Suppose I have two sets of vectors, $E_1=\{v_i\}_{i=1}^{k}$ and $E_2=\{u_i\}_{i=1}^{k}$, with each vector belonging to $\mathbb{C}^k$.

When is it possible to find a unitary matrix that maps $E_1$ to $E_2$? Is enough for the pairwise inner products between vectors in the set to be equal?

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    $\begingroup$ They are isomorphic. $\endgroup$ – Mhenni Benghorbal Nov 27 '13 at 16:44
  • $\begingroup$ @MhenniBenghorbal Is that equivalent to what I said about the pairwise inner products being equal? i.e. the structure of the sets remains unchanged or am I missing the point? $\endgroup$ – Stan Nov 27 '13 at 17:39
  • $\begingroup$ Isomorphic does not mean equal. Two different mathematical structures can be isomorphic. You should take some time to read about this concept. $\endgroup$ – Mhenni Benghorbal Nov 27 '13 at 17:43

There is a unitary matrix $Q$ that maps $E_1$ to $E_2$ iff $u_i^* u_j = v_i^* v_j$ for all $i,j$.

If there is a such unitary matrix, you must have $u_i^* u_j = (Qv_i)^* Qv_j = v_i^* Q^* Q v_j = v_i^* v_j$.

In the other direction, let $U = \operatorname{sp} \{ u_i \}$, and similarly for $V$. Note that $\|\sum_k \alpha_k u_k \|^2 = \sum_i \sum_j \overline{\alpha}_i \alpha_i u_i^* u_j = \sum_i \sum_j \overline{\alpha}_i \alpha_i v_i^* v_j = \|\sum_k \alpha_k v_k \|^2$, for any scalars $\alpha_k$.

Then if $\{ u_{n_k} \} $ are linearly independent, it follows that $\{ v_{n_k} \} $ are linearly independent, and vice versa. Hence $\dim U = \dim V$, and hence $\dim U^\bot = \dim V^\bot$.

Let $\{ \hat{u}_k \}$ be an orthonormal basis for $U^\bot$, and similarly, let $\{ \hat{v}_k \}$ be an orthonormal basis for $V^\bot$.

We define $Q$ on $U^\bot$ by $Q \hat{u}_k = \hat{v}_k$.

Define $Q$ on $U$ by $Q (\sum_k \alpha_k u_k) = \sum_k \alpha_k v_k$. To see that $Q$ is well defined, suppose $\sum_k \alpha_k u_k = \sum_k \beta_k u_k$. Then we have $\sum_k (\alpha_k-\beta_k) u_k = 0$, and the above shows that $\sum_k (\alpha_k-\beta_k) v_k = 0$. Hence $Q (\sum_k \beta_k u_k) = \sum_k \beta_k v_k = \sum_k \alpha_k v_k = Q (\sum_k \alpha_k u_k)$, as desired.

Finally, we check that $Q$ is unitary: Let $x,y \in \mathbb{C}^k$, and write $x = \sum_k \alpha_k u_k + \sum_k \hat{\alpha}_k \hat{u}_k$, $y = \sum_k \beta_k u_k + \sum_k \hat{\beta}_k \hat{u}_k$. We note that $Qx = \sum_k \alpha_k v_k + \sum_k \hat{\alpha}_k \hat{v}_k$, $Qy = \sum_k \beta_k v_k + \sum_k \hat{\beta}_k \hat{v}_k$. Expanding shows that $(Qx)^* (Qy) = x^* y$, from which follows that $Q^* Q = I$, hence $Q$ is unitary.


Let $T$ be the linear operator that takes the vector space $V_1 = \text{span}(E_1)$ to $V_2 = \text{span}(E_2)$ via linearly extending the map $T(u_i) = v_i$. Show that this map is well defined and an isometry. Let $U_1$ be the orthogonal complement of $V_1$, and $U_2$ be the orthogonal complement of $V_2$. Define any isometry $S$ you like taking $U_1$ to $U_2$ (making sure that they have the same dimension, and so an isometry can be found).

Then the map $u+v \mapsto Su + Tv$ ($u \in U_1$, $v\in V_1$) is an isometry on $\mathbb C^k$.

If you knew that the vectors in $E_1$ were linearly independent, that would shorten the proof, because you wouldn't have to mess around with the orthogonal complements.


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