Is the square root of a positive semi-definite matrix a unique result? I am trying to decompose a time series of $n$ observations  $\bf{\mathrm{v_c}}$ into the $n \times n$ variance-covariance structure $\sum$ and a random series $\bf{\mathrm{v}}$.
So, I can derive the variance-covariance matrix $\sum$ from the autocorrelation function of $\bf{\mathrm{v_c}}$.  This will be a Toeplitz matrix, which is positive semidefinite.  Therefore, I am able to compute a suitable matrix $\sum^{-\frac{1}{2}}$ to transform my correlated series into a random signal.   
$\bf{\mathrm{v}} = \sum^{-\frac{1}{2}}\bf{\mathrm{v_c}}$
I am able to do this using the sqrt(m) function in MATLAB, but can also find a Cholesky factorisation of the variance-covariance matrix and use this to induce the correlations.  However, I get different (but somewhat similar) results for the random series using the sqrtm and Cholesky methods.
I have read through several texts to determine how I might ascertain the square root of various matrices, and have looked at eigenvalue decomposition methods and so on.  I see there are only unique solutions under certain prescribed conditions - but I presume that these unique solutions are still only one of many roots?
My question is this:  is there any way to argue that one particular square root is preferable over another.  If not, is there a way to extract all possible solutions, such that all possible random functions may be obtained?   
 A: Let a matrix $\mathbb{V}$ have "square roots" $\mathbb{A}$ and $\mathbb{B}$; that is,
$$\mathbb{V = AA^\intercal = BB^\intercal}.$$
For simplicity, suppose the original matrix  $\mathbb{V}$ is invertible (which is equivalent to being positive definite under the assumptions).  Then $\mathbb{A}$, $\mathbb{B}$, and their transposes must also be invertible because
$$ \mathbb{I} = \mathbb{V^{-1}V} = \mathbb{V^{-1}AA^\intercal} = \mathbb{(V^{-1}A)A^\intercal}$$
exhibits a left inverse for $\mathbb{A}^\intercal$, implying $\mathbb{A}$ is invertible too; the same argument applies to $\mathbb{B}$, of course.  We exploit these inverses to write
$$\mathbb{(B^{-1}A)(B^{-1}A)^\intercal = B^{-1} (A A^\intercal) B^{-1}{^\intercal} =  B^{-1} (V) B^{-1}{^\intercal}=B^{-1}(BB^\intercal)B^{-1}{^\intercal} = I\ I = I},$$
showing that $\mathbb{O=B^{-1}A}$ is an orthogonal matrix: that is, $\mathbb{OO^\intercal=I}$.  The set of such matrices forms two smooth real manifolds of dimension $n(n-1)/2$ when $\mathbb{V}$ is $n$ by $n$. Geometrically, orthogonal matrices correspond to rotations or to a reflection followed by rotations, depending upon the sign of their determinant.
Conversely, when $\mathbb{A}$ is a square root of $\mathbb{V}$, similar (but easier) calculations show that $\mathbb{AO}$ is also a square root for any orthogonal matrix $\mathbb{O}$--and it does not matter here whether $\mathbb{A}$ is invertible or not.
It is also easy to see that multiplication by an orthogonal matrix (not equal to $\mathbb{I}$) really does alter the square root of an invertible matrix. After all, $\mathbb{AO = A}$ immediately implies $\mathbb{O = A^{-1}A = I}$.  This shows that the square roots of positive-definite matrices can be put into a one-to-one correspondence with the orthogonal matrices.
This demonstrates that square roots of positive-definite matrices are determined only up to multiplication by orthogonal matrices.  For the semi-definite case, the situation is more complicated, but at a minimum, multiplication by an orthogonal matrix preserves the property of being a square root.
If you wish to apply additional criteria to your square root you might be able to identify a unique one or at least narrow down the ambiguity: that will depend on your particular preferences.
