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Given an arbitrary symmetric N-by-N matrix A, how can its original values be calculated from $P$?

$$ P = A'A$$

Both $A$ and $P$ have \( \frac{N^2-N}{2}+N \) degrees of freedom.

Edit: added the constraint that A is symmetric

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closed as too localized by Will Jagy, Yemon Choi, Loop Space, S. Carnahan Feb 27 '12 at 10:10

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Look up Cholesky decomposition – Yoav Kallus Feb 27 '12 at 6:01
There can be many different looking symmetric matrices, each of whose square is the identity matrix. – Yemon Choi Feb 27 '12 at 6:37
There seem to be two unknown (google)s in this discussion. Could at least one of you please take a moment to choose a pseudonym, or better still your actual name? – Yemon Choi Feb 27 '12 at 7:35
To the OP: do you want all solutions, or just a possible solution? Are you perhaps also assuming P is positive? – Yemon Choi Feb 27 '12 at 7:37
I really do not know where this habit of using $A^T$ or $A'$ where $A$ is symmetric comes from... – Federico Poloni Feb 27 '12 at 9:16

Well, that depends on the ground field.

  • If the field is $\mathbb R$, there exists a unique positive semidefinite square root $\sqrt P$. If $P$ is positive definite, then $A$ is any matrix of the form $AD$ where $D$ is co-diagonal to $A$ and is a sign-matrix, i.e $D^2=I_n$. If $P$ is singular, there might be other solutions, but at least the ones above are valid.
  • If the field is $\mathbb C$, the situation is much worse. For instance, if $P=0_2$, there are non-zero solutions $A$, say $$z\begin{pmatrix} 1 & \pm i \\\\ \pm i & -1 \end{pmatrix},\qquad z\in\mathbb C.$$
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