Question

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

Answer 1

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 $$z\begin{pmatrix} 1 & \pm i \\ \pm i & -1 \end{pmatrix},\qquad z\in\mathbb C.$$

Answer 2

Since both A and P are symmetric,

$$P=A'A=AA$$ $$A=Q_1D_1Q_1'$$ $$P=Q_2D_2Q_2'$$ $$Q_2D_2Q_2'=Q_1D_1Q_1'Q_1D_1Q_1'$$ $$Q_2D_2Q_2'=Q_1D_1^2Q_1'$$ where $Q_1$ is the eigenvectors of $A$, $Q_2$ is the eigenvectors of $P$, and $D_1$ and $D_2$ are diagonal.

Thus,

$$Q=Q_1=Q_2$$ $$D_2 = D_1^2$$ $$A = Q\sqrt{D_2}Q'$$