# Thread: Proving the existence of a unitary operator...

1. ## Proving the existence of a unitary operator...

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!

2. Originally Posted by 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 $\displaystyle A= \alpha U$.

******

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 $\displaystyle V$.

******

But so far, I've had no luck in employing these to solve this problem, so would greatly appreciate your help!
i googled "unitary vector space" to see what exactly it is and then i found out that it's just an old name for a much more familiar name Hilbert space. anyway, first it's clear that we may assume

that $\displaystyle A \neq 0.$ let $\displaystyle \{e_1, e_2, \cdots , e_n \}$ be an orthonormal basis for $\displaystyle V.$ i claim that $\displaystyle ||Ae_1||=||Ae_2|| = \cdots = ||Ae_n||$ : we have $\displaystyle <e_i+e_j,e_i-e_j>=0,$ for all $\displaystyle i \neq j.$ thus, by the problem hypothesis, we

must also have $\displaystyle ||Ae_i||^2-||Ae_j||^2=<A(e_i+e_j),A(e_i-e_j)>=0,$ which proves the claim. let $\displaystyle \alpha=||Ae_1||$ and $\displaystyle U=\frac{1}{\alpha}A.$ it's easy to see that $\displaystyle <Ux,Uy>=<x,y>,$ for all $\displaystyle x,y \in V.$ Q.E.D.