Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

  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 $K$.
share|improve this answer
    
@ 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
show 4 more comments
  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, $A \neq B$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.