MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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 $$z\begin{pmatrix} 1 & \pm i \\\\ \pm i & -1 \end{pmatrix},\qquad z\in\mathbb C.$$
share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.