Let $A$ be an invertible $n \times n$ matrix over $\mathbb{C}$.

If the $k$-th power $A^k$ of $A$ and the $k$-th power  $A^{\circ k}$ of Schur product of $A$ equals

($k=1,2,\ldots,n+1$), then $A$ becomes diagonal. 

In the case that $A$ is an invertible bounded linear operator on an infinite dimensional Hilbert space $H$,

we can also define Schur product of operators,  and we can show that $A$ is diagonal,

if it satisfies $A^k = A^{\circ k}$ for any $k=1,2,\ldots .$