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.
The theorem you are trying to prove is not true:

Consider $A = B = \left(\begin{array}{cc}
1 &amp; 0\\
1 &amp; 0\end{array}\right)$.
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)