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Let $A:V \to V$ be a linear operator in inner product space $V$ and $$\forall x,y \in V \\ x\cdot y=0 \Rightarrow Ax \cdot Ay =0$$ Show there exists $\lambda \in \mathbb F$ such that $\lambda A$ is unitary operator.

I haven't done anything so far, and I would appreciate any hint you may have for me.

Thanks in advance

windircurse
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    As stated, this is untrue. Consider an $A$ that maps everything to the $0$ vector. Then all conditions of the hypothesis are true, but clearly $\lambda A$ is not unitary for any $\lambda$. – stochasticboy321 Jun 15 '16 at 17:48
  • Can this question be saved if we put restriction $A \not = 0$ – windircurse Jun 15 '16 at 17:54
  • I think this is true if $V$ has an orthonormal basis and if $A$ is surjective. – Dark Jun 15 '16 at 17:57

1 Answers1

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As commented, $A=0$ is a counterexample. If $A\ne0$ there exists $\lambda$ such that if $B=\lambda A$ then $$Bx\cdot By=x\cdot y$$for all $x,y$. That says $B$ is unitary, at least in the finite-dimensional case. (The right shift on $\ell^2(\Bbb N)$ is a counterexample in the infinite-dimensional case, if we're careful about the definition of "unitary"; if we assume as well that $A$ is surjective then $B$ is unitary.)

Suppose $||x||=||y||=1$. Then $$y=(x\cdot y)x+(y-(x\cdot y)x)=z+w.$$

Now $x\cdot w=0$, hence $Ax\cdot Aw=0$. Multiplying this out shows that $$Ax\cdot Ay=(x\cdot y)||Ax||^2.$$ Similarly $$Ax\cdot Ay=(x\cdot y)||Ay||^2.$$

So far we've shown this: If $||x||=||y||=1$ and $x\cdot y\ne0$ then $||Ax||=||Ay||$.

But for any $x,y$ with $||x||=||y||=1$ there exists $z$ such that $x\cdot z\ne 0$ and $y\cdot z\ne 0$. So in fact $||Ax||=||Ay||$ whnever $||x||=||y||=1$. If $A\ne0$ this shows that there exists $\lambda$ so that if $B=\lambda A$ then $||Bx||=||x||$ for all $x$.

Hence $Bx\cdot By=x\cdot y$ for all $x,y$.

David C. Ullrich
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  • Could you please explain the part "if $B= \lambda A$ then...", I didn't quite get it . – windircurse Jun 15 '16 at 18:16
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    @windircurse Since $A\ne0$ there exists $x_0$ with $||x_0||=1$ and $Ax_0\ne0$. Let $\lambda=1/||Ax_0||$. Then $||Bx_0||=1$, and since $||Ax||=||Ay||$ whenever $||x||=||y||=1$ this shows that $||Bx||=1$ whenever $||x||=1$, and hence $||Bx||=||x||$ for all $x$. – David C. Ullrich Jun 15 '16 at 20:43