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The vector space $V$ is equipped with a Hermitian scalar product and an orthonormal basis $\{e_1,\ldots,e_n\}$. A second orthonormal basis $\{e_1',\ldots,e_n'\}$ is related to the first one by $e_j'=\sum U_{ij}e_i$, where $U_{ij}$ are complex numbers. Show that $U_{ij}=\langle e_i,e_j'\rangle$ and that the matrix $U$ with entries $U_{ij}$ is unitary.

Need help particularly with the second part!!

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Hint #1

Look at the definition of $e_j'$ and take its inner product with $e_i$. In other words, compute

$$ \langle e_i, e_j' \rangle \;\; =\;\; \left \langle e_i, \; \sum_{k=1}^n U_{kj} e_k \right \rangle. $$

Hint #2

Recall that for $U$ to be unitary we must have that $UU^* = U^*U = I$. Equivalently this can be stated that $\langle Uv, Uw\rangle = \langle u,w\rangle$ for all vectors $u$ and $w$ in your vector space. Rewrite $u$ and $w$ in the basis $\{e_1, \ldots, e_n\}$ and this should be relatively easy to demonstrate.

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I assume you can determine the matrix entries of $U$ by using the fact that the bases are orthonormal. In what follows, I'm using $(\cdot,\cdot)$ to denote the inner product, which is the old way and saves typing "langle", "rangle" all over the place. This inner product is linear in the first coordinate and conjugate linear in the second.

If $\{ e_1',\cdots,e_n'\}$ is an orthonormal basis of $V$, then every $v$ can be written as $$ v = \sum_{k=1}^{n}(v,e_k')e_k'. $$ Automatically $$ \|v\|^{2} = \sum_{k=1}^{n}|(v,e_k')|^{2},\;\;\; v \in V. $$ More generally, $$ (v,w) = \sum_{k=1}^{n}(v,e_k')(e_k',w) =\sum_{k=1}^{n}(v,e_k')\overline{(w,e_k')},\;\;\; v,w\in V. $$ So this gives you a map $L : V \mapsto \mathbb{C}^{n}$ defined by $$ Lv = \left[\begin{array}{c}(v,e_1')\\(v,e_2')\\ \vdots \\(v,e_n')\end{array}\right] $$ This map is isometric and preserves inner products. That is, $$ (v,w) = (Lw)^{\star}(Lv), $$ where $(Lw)^{\star}$ is the conjugate transpose of the $n\times 1$ matrix $Lw$.

If you have another orthonormal basis $\{ e_1,\cdots,e_n \}$, then $\{ Le_1, Le_2,\cdots,Le_n \}$ is a orthonormal basis of $\mathbb{C}^{n}$ because $(Le_k)^{\star}(Le_j)=(e_j,e_k)=\delta_{j,k}$. Therefore, the matrix $U$ with $j$-th column $Le_j$ for $1 \le j \le n$ is a unitary matrix. (NOTE: Your matrix $[U_{i,j}]=[(e_i,e_j')]$ is the the transpose of this one because of the way that $U$ is defined in your problem.)

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