I know that, if a matrix commutes with all matrices, then it is a multiple of the identity; see here. The same conclusion holds if a (special) orthogonal matrix commutes with all (special) orthogonal matrices, as shown here and here.
In this context, I wonder if the following claim is true.
Claim: If a symmetric matrix $A$ commutes with every other symmetric matrix, does it then follow that $A = \lambda I$ for some $\lambda \in \mathbb{R}$?
Best Answer
Let $E_{ij}$ denote the matrix with a $1$ in the $(i,j)$-th place and $0$'s everywhere else. Then $E_{ii}$ is symmetric, and so from $$E_{ii}A=AE_{ii},$$ it follows that the $i$-th row and column are all zeros except in the $i$-th place. This shows that $A$ is diagonal.
Also $E_{ij}+E_{ji}$ is symmetric, and so from $$(E_{ij}+E_{ji})A=A(E_{ij}+E_{ji}),$$ it follows that the $i$-th and $j$-th diagonal entries of $A$ are the same. This shows that $A=cI$ for some constant $c$.
Note that none of this assumes that the matrices are real; this hold for matrices over any commutative ring.