Property of $Q$ such that $frac{x^TQx}{|x|^2} = text{const}, forall xin mathbb{R}^n$












0












$begingroup$


I am curious about the following problem:




  1. suppose $Q^TQ = I$, i.e., $Q$ is orthogonal

  2. we want $$frac{x^TQx}{|x|^2} = text{const}, forall xinmathbb{R}^n$$


My question is what properties of $Q$ to let this equality hold?



I think the only property is $Q=I$. Are there any other properties?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
    $endgroup$
    – Damien
    Dec 3 '18 at 23:16










  • $begingroup$
    @Damien I think even if no symmetric, it is an orthogonal base change
    $endgroup$
    – sleeve chen
    Dec 3 '18 at 23:24






  • 1




    $begingroup$
    You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
    $endgroup$
    – Damien
    Dec 4 '18 at 6:05
















0












$begingroup$


I am curious about the following problem:




  1. suppose $Q^TQ = I$, i.e., $Q$ is orthogonal

  2. we want $$frac{x^TQx}{|x|^2} = text{const}, forall xinmathbb{R}^n$$


My question is what properties of $Q$ to let this equality hold?



I think the only property is $Q=I$. Are there any other properties?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
    $endgroup$
    – Damien
    Dec 3 '18 at 23:16










  • $begingroup$
    @Damien I think even if no symmetric, it is an orthogonal base change
    $endgroup$
    – sleeve chen
    Dec 3 '18 at 23:24






  • 1




    $begingroup$
    You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
    $endgroup$
    – Damien
    Dec 4 '18 at 6:05














0












0








0





$begingroup$


I am curious about the following problem:




  1. suppose $Q^TQ = I$, i.e., $Q$ is orthogonal

  2. we want $$frac{x^TQx}{|x|^2} = text{const}, forall xinmathbb{R}^n$$


My question is what properties of $Q$ to let this equality hold?



I think the only property is $Q=I$. Are there any other properties?










share|cite|improve this question









$endgroup$




I am curious about the following problem:




  1. suppose $Q^TQ = I$, i.e., $Q$ is orthogonal

  2. we want $$frac{x^TQx}{|x|^2} = text{const}, forall xinmathbb{R}^n$$


My question is what properties of $Q$ to let this equality hold?



I think the only property is $Q=I$. Are there any other properties?







linear-algebra matrices orthogonal-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 23:07









sleeve chensleeve chen

3,07141852




3,07141852








  • 1




    $begingroup$
    $Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
    $endgroup$
    – Damien
    Dec 3 '18 at 23:16










  • $begingroup$
    @Damien I think even if no symmetric, it is an orthogonal base change
    $endgroup$
    – sleeve chen
    Dec 3 '18 at 23:24






  • 1




    $begingroup$
    You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
    $endgroup$
    – Damien
    Dec 4 '18 at 6:05














  • 1




    $begingroup$
    $Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
    $endgroup$
    – Damien
    Dec 3 '18 at 23:16










  • $begingroup$
    @Damien I think even if no symmetric, it is an orthogonal base change
    $endgroup$
    – sleeve chen
    Dec 3 '18 at 23:24






  • 1




    $begingroup$
    You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
    $endgroup$
    – Damien
    Dec 4 '18 at 6:05








1




1




$begingroup$
$Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
$endgroup$
– Damien
Dec 3 '18 at 23:16




$begingroup$
$Q=Q^T$ i.e. Q symmetric seems sufficient, in addition to $Q$ orthogonal. $Q$ then corresponds to an orthogonal base change
$endgroup$
– Damien
Dec 3 '18 at 23:16












$begingroup$
@Damien I think even if no symmetric, it is an orthogonal base change
$endgroup$
– sleeve chen
Dec 3 '18 at 23:24




$begingroup$
@Damien I think even if no symmetric, it is an orthogonal base change
$endgroup$
– sleeve chen
Dec 3 '18 at 23:24




1




1




$begingroup$
You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
$endgroup$
– Damien
Dec 4 '18 at 6:05




$begingroup$
You are right of course. It was too late yesterday when I posted it. But the symmetry condition seems sufficient. Do you agree ?
$endgroup$
– Damien
Dec 4 '18 at 6:05










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