I have a question on an exercise 2.67 in Nielson and Chuang. This is the exercise:
Exercise 2.67 Suppose $\:V\:$ is a Hilbert space with a subspace $\:W$. Suppose $\: U : W \rightarrow V\:$ is a linear operation which preserves inner products, that is, for any $\:|w_{1}\rangle\:$ and $\:|w_{2}\rangle\:$ in $\:W$, $$ \langle w_{1} |U^{\dagger}U|w_{2}\rangle= \langle w_{1} |w_{2}\rangle. $$ Prove that there exists a unitary operator $ U' : V \rightarrow V$ which extends $\:U$. That is, $\:U'|w\rangle=U|w\rangle\:$ for all $\:|w\rangle\:$ in $\:W\:$, but $\:U'\:$ is defined on the entire space $\:V$. Usually we omit the prime symbol $\:'$ and just write $\:U\:$ to denote the extension.
I'm not being able to figure out how to a) prove it b) construct the extension U' for a given U in a specific problem.
Everywhere I search people just refer to the problem or say it's trivial, without giving any direction.
$U|w_{1,2}\rangle=|v_{1,2}\rangle$ so $\langle v_2|v_1\rangle=\langle w_2|U^{\dagger}U|w_1\rangle=\langle w_2|w_1\rangle$. And so the unitary operator U' that preserves the inner product for the v's ( $\langle w_2|U^{'\dagger}U'|w_1\rangle=\langle w_2|w_1\rangle$) should be related to U.
But what this relationship is eludes me. Or how to find a U' given a U in a subsystem (ie. what $\sigma_x\otimes\sigma_x$ for $2\times 2$ subsystem might translate to as a U' in an $N\times N$ system).
We already settled in comments, that this is only true for finite dimensions.
Now people say it's trivial, which either means they actually don't know how to do, or it's not very hard but quite technical to write down exaclty and people are lazy, or it is really simple and you should see why.
Here, it's a combination of the second and the third. The idea is the following: The condition on $U$ ensures that it sends an orthonormal basis of $W$ to an orthonormal basis of $V$. Hence we can extend it to a unitary map by mapping an orthonormal basis of the complement of $W$ in V, $W^{\perp}$ to an orthonormal complement of the image of $U$, i.e. $\operatorname{im}(U)^{\perp}$. Then the extended $U$ sends an orthonormal basis to an orthonormal basis, which means that it's a unitary.
To make this more precise, we know that $W$ is some $m$-dimensional subspace of the $n$-dim. space $V$, hence $W$ too has an orthonormal basis. Let's call this $\{w_1,\ldots,w_m\}$ and remember that it can be extended to an orthonormal basis of $V$ by adding some vectors $w^{\prime}_{m+1},\ldots, w^{\prime}_n$.
Now what you need to see is that $U$ maps the orthonormal basis of $W$ to an orthonormal set in $V$. In other words: If $Uw_i=v_i$, then we have
$$ \langle v_i|v_j\rangle = \langle Uw_i |Uw_j\rangle = \langle w_i|w_j \rangle =\delta_{ij} $$
In the first step, we used the definition of $v_i$, in the second step we used that $U$ keeps inner products the same and in the third step, we used that the $w_i$ are orthonormal ($\delta_{ij}$ is the Kronecker delta).
So $\{v_1,\ldots,v_m\}$ are an orthonormal set, hence it too can be extended to an orthonormal basis of $V$ by adding some orthonormal vectors $\{v^{\prime}_{m+1},\ldots,v^{\prime}_n\}$.
But then you can just write down the unitary extension $U^{\prime}$ by setting $U^{\prime}|w^{\prime}_j\rangle=|v^{\prime}_j\rangle$ for $j=m+1,\ldots,n$.