Uniqueness of symmetric positive definite matrix decomposition 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.
 A: *

*If $K=AA^\mathrm{T}$ then $K=AUU^\mathrm{T}A^\mathrm{T}$ where $U$ is an arbitrary orthogonal matrix. Permutation of the columns of $A$ and changing the sign of the columns of $A$ are examples of this transform. If you disregard the dimensionality of $A$ you can also use $A'=\left[ A\ 0_{n\times m}\right]U$ with an orthogonal $U$ and obtain the same $K$.

*Positive-definite or positive-semidefinite doesn't make a difference.

*Fixing the number of columns is not enough because of the examples I mentioned in 1.

*Assuming that square root of $K$ is defined as a matrix $M$ such that $K=M^2=M\,M$, in general $A$ is not a square root. In fact $M=U\Lambda^{1/2} U^\mathrm{T}$ where $K=U\Lambda U^\mathrm{T}$ is the eigen-decomposition of $K$.

A: *

*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, $A \neq B$.
