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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.

2 Answers 2

2
  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?2012-10-16
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    @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.2012-10-16
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    @ 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 $MM^T=25$.2012-10-16
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    @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.2012-10-16
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    How about $A=[3, 4]$, $B=[ \sqrt{24}, 1]$? $AA^T=BB^T$, and the dimension is the same.2012-10-16
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    @user25004: This happens because you allow a dimension of $A$ (or $B$) to be larger than $n$ the dimension of $M$. These examples do not give what are commonly known as the decomposition of a PSD matrix.2012-10-16
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    @user25004: You can generalize your examples by setting $A=CU$ and $B=CV$, where $U$ and $V$ are fat matrices with $UU^\mathrm{T}=VV^\mathrm{T}=I$.2012-10-16
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    Yes. Your generalization actually works. $U$ does not need to be fat. Let it be a square matrix. then it does not change the dimension, but still gives a new matrix.2012-10-16
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    @user25004: Now, I see the point. So, if $A$ is a solution so is $AU$ where $U$ is a orthogonal matrix. I'll correct the answer above.2012-10-16
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  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.2012-10-17