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?


1 Answer 1


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.

  • 6
    $\begingroup$ (+1) @hydrologist: As a complement to whuber's answer: One possible criterion that will lead to uniqueness is to insist on the square-root itself being positive semidefinite. Uniqueness of $\mathbb A$ then holds under the weaker condition that $\mathbb V$ is positive semidefinite. An instructive example to see what can "go wrong" is to look at the possible square roots of $\mathbb I$, even just the diagonal ones! :) $\endgroup$
    – cardinal
    Commented Mar 5, 2013 at 19:51
  • 1
    $\begingroup$ @cardinal: Thank you for your responses, which are most helpful and very much appreciated! $\endgroup$ Commented Mar 5, 2013 at 22:45
  • $\begingroup$ @whuber: Thank you again for your help. This has been most useful. $\endgroup$ Commented Mar 5, 2013 at 22:51
  • 1
    $\begingroup$ This is known as unitary freedom of square roots $\endgroup$ Commented Nov 24, 2018 at 2:14

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