Prove that $A \circ B = AB$ if and only if both $A$ and $B$ are diagonal Definition. Hadamard product. Let $A,B \in \mathbb{C}^{m \times n}$. The Hadamard product of $A$ and $B$ is defined by $[A \circ B]_{ij} = [A]_{ij}[B]_{ij}$ for all $i = 1, \dots, m$, $j = 1, \dots, n$.
Remark. See details in this Hadamard product wiki article.
There is the following remark in Million's paper in Chapter 2:

We can relate the Hadamard product with matrix multiplication via
  considering diagonal matrices, since $A \circ  B = AB$ if and only if
  both $A$ and $B$ are diagonal.

So there is a theorem, that $A \circ  B = AB$ if and only if both $A$ and $B$ are diagonal, but I don't know how to prove it, and I didn't find it in the literature, because not many books have written in this topic.
Edit. In this theorem probably $m=n$. Million didn't write about it.

Proof: (assuming $n=m$)
($\Rightarrow$) by assumption, $A$ and $B$ are diagonal.  Then $AB = diag(a_{11}b_{11}, a_{22}b_{22}, \dots, a_{nn}b_{nn})$ and $[A\circ B]_{i,j} = [A]_{i,j} [B]_{i,j}$.  Since $A$ and $B$ are diagonal, then $[A\circ B]_{i,j} = 0$ for all $i\neq j$, and $[A\circ B]_{i,i} = a_{ii}b_{ii}$ $\Rightarrow$ $[A\circ B]_{i,j} = diag(a_{11}b_{11}, a_{22}b_{22}, \dots, a_{nn}b_{nn})=AB$
($\Leftarrow$) (retracted)
