Constructing a unitary operator for a larger vector space with some constraints on subspace I am working through Nielsen and Chuang (self-study), and I came across a particular problem that I have no clue how to proceed forward (Exercise 2.67).
Some googling led me to 2 sites which have solutions, but it's not clear to me how that approach was chosen - 1 and 2.
I am looking for guidance on how one should think about such problems. It's not immediately apparent to me why the larger space $V$ is thought to be a space that is composed of $W$ and $W^\perp$. Why can the larger space $V$ not contain other subspaces that are not orthogonal to $W$ (like planes intersecting at an angle if we think of $W$ and $V$ in ${\rm I\!R}^2$)? Is the reasoning here that linear combinations of $W$ and $W^\perp$ can construct every other subspace of $V$ (that seems a little dubious to me - esp. if $V$ is a higher-dimensional space).
Any tips and pointers will be greatly appreciated.
Problem statement (verbatim):
Suppose $V$ is a Hilbert space with a subspace $W$. Suppose $U : W → V$ is a linear operator which preserves inner products, that is, for any $|w_1\rangle$ and $|w_2\rangle$ in $W$,
$\langle w_1|U^†U|w_2\rangle = \langle w_1|w_2\rangle$
Prove that there exists a unitary operator $U' : V → V$ 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.
 A: Whenever we have a vector space, there is a complete orthogonal basis that can be constructed for the space. And that basis can be transformed to arrive at another basis which is also complete and orthogonal. This can be done in many ways, one of which is Schmidt decomposition. Given a vector $\phi$, you can find a whole basis using Schmidt decomposition and that contains all vectors orthogonal to $\phi$, as well as each other. So if you have $|w_i\rangle \in W$, then taking all vectors orthogonal to $|w_i\rangle $, which belong to $W^\perp$, you should be able to form a basis which can be used to construct any vector in the vector space. For example, consider vectors in 3D space. Let's say we have $W$ denoting vectors in a 2D plane, i.e., the $XY$ plane in the Cartesian system. Then $W^\perp$ denotes all vectors perpendicular to the $XY$ plane. So now, you have essentially all vectors in 3D space = $W\cup W^\perp$. Similarly, if you consider $W$ to be just one vector. Say it points along $\hat{x}$. Then, $W^\perp = \{\hat{y},\hat{z}\}$. And thus, $W\cup W^\perp = \{\hat{x},\hat{y},\hat{z}\}$, which forms a complete basis for vectors in 3D. I hope this analogy helps in making the point clear. 
