) \langle u,w\rangle u\rangle$$

$$=\langle v,w \rangle + (\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle+(\mu-1)\overline{\langle u,v\rangle}\langle u,w\rangle+(\bar{\mu}-1)(\mu-1)\overline{\langle u,v\rangle}\langle u,w\rangle=0$$ so $\mu \bar{\mu}-1=0$.

The part that is puzzling me: This property that was used (although it is not shown):$$\langle v,(\mu -1)\langle u,w\rangle u\rangle =(\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle$$

However, I do not understand this. I understand how the $(\bar{\mu}-1)$ is taken out of the inner product, but I don't understand why $\langle u,w \rangle$ isn't complex conjugate and why $\langle v,u \rangle$ is. I've tried using conjugate symmetry on the whole inner product and then taking the other inner product out, but I still don't get the given result.

Note: $u$ is a unit vector.

Edit: I've posted the whole question and solution to try and make it more clear.

$\endgroup$
  • $\begingroup$ inner product space $V$ $\endgroup$ – martini Jan 3 '17 at 19:32
  • $\begingroup$ Looks odd to me. The $\langle v, u \rangle$ should not be conjugated. $\endgroup$ – copper.hat Jan 3 '17 at 19:41
  • $\begingroup$ @copper.hat I've posted the original question, maybe I forgot to mention a certain property. $\endgroup$ – user390960 Jan 3 '17 at 19:46
  • $\begingroup$ use \in for $x\in X$ $\endgroup$ – juan arroyo Jan 3 '17 at 19:46
1
$\begingroup$

It appears that this computation is done with the assumption that the inner product is conjugate-linear in the first variable, rather than the second variable. So $\langle \lambda u,v\rangle=\bar{\lambda}\langle u,v\rangle$ and $\langle u,\lambda v\rangle=\lambda\langle u,v\rangle$ instead of the other way around. The second term $(\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle$ then comes not from $\langle v,(\mu -1)\langle u,w\rangle u\rangle$ but from $\langle (\mu-1)\langle u,v\rangle u,w\rangle$ (and it is instead the third term that comes from $\langle v,(\mu -1)\langle u,w\rangle u\rangle$).

Indeed, this assumption is necessary for the statement to make sense: notice that $\phi$ isn't even linear if the inner product is conjugate-linear in the second variable. You need to know that if you replace $v$ by $\lambda v$, the term $(\mu-1)\langle u,v\rangle u$ gets multiplied by $\lambda$ (rather than $\bar{\lambda}$).

$\endgroup$
  • $\begingroup$ There are two competing definitions of "inner product": one that is conjugate-linear in the first variable and one that is conjugate-linear in the second variable. The two definitions are essentially equivalent, since you can take an inner product of one type and just swap the order of its inputs to get an inner product of the other type. But you can't just freely assume one or the other: when someone says "$\langle \cdot,\cdot\rangle$ is an inner product", they always mean one and not the other, and it may make a difference for some computations. $\endgroup$ – Eric Wofsey Jan 3 '17 at 20:20
  • $\begingroup$ Thank you very much! Makes perfect sense now. $\endgroup$ – user390960 Jan 3 '17 at 20:22
0
$\begingroup$

The formula in your notes does not work:

Consider $u=w=1$ and $v=i$.

$\endgroup$
  • $\begingroup$ I've posted the original question, maybe I forgot to mention a certain property. $\endgroup$ – user390960 Jan 3 '17 at 19:46
  • $\begingroup$ That line still doesn't work. Maybe you could provide the whole solution to see what is really going on. $\endgroup$ – Jack Jan 3 '17 at 19:50
  • $\begingroup$ Sure, will add it. $\endgroup$ – user390960 Jan 3 '17 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy