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

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

Let $A$ be a linear operator on a finite-dimensional unitary space $V$ such that, for every $x,y \in V$ the following implication holds:
$(x \perp y) \Rightarrow (Ax \perp Ay)$.
Prove that then there exists a scalar $\alpha$ and a unitary operator $U$ (where $U:V \rightarrow V$) such that $A= \alpha U$.

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I'm aware of the following facts:
if an operator $U$ is unitary, and $U:V \rightarrow V$, then for every $x, y \in V$ we have $=$

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

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

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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 $A$ be a linear operator on a finite-dimensional unitary space $V$ such that, for every $x,y \in V$ the following implication holds:
$(x \perp y) \Rightarrow (Ax \perp Ay)$.
Prove that then there exists a scalar $\alpha$ and a unitary operator $U$ (where $U:V \rightarrow V$) such that $A= \alpha U$.

******

I'm aware of the following facts:
if an operator $U$ is unitary, and $U:V \rightarrow V$, then for every $x, y \in V$ we have $=$

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

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

******

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

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