Existence of an unitary operator that transforms a set of orthogonal vectors to another set of orthogonal vectors. This is part of my research project, where I wanted to find some form of unitary transformation that links between bases of two different subspaces of the same vector space, which are some common properties. Bellow I have tried to explain my doubt in a more mathematical form.
Consider all the following vectors belonging to $\mathbb{C}^m$ where $m>n$.
Suppose $A:=\{w_1,w_2,\cdots,w_i,\alpha_{i+1},\cdots,\alpha_{n}\}$ and $B:=\{w_1,w_2,\cdots,w_i,\beta_{i+1},\cdots,\beta_{n}\}$ are two sets of orthogonal vectors.

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*How to show that there exists an operator that transforms the elements of the set $A$ to elements of the set $B$ (in particular $U \alpha_j = \beta_j$ for $i+1\leq j\leq n$).
(I was thinking as both set $A$ and $B$ constitute orthogonal vectors, so they form basis of a subspace, and there exists some basis transform operator. But needed some explicit proof.)


*Is there a way to find the explicit form of $U$ (maybe in the matrix form).
Specifically to show that $U$ is a unitary operator.
 A: I assume that all the vectors $w_i,\alpha_j,\beta_k$ are unit length.
First select $a_{n+1},...,a_{m} \in \mathbb{C}^m$ to be unit length vectors such that $A \cup \{a_{n+1},...,a_{m}\}$ consists of $m$ orthogonal vectors (and is therefore an orthogonal basis of $\mathbb{C}^m$) and do the same for $b_{n+1},...,b_{m}$. (The only reason we consider these vectors is to ensure that the matrix $U$ is unitary.)
Having collected these vectors you can use the unitary operator
$$U = \sum_{j=1}^i w_jw_j^T + \sum_{j=i+1}^n \beta_j\alpha_j^T + \sum_{j=n+1}^m b_ja_j^T$$
By orthogonality you will have
$$Uw_k = \sum_{j=1}^i w_jw_j^Tw_k + \sum_{j=i+1}^n \beta_j\alpha_j^Tw_k + \sum_{j=n+1}^m b_ja_j^Tw_k = w_kw_k^Tw_k = w_k(1)=w_k$$
$$U\alpha_k = \sum_{j=1}^i w_jw_j^T\alpha_k + \sum_{j=i+1}^n \beta_j\alpha_j^T\alpha_k + \sum_{j=n+1}^m b_ja_j^T\alpha_k = \beta_k\alpha_k^T\alpha_k = \beta_k(1)=\beta_k$$
$$Ua_k = \sum_{j=1}^i w_jw_j^Ta_k + \sum_{j=i+1}^n \beta_j\alpha_j^Ta_k + \sum_{j=n+1}^m b_ja_j^Ta_k = b_ka_k^Ta_k = b_k(1)=b_k$$
since everything is orthogonal to everything but itself and we assume unit length.
To show that it is a unitary matrix, note that since $A\cup \{a_{n+1},...,a_m\}$ is a basis of $\mathbb{C}^m$ we can write any $v\in \mathbb{C}^m$ as a linear combination
$$v = \sum_{j=1}^i x_jw_j + \sum_{j=i+1}^n x_j\alpha_j + \sum_{j=n+1}^m x_ja_j $$
and by the pythagorean theorem
\begin{align}
||v||_2^2 &= ||\sum_{j=1}^i x_jw_j + \sum_{j=i+1}^n x_j\alpha_j + \sum_{j=n+1}^m x_ja_j ||_2^2 \\
&= \sum_{j=1}^i ||x_jw_j||_2^2 + \sum_{j=i+1}^n ||x_j\alpha_j||_2^2 + \sum_{j=n+1}^m ||x_ja_j||_2^2 & (\text{Pythagora}) \\
&= \sum_{j=1}^i x_j^2 + \sum_{j=i+1}^n x_j^2 + \sum_{j=n+1}^m x_j^2 & (\text{Unit-Length}) \\
||Uv||_2^2 &= ||\sum_{j=1}^i x_jUw_j + \sum_{j=i+1}^n x_jU\alpha_j + \sum_{j=n+1}^m x_jUa_j||_2^2 & (\text{Linearity}) \\
&= ||\sum_{j=1}^i x_jw_j + \sum_{j=i+1}^n x_j\beta_j + \sum_{j=n+1}^m x_jb_j||_2^2 & (\text{Shown Above}) \\
&= \sum_{j=1}^i ||x_jw_j||_2^2 + \sum_{j=i+1}^n ||x_j\beta_j||_2^2 + \sum_{j=n+1}^m ||x_jb_j||_2^2 & (\text{Pythagora})\\
&= \sum_{j=1}^i x_j^2 + \sum_{j=i+1}^n x_j^2 + \sum_{j=n+1}^m x_j^2 & (\text{Unit-Length})
\end{align}
This shows that $||Uv||_2 = ||v||_2$ for all $v$ and which means that $U$ is  unitary.
