<s>[INST] Product of a matrix and its Hermitian transpose? Suppose I need to find a matrix B such that 
$B^H B  = A$
and $A = \begin{bmatrix}4 &0& 0\\  0 &1 &i\\ 0 &-i& 1\\\end{bmatrix}$
How do I proceed with a Product of a matrix and its Hermitian transpose ?
 [/INST] This will illustrate how to handle your specific, small matrix, but the other post points to a general and practical method.
Diagonalize A. It must have non-negative eigenvalues or $B^{H}B=A$ is not possible. The eigenvalues of $A$ are $4,2,0$. Check that $A$ has a basis of eigenvectors:
$$
    A\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right] = 4\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right],\;\;
    A\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right] = 2\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right],\;\;
    A\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right] = 0\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right]
$$
The eigenvectors are automatically orthogonal for distinct eigenvalues. Therefore, the normalized eigenvectors give a unitary transition matrix
$$
     U = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array}\right].
$$
That is $U^{H}U=UU^{H}=I$, and
$$
       U^{H}AU = \left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right],\;\;\; A = U\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} =
 U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
So one matrix that will work for $B$ is
$$
B=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
However, the better choice for $B$ may be the unique Hermitian matrix ($B=B^{H}$) with non-negative eigenvales for which $B^{2}=A$:
$$
B=U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
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