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:
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Up to what transformation the above matrix decomposition is unique.
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Is positive definiteness (PD) and positive-semi definiteness (PSD) of $K$ makes difference in uniqueness of this decomposition?
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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?
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$A$ 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.
Best Answer