Rotation operator is positive.
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0
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I want to know if my proof is right.
the problem is:
Let V be $mathbb{R^2}$, with the standard inner product. If $theta$ is a real number, let T
be the linear operator 'rotation through $theta$,
$T_{theta}(x_1, x_2) = (x_l cos theta - x_2 sin theta, x_1 sin theta + x_2 cos theta).$
For which values of $theta$ is $T_{theta}$ a positive operator?
The conditions for $T_theta$ to be a positive operator are $T=T^*$ and $langle Talpha,alpharangle >0$.
First of all, I computed the associated matrix $[T]_b=begin{bmatrix} costheta & -sintheta \ sintheta & costheta
end{bmatrix}$.
The only way that $T=T^*$ is $sintheta=0$ or $theta=pi k$ with $kin mathbb{Z}$.
Then, $langle Talpha,alpharangle=costheta(x_1^2+x_2^2)$, to $langle Talpha,alpharangle$ be positive $theta$ must be in the first or fourth quadrant.
Hence, $theta=2pi k$ with $kin mathbb{Z}$.
Is anything wrong?
linear-algebra
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
add a comment |
up vote
0
down vote
favorite
I want to know if my proof is right.
the problem is:
Let V be $mathbb{R^2}$, with the standard inner product. If $theta$ is a real number, let T
be the linear operator 'rotation through $theta$,
$T_{theta}(x_1, x_2) = (x_l cos theta - x_2 sin theta, x_1 sin theta + x_2 cos theta).$
For which values of $theta$ is $T_{theta}$ a positive operator?
The conditions for $T_theta$ to be a positive operator are $T=T^*$ and $langle Talpha,alpharangle >0$.
First of all, I computed the associated matrix $[T]_b=begin{bmatrix} costheta & -sintheta \ sintheta & costheta
end{bmatrix}$.
The only way that $T=T^*$ is $sintheta=0$ or $theta=pi k$ with $kin mathbb{Z}$.
Then, $langle Talpha,alpharangle=costheta(x_1^2+x_2^2)$, to $langle Talpha,alpharangle$ be positive $theta$ must be in the first or fourth quadrant.
Hence, $theta=2pi k$ with $kin mathbb{Z}$.
Is anything wrong?
linear-algebra
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
2
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to know if my proof is right.
the problem is:
Let V be $mathbb{R^2}$, with the standard inner product. If $theta$ is a real number, let T
be the linear operator 'rotation through $theta$,
$T_{theta}(x_1, x_2) = (x_l cos theta - x_2 sin theta, x_1 sin theta + x_2 cos theta).$
For which values of $theta$ is $T_{theta}$ a positive operator?
The conditions for $T_theta$ to be a positive operator are $T=T^*$ and $langle Talpha,alpharangle >0$.
First of all, I computed the associated matrix $[T]_b=begin{bmatrix} costheta & -sintheta \ sintheta & costheta
end{bmatrix}$.
The only way that $T=T^*$ is $sintheta=0$ or $theta=pi k$ with $kin mathbb{Z}$.
Then, $langle Talpha,alpharangle=costheta(x_1^2+x_2^2)$, to $langle Talpha,alpharangle$ be positive $theta$ must be in the first or fourth quadrant.
Hence, $theta=2pi k$ with $kin mathbb{Z}$.
Is anything wrong?
linear-algebra
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
I want to know if my proof is right.
the problem is:
Let V be $mathbb{R^2}$, with the standard inner product. If $theta$ is a real number, let T
be the linear operator 'rotation through $theta$,
$T_{theta}(x_1, x_2) = (x_l cos theta - x_2 sin theta, x_1 sin theta + x_2 cos theta).$
For which values of $theta$ is $T_{theta}$ a positive operator?
The conditions for $T_theta$ to be a positive operator are $T=T^*$ and $langle Talpha,alpharangle >0$.
First of all, I computed the associated matrix $[T]_b=begin{bmatrix} costheta & -sintheta \ sintheta & costheta
end{bmatrix}$.
The only way that $T=T^*$ is $sintheta=0$ or $theta=pi k$ with $kin mathbb{Z}$.
Then, $langle Talpha,alpharangle=costheta(x_1^2+x_2^2)$, to $langle Talpha,alpharangle$ be positive $theta$ must be in the first or fourth quadrant.
Hence, $theta=2pi k$ with $kin mathbb{Z}$.
Is anything wrong?
linear-algebra
linear-algebra
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
edited Nov 18 at 2:44
Paul Sinclair
18.9k21440
18.9k21440
asked Nov 17 at 18:22
Bayesian guy
4710
asked Nov 17 at 18:22
Bayesian guy
4710
asked Nov 17 at 18:22
Bayesian guy
4710
4710
2
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47
add a comment |
2
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47
2
2
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47
add a comment |
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2
That's perfect. In other words, only the identity is positive among the rotations.
– Berci
Nov 17 at 18:47
Nice, thank you I didn't realize that!!
– Bayesian guy
Nov 18 at 3:47