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:
- suppose $Q^TQ = I$, i.e., $Q$ is orthogonal
- 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
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
$endgroup$
add a comment |
0
$begingroup$
I am curious about the following problem:
- suppose $Q^TQ = I$, i.e., $Q$ is orthogonal
- 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
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
$endgroup$
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
add a comment |
0
0
0
$begingroup$
I am curious about the following problem:
- suppose $Q^TQ = I$, i.e., $Q$ is orthogonal
- 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
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
$endgroup$
I am curious about the following problem:
- suppose $Q^TQ = I$, i.e., $Q$ is orthogonal
- 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
linear-algebra matrices orthogonal-matrices
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
asked Dec 3 '18 at 23:07
sleeve chensleeve chen
3,08141852
3,08141852
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
add a comment |
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
add a comment |
0
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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