Finding a block matrix factorization

Suppose it is known that a matrix $B$ of size $(ns)\times(ns)$ has the form $B=\begin{bmatrix} A_1 \\ \vdots \\ A_n \end{bmatrix}\begin{bmatrix} A_1^* & \cdots & A_n^* \end{bmatrix}$, where $A_1, \cdots, A_n$ are block matrices of size $s \times s$. How is it possible to determine these $A_i$'s?

It is stated in the book Positive Trigonometric Polynomials and Signal Processing Applications, 2nd Edition, by Bogdan Dumitrescu on page 270 that this can be done by eigenvalue decomposition or Cholesky factorization with pivoting. How to do this explicitly is not evident to me. I would be very grateful for assistance in understanding this. Thank you.

Thank you.

• Does $A_i^*$ mean the complex conjugate of $A_i$, the Hermitian adjoint of $A_i$, the transpose of $A_i$, or something else? – user5713492 Mar 12 '18 at 5:02
• $A_i^*$ is the conjugate transpose of $A_i$. – user49097 Mar 12 '18 at 5:29
• @Rodgrio de Azevedo Yes, I see that the rank is equal to the maximum rank of the block matrices. (Please see my updated post.) How exactly can this be done with Cholesky decomposition? Thank you. – user49097 Mar 14 '18 at 10:15
• Cholesky decomposition always yields a solution in which $n=1$, i.e. there exists a square matrix $A$ such that $B=AA^*$. However, this solution seems trivial for your requirement. – Saad Mar 14 '18 at 13:26
• @Alex Francisco Thanks, yes, I've now clarified this in the question statement. – user49097 Mar 14 '18 at 17:58

$\DeclareMathOperator{\diag}{diag}$ More generally, suppose it is known that $B$ is $N$-by-$N$ and positive semi-definite of rank $\le s$ (where $N$ is not necessarily a multiple of $s$.) Then there exists a unitary $N$-by-$N$ matrix $U$ such that $U^* B U = \diag(d_1, \dots, d_s, 0, 0, \dots)$, with $d_i \ge 0$. So $$B = U \diag(\sqrt{d_1}, \dots, \sqrt{d_s}, 0, \dots, 0)^2 U^*.$$ Take $A$ to be the first $s$ columns of $U \diag(\sqrt{d_1}, \dots, \sqrt{d_s}, 0, \dots, 0)$. Then $B = A A^*$.
Finally, the computation of $U$ is equivalent to computing an orthonormal basis of eigenvectors of $B$.
$A$ is not unique as it could be replaced by $A V$ where $V$ is $s$-by- $s$ unitary.