Proof: F is isometric if and only if its matrix is orthogonal/unitary I'd like to show that $F \in End(V)$ isometric $<=> M_{\beta \beta} (F)$ orthogonal/unitary
But it seems as if I still have some trouble doing that ;/
"=>"
$<v_i, v_j> = <f(v_i),f(v_j)> = <K_{\beta}^{-1} (f(v_i)), K_{\beta}^{-1} (f(v_j))> $.
So, the rows of $M_{\beta \beta}$ form an orthogonal basis of $K^n$.
"<="
The rows of the matrix form an orthogonal basis, so we have $\delta_{ij} = <f(v_i), f(v_j)>$ = ??
 A: Hint:
1) Let $M$ be a square matrix and $v$ be any vector. Then $||Mv||_2^2=v^TM^TMv$. What happens if $M$ is orthogonal?
2) If $M$ is not a orthogonal matrix, then $M$ have a eigenvalue $\lambda$ such that $|\lambda|\neq 1$. Let $b$ be the corresponding eigenvector, then $||Mb||_2^2=||\lambda b||_2^2=|\lambda|^2||b||_2^2$.
A: The answer provided by dineshdileep is only half-true. for the converse part you have to prove the norm of every column is equal to  1, and that the columns are perpendicular to each other.
Hints:
1.for the norm part you can use vectors like $x^T$ = $[1 \: 0 \: ...  \: 0]$
2.to show each two columns are perpendicular to each other you can use vectors like $x^T$ = $[1 \: 1 \: ... \: 0]$
(the vector I just mentioned can be used to show columns $q_1$ and $q_2$ are perpendicular.)
A: I only prove the converse part. first consider the lemma below:
Lemma.
if $T$ is a linear mapping over the field $\mathbb{C}$, and we know that:
\begin{align*}
\langle v,Tv \rangle = 0
\end{align*}
then $T = 0$
proof
simply write:
\begin{align*}
\langle Tu,w \rangle = \frac{\langle T(u+w), u+w \rangle - \langle T(u - w), u - w \rangle}{4} + \frac{\langle T(u + iw), u + iw \rangle - \langle T(u - iw) , u-iw \rangle}{4}i
\end{align*}
note that each term on the right hand side is in the form $\langle Tv,v \rangle$ and is therefore $0$. so we have that $\langle Tu,v \rangle = 0$ for every $u,v$.this yields:
\begin{align*}
\langle Tu,Tu \rangle = ||Tu||^2 = 0 \Rightarrow Tu = 0 \hspace{0.5cm}(\forall u)
\end{align*}
now! for the main problem. Assume that $U$ is a square isometry matrix:
\begin{align*}
 &\langle v,v \rangle = ||v||^2 = ||Uv||^2 = \langle Uv,Uv\rangle = \langle U^*Uv,v \rangle \\
 &\Rightarrow \langle v,v \rangle = \langle U^*Uv,v \rangle \Rightarrow \langle (U^*U-I)v,v \rangle = 0 \hspace{0.5cm}(\forall v)
\end{align*}
then using the proven lemma, you can deduce the result.
