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Post Closed as "Not suitable for this site" by Venkataramana, Ricardo Andrade, Willie Wong, Stefan Kohl♦, Jeremy Rickard
replaced inappropriate tag (since question will not be deleted)
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Ricardo Andrade
Ricardo Andrade
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Is the following assertion true? :

Suppose $p, q \geq 3$. Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplicationImultiplication. I want to show that $MA = A $ $MA = A$, where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course, showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?.

Is the following assertion true? :

Suppose $p, q \geq 3$. Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplicationI want to show that $MA = A $ , where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?.

Is the following assertion true?

Suppose $p, q \geq 3$. Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplication. I want to show that $MA = A$, where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course, showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?

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Vanya
Vanya
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Is the following assertion true? :

Suppose $p, q \geq 3$. Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplication. ImultiplicationI want to show that $MA = A $ , where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?.

Is the following assertion true? :

Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplication. I want to show that $MA = A $ , where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?

Is the following assertion true? :

Suppose $p, q \geq 3$. Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplicationI want to show that $MA = A $ , where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?.

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Vanya
Vanya
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Action of rotation group on Matrices

Is the following assertion true? :

Consider the action of $SO(p,\mathbb{R})$ on $p \times q$ matrices by left multiplication. I want to show that $MA = A $ , where $M \in SO(p,\mathbb{R})$ and $A \in M_{p \times q}(R)$, implies A = 0. Equivalently, if I show that the fiber over a non-zero matrix is a singleton, it seems okay. Of course showing it by choosing various matrices $M$ in $SO$ would do the job, but is there another proof using group actions?