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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$\begingroup$ Look up Cholesky decomposition $\endgroup$– Yoav KallusFeb 27, 2012 at 6:01
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2$\begingroup$ There can be many different looking symmetric matrices, each of whose square is the identity matrix. $\endgroup$– Yemon ChoiFeb 27, 2012 at 6:37
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3$\begingroup$ 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? $\endgroup$– Yemon ChoiFeb 27, 2012 at 7:35
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1$\begingroup$ To the OP: do you want all solutions, or just a possible solution? Are you perhaps also assuming P is positive? $\endgroup$– Yemon ChoiFeb 27, 2012 at 7:37
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4$\begingroup$ I really do not know where this habit of using $A^T$ or $A'$ where $A$ is symmetric comes from... $\endgroup$– Federico PoloniFeb 27, 2012 at 9:16
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1 Answer
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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 $$z\begin{pmatrix} 1 & \pm i \\\\ \pm i & -1 \end{pmatrix},\qquad z\in\mathbb C.$$
answered Feb 27, 2012 at 7:35