Show that there exists a $\alpha \in \mathbb{C}$ for which $\alpha f: V \rightarrow V$ is an isometry.

Question

Task: Let $(V,\langle.,.\rangle)$ be a unitary vector space, $f \in \operatorname{Hom}_{\mathbb{C}}(V, V)=\operatorname{End}_{\mathbb{C}}(V) $ an endomorphism $ \neq 0 $ with the property that for all $v, w \in V $with $\langle v, w\rangle=0$ also $\langle f(v ), f(w)\rangle=0 $ holds. Show that there is a $\alpha \in \mathbb{C} $ for which $\alpha f\colon V \rightarrow V $ is an isometry.

Problem: I've already recorded this visually, but only for $\mathbb{R}^{2}$, for $\mathbb{C}$ I'm not so sure what it looks like. I know that every isometry is either a reflection or more precisely a rotation, but unfortunately I am at the end of my ideas how to show it. Perhaps someone can give me a concrete example of how such a $\alpha$ for $\mathbb{C}$ works, or give me the name of the problem I can look up online. Thanks very much!

Answer 1

An isometry $g\colon V\rightarrow V$ is a map, that preserves the scalar product, so $\langle g(v),g(w)\rangle=\langle v,w\rangle$ for all $v,w\in V$.

Hint: Consider your map $f\colon V\rightarrow V$, which maps orthogonal vectors to orthogonal vectors, so $\langle v,w\rangle=0\Rightarrow\langle f(v),f(w)\rangle=0$ for all $v,w\in V$. For any vectors $v,w\in V$ (even if $\langle v,w\rangle\neq 0$), we have: $$\left\langle v,w-\frac{\langle v,w\rangle}{\|v\|^2}v\right\rangle=0,$$ so this property can be applied and we get: $$\langle f(v),f(w)\rangle-\frac{\|f(v)\|^2}{\|v\|^2}\langle v,w\rangle =\left\langle f(v),f\left(w-\frac{\langle v,w\rangle}{\|v\|^2}v\right)\right\rangle=0.$$ So $\alpha=\|v\|/\|f(v)\|$ fits, if we can show, that this value is independent of $v$. Now go back and try, if you could also change the first entry of the scalar product to make it vanish and apply your property.

Answer 2

Is $V$ finite-dimensional? If so, here is a hint.

I assume that the inner product is linear in the second slot.

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Suppose that $e_1$, $e_2$, $\dots$, $e_n$ is an orthonormal basis of $V$.

Suppose that $i \neq j$. Suppose that $k$ is a complex number such that $|k| = 1$. Then $$ \begin{aligned} & \langle {e_i + ke_j, e_i - ke_j} \rangle \\ = {} & \langle {e_i + ke_j, e_i} \rangle - k \langle {e_i + ke_j, e_j} \rangle \\ = {} & \langle {e_i, e_i} \rangle + \overline{k} \langle {e_j, e_i} \rangle - k \langle {e_i, e_j} \rangle - k\overline{k} \langle {e_j, e_j} \rangle \\ = {} & 1 - k\overline{k} = 0. \end{aligned} $$ Hence $$ \begin{aligned} & 0 \\ = {} & \langle {f(e_i + ke_j), f(e_i - ke_j)} \rangle \\ = {} & \langle {f(e_i) + kf(e_j), f(e_i) - kf(e_j)} \rangle \\ = {} & \langle {f(e_i), f(e_i)} \rangle + \overline{k} \langle {f(e_j), f(e_i)} \rangle - k \langle {f(e_i), f(e_j)} \rangle - \langle {f(e_j), f(e_j)} \rangle. \\ \end{aligned} $$

Hence (in which we take $k = 1$, $-1$, $\mathrm{i}$, $-\mathrm{i}$, respectively) $$ \begin{aligned} & 0 = \langle {f(e_i), f(e_i)} \rangle + \langle {f(e_j), f(e_i)} \rangle - \langle {f(e_i), f(e_j)} \rangle - \langle {f(e_j), f(e_j)} \rangle, \\ & 0 = \langle {f(e_i), f(e_i)} \rangle - \langle {f(e_j), f(e_i)} \rangle + \langle {f(e_i), f(e_j)} \rangle - \langle {f(e_j), f(e_j)} \rangle, \\ & 0 = \langle {f(e_i), f(e_i)} \rangle - \mathrm{i} \langle {f(e_j), f(e_i)} \rangle - \mathrm{i} \langle {f(e_i), f(e_j)} \rangle - \langle {f(e_j), f(e_j)} \rangle, \\ & 0 = \langle {f(e_i), f(e_i)} \rangle + \mathrm{i} \langle {f(e_j), f(e_i)} \rangle + \mathrm{i} \langle {f(e_i), f(e_j)} \rangle - \langle {f(e_j), f(e_j)} \rangle. \end{aligned} $$

We learn from the system of linear equations that $$ \langle {f(e_i), f(e_i)} \rangle = \langle {f(e_j), f(e_j)} \rangle $$ and that $$ \langle {f(e_i), f(e_j)} \rangle = \langle {f(e_j), f(e_i)} \rangle = 0. $$

Since $f \neq 0$, one of $f(e_1)$, $f(e_2)$, $\dots$, $f(e_n)$ must be nonzero. Without loss of generality, let $f(e_1)$ be nonzero. Let $\ell = \sqrt{\langle {f(e_1), f(e_1)} \rangle} > 0$. Let $$ u_i = \frac{1}{\ell} f(e_i), \quad i = 1, 2, \dots, n. $$ It is not hard to check that $$ \langle {u_i, u_i} \rangle = 1 $$ and that for $i \neq j$, $$ \langle {u_i, u_j} \rangle = 0. $$

I leave it as an exercise for you that you can take $\alpha$ to be $1/\ell$ so that $\alpha f$ is an isometry.