[Math] Uniqueness of symmetric positive definite matrix decomposition

hilbert-spaceslinear algebramachine learningmatricesnumerical linear algebra

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:

  1. Up to what transformation the above matrix decomposition is unique.

  2. Is positive definiteness (PD) and positive-semi definiteness (PSD) of $K$ makes difference in uniqueness of this decomposition?

  3. 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?

  4. $A$ 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

  1. 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 $K$.
  2. Positive-definite or positive-semidefinite doesn't make a difference.
  3. Fixing the number of columns is not enough because of the examples I mentioned in 1.
  4. 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 $K$.
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