TITLE: Uniqueness of symmetric positive definite matrix decomposition
QUESTION [6 upvotes]: We know that any symmetric positive semi-definite matrix $K$ can be written as $K= AA^T$, where $A$ has real components. 
One way to get to $A$ is to compute eigen value decomposition of $K= P^T DP$ and define $A= P^T \sqrt{D}$, where $\sqrt{D}$ simply computes the square roots of diagonal elements.
Now, I wonder to what extent such a decomposition is unique. Of course if $AA^T=K$ then $-A$ also works. 
My questions are:

Up to what transformation the above matrix decomposition is unique.
Is positive definiteness (PD) and positive-semi definiteness (PSD) of $K$ makes difference in uniqueness of this decomposition? 
To have a unique solution, do we need to fix the number of columns of $A$ (for a PSD or PD matrix)? Is the decomposition unique only if we are given this dimension?
$A$ is different from square root of $K$, right? Because square root does not have to be symmetric?!

Answering any part will be useful for me. Specially part 2.

REPLY [2 votes]: The Cholesky decomposition $K=AA^T$ of a positive definite matrix $K$ is unique when $A$ has positive diagonal entries.
In general, Cholesky decomposition of positive semi-definite matrix $K$ is not unique.
I don't understand question 3. Does't $A$ have the same size as $K$?
The square root of a positive (semi-)definite matrix $K$ is defined as a Hermitian matrix $B$, such that $K=BB$, so in general, $A \neq B$.