Action of rotation group on Matrices [closed]

Question

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?

Answer 1

The only vector fixed by all rotations in $\mathbb{R}^p$ for $p\geq 2$ is the zero vector. This then implies your result (for $p\geq 2$ and $q \geq 1$), since the action of $SO(p)$ on the matrices is such that every column transforms as a vector. Hence if $A$ is fixed, every column of $A$ is fixed, so every column must be zero.