# 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:

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.

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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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@ S.B. Thanks a lot. Is it easy to prove that in part 1, the only other possibility is multiplication by -1? Why it is not possible for $AA^T =BB^T$, while dimension of $A, B$ are different? – user25004 Oct 16 '12 at 18:19
@user25004: To clarify that the multiplication by -1 and the column permutation I mentioned are the only options assuming that we consider matrices with fixed dimensions. Obviously, you can always put all-zero columns in $A$ which don't change the $AA^\mathrm{T}$. The simplest way to show this that I know is to use uniqueness of the eigen-decomposition up-to the permutation and the sign change. – S.B. Oct 16 '12 at 18:35
@ S.B. Now after reading more I think it is not completely true. Let $A=[3, 4]$, then $AA^T= 25$. Also if $B = [5]$, $BB^T = 25$. There can be infinitely many other matrices $M$ such that $MM^T=25$. – user25004 Oct 16 '12 at 23:03
@user25004: As I said in the comment above I assumed that we look at matrices of a given dimension. You're example compares a $2\times 1$ matrix with a $1\times 1$ matrix. – S.B. Oct 16 '12 at 23:15
How about $A=[3, 4]$, $B=[ \sqrt{24}, 1]$? $AA^T=BB^T$, and the dimension is the same. – user25004 Oct 16 '12 at 23:17
1. The Cholesky decomposition $K=AA^T$ of a positive definite matrix $K$ is unique when $A$ has positive diagonal entries.

2. In general, Cholesky decomposition of positive semi-definite matrix $K$ is not unique.

3. I don't understand question 3. Does't $A$ have the same size as $K$?

4. The square root of a positive (semi-)definite matrix $K$ is defined as a Hermitian matrix $B$, such that $K=BB$, so in general, $A \neq B$.

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Thanks. Cholesky decomposition returns lower triangular $A$. Apparently this constraint can make the solution unique for PSD matrices. – user25004 Oct 17 '12 at 16:00