Proving the existence of a unitary operator...

**gusztav** · June 10th 2009, 11:19 PM

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 $<Ux, Uy>=<x,y>$

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

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!

**NonCommAlg** · June 11th 2009, 05:15 PM

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 $<Ux, Uy>=<x,y>$

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

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 $<e_i+e_j,e_i-e_j>=0,$ for all $i \neq j.$ thus, by the problem hypothesis, we

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

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