Hadamard product and being unitary

Question

Let $A\in\mathbb{C}^{n\times n}$ and $A=B\circ B$ where $B$ is a unitary matrix and $\circ$ accounts for the Hadamard product. Can we say any thing about $A$ to be unitary or not?

Answer 1

No.

$B=\frac {1}{2}\begin{pmatrix}1+i&1-i\\1-i&1+i\end{pmatrix} $ is unitary .

It is your task to compute $A=B\circ B$ and prove that $\det(A)=0.$ Hence $A$ is not unitary.

Answer 2

Let $x$ a vector of $\mathbb{C}^n$. Suppose :

$$\| x \circ x \| = 1 = \|x\|.$$

With $\| \cdot \|$ the euclidean norm.

Note that $1 = \|x\| \implies |x_i| \le 1$

Hence :

$$\sum_i |x_i|^4=\sum_i |x_i^2|^2 = 1 = \sum_i |x_i|^2.$$

We deduce :

$$\sum_i |x_i|^2-\sum_i |x_i|^4=0.$$

But :

$$\sum_i |x_i|^2-\sum_i |x_i|^4= \sum_i |x_i|^2(1-|x_i|^2)=0.$$

Since $0 \le |x_i| \le 1$, the only solutions for $x$ is to have only one non-null component being $1$.

Hence, $A$ is unitary iff $B$ is the identity matrix (up to a column permutation).