Prove the matrix of an orthogonal linear transformation relative to an orthonormal basis is orthogonal.

Question

Let $V$ be an $n$-dimensional inner product space (not necessarily the standard inner product), and let $L:V \rightarrow V$ be an orthogonal linear transformation. Let $B=\{v_1,...,v_n\}$ be an orthonormal basis for $V$. Let $A$ be the matrix of $L$ relative to this basis. Prove that $A$ is orthogonal.

My attempt:

To prove that $A$ is orthogonal, I could prove that $A^TA=I$, or I could prove that the inner product of any two columns of $A$ is zero and the length of any column of $A$ is one. (I use standard the inner product for the first proof but not the second, right?)

I know how to prove that $A$ is orthogonal if $V= \mathbb R^n$, but I am not sure how to generalize to any orthogonal basis. This is how I would do it for $V= \mathbb R^n$:

1. Column $i$ of $A$ is simply $T(e_i)$, where $e_i$ is the $i$^th vector in the $n$-dimensional standard basis.
2. Since $T$ is an orthogonal transformation and therefore preserves length, $||T(e_i)||=||e_i||=1$ since the basis is orthogonal.
3. Since $T$ is an orthogonal transformation, for all $x,y$ in $V$, $<Tx,Ty>=<x,y>$. Moreover, Since the basis $B$ is orthogonal, the inner product of any two basis vectors is $0$. Therefore, the inner product of any two columns of $A$ is $0$.
4. Since the length of every column of $A$ is $1$ and the inner product of any two columns of $A$ is $0$, $A$ is orthogonal.

My problem with generalizing to any $V$ is step 1: Specifically, what would $A$ look like for an arbitrary $V$?

Answer 1

Hint: For a general $\ V\ $: $$ T\left(v_j \right)= \sum_{i=1}^na_{ij}v_i\ , $$ so $$ \left\langle v_j, v_k\right\rangle= \left\langle T\left(v_j\right), T\left(v_k\right)\right\rangle = \left\langle \sum_{i=1}^na_{ij}v_i, \sum_{l=1}^na_{lk}v_l\right\rangle\ . $$ What happens if you use the bilinearity of the inner product to pull the coefficients of $\ v_i\ $ and $\ v_l\ $ in the rightmost term of these equations out of it?