Prove an orthogonal map has orthogonal matrix with respect to orthonormal basis Let $V$ be a vector space with the orthonormal basis $Q = \{ \vec{q_1},\ldots, \vec{q_n} \}$ and let $\ell:V\to V$ be an orthogonal map. Prove that the matrix $L$ of of $\ell$ with respect to $Q$ is orthogonal.
Note: By orthogonal map I mean that $\ell$ is linear and satisfies $\left\Vert \ell(\vec{x}) \right\Vert = \left\Vert \vec{x} \right\Vert$ for all $x \in V$. By orthogonal matrix I mean that $L$ has orthonormal columns.
I have that
$$
L=\left[
\begin{array}{ccc}
[\ell(\vec{q_1})]_Q & \cdots & [\ell(\vec{q_n})]_Q
\end{array}
\right]
$$
but I don't have any idea how to proceed.
 A: A proof sketch. 


*

*Show that $\langle \ell(\vec x), \ell(\vec {y}) \rangle = \langle \vec x , \vec y \rangle$ holds for all $\vec x, \vec y \in V$. For this step, you may need the following hint: $$\langle \vec x, \vec y \rangle = \frac{1}{2} (\| \vec x+\vec y \|^2 - \| \vec x \|^2 - \| \vec y \|^2) .$$ 

*Show that if the $i^{\rm th}$ column of $L$ is $c_i \in \mathbb R^n$, then 
$$
\ell(\vec {q_i}) = \sum_{k = 1}^n c_{i,k} \cdot \vec {q_k}. 
$$

*Show that for $1 \leq i, j \leq n$, we have
$$
\langle \ell(\vec {q_i}) , \ell(\vec {q_j}) \rangle = \langle c_i, c_j \rangle. 
$$

*Using (1.), what can you say about $\langle \ell(\vec {q_i}) , \ell(\vec {q_j}) \rangle$? What does this mean about $\langle c_i, c_j \rangle$? 
A: Suppose $T: V \to V$ is an isometry, i.e. $\|Tx\| = \|x\|$ for all $x \in V$. Then $T$ is clearly injective, so since $V$ is finite-dimensional, it is also surjective. This means $T^{-1}$ exists. By the polarization identity we have
$$\langle Tx, Ty \rangle = \frac{1}{2}(\|T(x+y) \|^2- \|Tx\|^2 - \|Ty\|^2)$$
$$ = \frac{1}{2}(\|x+y \|^2- \|x\|^2 - \|y\|^2) = \langle x, y \rangle$$
which implies
$$\langle Tx, y \rangle = \langle x, T^{-1}y\rangle$$
Hence $T^{-1} = T^*$. But the matrix for $T^*$ with respect to an orthornormal basis $Q = \{q_1, ..., q_n\}$ is Hermitian transpose of the matrix for $T$:
$$[T]_Q^* = [T^*]_Q =  [T^{-1}]_Q = [T]_Q^{-1}$$
So
$$[T]_Q[T^*]_Q = [T]_Q[T]^{-1}_Q = I$$
Let $(A)_i$ and $(A)^j$ denote the $i$-th row and $j$-th column of $A$ respectively. Then by the definition of matrix multiplication and Hermititan transpose
$$[T]_Q{[T^*]_Q}_{ij} = \langle ([T]_Q)^i, {([T]^*_Q)}_j \rangle = \langle ([T]_Q)^i, {([T]_Q)}^j \rangle = \delta_{ij}$$
So the columns of $[T]_Q$ are orthonormal.
