If we suppose that:
$AR=RA$, for any rotation matrix $R$.
Then can we say that $A$ is invariant under rotation?
If so, why?
I tried to do it myself, but I can't seem to get there.
I know what invariant under rotations means, I don't know how I get from what I assumed to invariant under rotation. Or if it is even possible..
Let $A\in M_n(\mathbb{R})$ be s.t., for every $R\in O^+(n)$, $AR=RA$.
$\textbf{Proposition.}$ If $n=2$, then $A$ is in the form $\lambda S$ where $S\in O^+(2)$.
If $n=3$, then $A$ is a scalar matrix.
$\textbf{Proof.}$ $\bullet$ If $n=2$, then $A$ commute with the matrices $\begin{pmatrix}\cos(u)&-\sin(u)\\\sin(u)&\cos(u)\end{pmatrix}$, that is equivalent to $A$ commute with $U=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, that is equivalent to $A\in span(I_2,U)$.
$\bullet$ If $n=3$, let $v$ be a unitary vector. Since $A$ commute with the matrix $Rot(v,\pi/2)$, there is $\lambda$ s.t. $Av=\lambda v$. Thus, any non-zero vector is an eigenvector of $A$ and consequently, $A$ is a scalar matrix.