Proof: F is isometric if and only if its matrix is orthogonal/unitary

Question

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 $\delta_{ij} = <f(v_i), f(v_j)>$ = ??

Answer 1

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 $||Mb||_2^2=||\lambda b||_2^2=|\lambda|^2||b||_2^2$.

Answer 2

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.)

Answer 3

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.