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**gusztav** Let $\displaystyle A$ be a linear operator on a finite-dimensional unitary space $\displaystyle V$ such that, for every $\displaystyle x,y \in V$ the following implication holds:

$\displaystyle (x \perp y) \Rightarrow (Ax \perp Ay)$.

Prove that then there exists a scalar $\displaystyle \alpha$ and a unitary operator $\displaystyle U$ (where $\displaystyle U:V \rightarrow V$) such that $\displaystyle A= \alpha U$.

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I'm aware of the following facts:

if an operator $\displaystyle U$ is unitary, and $\displaystyle U:V \rightarrow V$, then for every $\displaystyle x, y \in V$ we have $\displaystyle <Ux, Uy>=<x,y>$

Also, on a unitary space $\displaystyle V$ the following is true for every $\displaystyle x, y \in V$: $\displaystyle x \perp y \Leftrightarrow <x,y>=0$

And finally, if $\displaystyle U$ is a unitary operator, then if $\displaystyle \{ e_1,...e_n\}$ is an orthonormal basis for $\displaystyle V$, then $\displaystyle \{ U e_1,...U e_n\}$ is also an orthonormal basis for $\displaystyle V$.

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But so far, I've had no luck in employing these to solve this problem, so would greatly appreciate your help!