# Finding a matrix by using hermitian

$A=\left[ \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 1 & i \\ 0 & -i & 1 \end{array} \right]$

Find a matrix $B$ such that $B^*B$ =$A$ (Star means conjugate transpose of $B$)

I think that $A$ is hermitian, and so $A^* =A$ Also, We can edit this as $BB^*=A$. But I couldn't solve it completely.. Thank you

$$(B^*B)^*=B^*B\neq BB^*$$
You can examine the $4$ and $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ separately since the off diagonal elements are zero. I was able to simply guess which matrix would give $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ when multiplied with its conjugate transpose.
$$B=\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & -i & 1 \\ \end{array} \right)$$