Question: Let $\mathcal{B}$ be an orthonormal basis for a finite dimensional complex vector space $V$. Let $U$ be an operator on that space. Show that $U$ is a unitary operator if and only if $[U]\_{\mathcal{B}}$ is a unitary matrix.

I'm stuck on how to even approach this problem.

Answer: 
There are a couple of things going on here. You have some abstract $\mathbb{C}$-vector space, equipped with an inner product $(-, -)\_V: V \times V \to \mathbb{C}$, which I will consider to be conjugate-linear in the first argument, and $\mathbb{C}$-linear in the second: $(\lambda v, \mu w)\_V = \overline{\lambda} \mu (v, w)\_V$. A *unitary operator* $U: V \to V$ is one which preserves the inner product, in that $(Uv, Uw)\_V = (v, w)\_V$ for all $u, w \in V$. Note that it makes no sense to define the conjugate-transpose of $U$, since it is not a matrix: let's forget about $V$ and $U$ for the moment.

There is another well-known inner product space, which is $\mathbb{C}^n$, the space of length-$n$ column vectors over $\mathbb{C}$, with inner product given by $(x, y)\_{\mathbb{C}^n} := x^\* y$, where $x^\*$ is the conjugate-transpose of $x$. The nice thing about $\mathbb{C}^n$ is that it comes with a distinguished basis, so that every linear transformation $\mathbb{C}^n \to \mathbb{C}^n$ is exactly some $n \times n$ complex matrix. In $\mathbb{C}^n$, we have the same notion of a unitary map $T: \mathbb{C}^n \to \mathbb{C}^n$ (it must satisfy $(Tx, Ty)\_{\mathbb{C}^n} = (x, y)\_{\mathbb{C}^n}$ for all $x, y \in \mathbb{C}^n$), but we may now characterise such matrices:

**Show:** $T \in \operatorname{Mat}\_n(\mathbb{C})$ is unitary if and only if $T T^\* = I$, where $T^\*$ is the conjugate-transpose of the matrix $T$.

It will also be handy to have the following facts:

1. (Definition) An *isomorphism of inner product spaces* $S: V \to W$ is a linear map such that $(Sv\_1, S v\_2)\_U = (v\_1, v\_2)\_V$ for all $v\_1, v\_2 \in V$.
2. (Prove:) If $S$ is an isomorphism of inner product spaces, so is $S^{-1}$.
3. Note that a unitary operator on $V$ is an isomorphism of inner product spaces $V \to V$.

Returning to the original inner product space $V$, a choice of basis $\mathcal{B} = (b\_1, \ldots, b\_n)$ on $V$ amounts to a linear isomorphism $B: \mathbb{C}^n \to V$ (where $n = \dim V$), given by $B e\_i = b\_i$, where by $e\_i$ I mean the column vector in $\mathbb{C}^n$ having a $1$ in the $i$th position and zeros elsewhere. 

**Show:** If the basis $\mathcal{B}$ is orthonormal, then $B$ is an isomorphism of inner product spaces.

After all the pieces are in place, the result is easy: the matrix of the operator $U: V \to V$ in the basis $\mathcal{B}$ is $B^{-1} U B \in \operatorname{Mat}\_n(\mathbb{C})$, and all we have to do is verify that this is a unitary operator: for any $x, y \in \mathbb{C}^n$, we have
$$ (B^{-1}UBx, B^{-1}UBy)\_V = (UBx, UBy)\_{\mathbb{C}^n} = (Bx, By)\_V = (x, y)\_{\mathbb{C}^n}$$
where the equalities follow from repeatedly applying the facts that $B$, $U$, and $B^{-1}$ are isomorphisms of inner product spaces. Now the result follows from the characterisation of unitary matrices.
Given $\mathcal B$ an orthonormal basis of $V$

$$\begin{align}&[UU^\*]\_\mathcal{B}=[U^\*U]\_\mathcal{B}=[I]\_\mathcal{B}=I\_n\\
\iff &[U]\_\mathcal{B}[U^\*]\_\mathcal{B}=[U^\*]\_\mathcal{B}[U]\_\mathcal{B}=I\_n\\
\iff &[U]\_\mathcal{B}[U]\_\mathcal{B}^\*=[U]\_\mathcal{B}^\*[U]\_\mathcal{B}=I\_n
.\end{align}$$

notice that it works for both direction so this complete the proof.