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Given a symmetric matrix $B \in \mathbb{C}^{n\times n}$.

How many coefficients of $A \in \mathbb{C}^{n\times n}$ can you obtain from the following equation? $$A^\top A=B$$

I think this problem is under determined? Isn't it? Sure $B$ must be symmetric. Thus only $\frac{n(n+1)}{2}$ complex equations remain for $n\times n$ coefficients.

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  • $\begingroup$ I'd strongly suggest using capital letters for matrices. Else $a^\top a=b$ looks like $a$ is a (column) vector and $b$ is a scalar. $\endgroup$
    – Leo
    Dec 11, 2014 at 14:33
  • $\begingroup$ @Leo Done!Is it OK now? Is $A^\top A$ clear? Means transposed matrix $A$ multiplied with matrix $A$. $\endgroup$
    – Matthias
    Dec 11, 2014 at 14:34
  • $\begingroup$ is $A$ unique? if $A$ is a solution isn't $AQ,$ for any orthogonal matrix $Q$ is also a solution? $\endgroup$
    – abel
    Dec 14, 2014 at 1:39
  • $\begingroup$ @abel seems to be so. $\endgroup$
    – Matthias
    Dec 14, 2014 at 17:03
  • $\begingroup$ i should have said if $A$ is a solution so is $QA$ for any orthogonal matrix $Q.$ not $AQ$ as i said in the previous comment. $\endgroup$
    – abel
    Dec 14, 2014 at 17:10

3 Answers 3

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Not completely sure about complex numbers, but... If $A^\top A=B$ then, $B$ is symmetric positive semi-definite. So you can take the eigenvalue/vector decomposition of $B$, such that $B=U\Lambda U^\top$. Then $A=(U\Lambda^{1/2})^\top$.

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    $\begingroup$ Nothing changes in $\mathbb{C}$, because the eigenvalues of $B$ are all real and nonnegative, so their square roots are real and nonnegative. $\endgroup$
    – Ian
    Dec 11, 2014 at 15:01
  • $\begingroup$ what about uniqueness? $\endgroup$
    – abel
    Dec 14, 2014 at 2:20
  • $\begingroup$ @abel That means that $A=U \left(\Lambda^{1/2}\right)^\top$ with arbitrary $U\in \text U(n)$ is the solution? $\endgroup$
    – Matthias
    Dec 14, 2014 at 19:46
  • $\begingroup$ you are using the same letter $U$ in two different ways. the solution is not unique. $\endgroup$
    – abel
    Dec 14, 2014 at 19:53
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From matrix analysis, if $B$ is complex-symmetric, there is a unitary matrix $U$ such that $B=UDU^{T}$ where the columns of $U$ are eigenvectors of $BB^{*} = B \bar{B}$ and $D$ is diagonal and the entries are the positive square roots of the corresponding eigenvalues. Now, if the columns of $U$ are real, then $U$ is orthogonal and so $B$ is orthogonally diagonalisable, and what I wrote then follows. So, if the eigenvectors of $B \bar{B}$ are real, then the eigenvalues of $B$ are real. The last part here may not be of any use but I think the former is.

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The davcha and Ian posts are false when $\mathbb{C}$ is the underlying field. Of course, the OP gave a green chevron to davcha, although his solution is valid only on R....

  1. Any complex matrix is similar to a symmetric matrix. Then a symmetric matrix may be non-diagonalizable !

  2. The real case is treated in: General Cholesky-like decomposition . cf. my post (which has not been read) where all solutions are determined.

  3. If $A$ is a complex matrix , then the eigenvalues of $A^TA$ have no special property and $B$ is certainly not symmetric positive semi-definite.

  4. About the OP's question. Using the Autolatry's post, we can find particular solutions. Indeed there is a SVD of $B$ in the form $B=UDU^T$ (beware $U^T\not= U^{-1}$). Let $\Delta$ be one of the square roots of $D$. Then the $A=\Delta U^T$ are particular solutions. Yet, there are many other solutions. For example, for a generic symmetric $B$, if $n=3$ (resp $n=4$) then the degree of freedom of the solutions is $3$ (resp. $6$).

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