I recently created a related topic about the square root matrix, in case you'd like to refer to that one.

Here's what we want: Consider the matrix $\Omega=E(\mathbf{u}^{\top}\mathbf{u})$, where $\mathbf{u}$ is $n \times 1$, so that $\Omega$ is $n \times n$. Notice $\mathbf{u}^{\top}\mathbf{u}$ is symmetric and positive definite; therefore, $\Omega=E(\mathbf{u}^{\top}\mathbf{u})$ should also be symmetric and positive definite. Since it is symmetric and positive definite, therefore it is invertible, so $\Omega^{-1}$ exists. Moreover, $\Omega^{-1}$ is also symmetric and positive definite.

Next, we use the fact of the symmetry and positive definiteness of $\Omega^{-1}=\left[E(\mathbf{u}^{\top}\mathbf{u})\right]^{-1}$, to decompose it as \begin{equation} \Omega^{-1}=\Omega^{-1/2}\Omega^{-1/2} \end{equation} where the square root $\Omega^{-1/2}$ is also symmetric, and with the additional property that \begin{equation} \Omega^{-1/2}\Omega\Omega^{-1/2}=I \end{equation} It was this particular property that inspired my previous question.

My new question is this: what kind of matrix decomposition are we using here? From looking at the Wikipedia page for matrix decomposition, I think it is either

  • Cholesky Decomposition
  • Eigendecomposition
  • Jordan Normal Form Decomposition

which one, however, I am not sure.

By the way: this matrix decomposition into square root matrices is an application used in Generalized Least Squares, and there seems to be a lot of misunderstanding among those using it. The reason I say that is because a quick Google search shows that each author refers to what they're doing here as something different.

Some authors say they're doing Cholesky Decomposition; others talk about eigenvalues. Some authors even decompose $\Omega^{-1}=\Psi\Psi^{\top}$ (with transposes!), which seems quite unnecessary to me, since I believe the square root should be symmetric, also. The author I am choosing to follow writes $\Omega^{-1}=\Omega^{-1/2}\Omega^{-1/2}$ (without transposes), and I think this is optimal, since again, I believe $\Omega^{-1/2}$ is symmetric (though I don't know how to prove that).

So, can anyone set this straight? What kind of decomposition are we actually using here for this problem? Thank you for your thoughts!

  • $\begingroup$ Your decomposition, or the equivalent decomposition $\Omega = \Omega^{1/2} I \Omega^{1/2}$ is not a Cholesky decomposition, not an eigendecomposition, nor a Jordan decomposition. $\endgroup$ – Omnomnomnom Oct 24 '14 at 4:12
  • $\begingroup$ Thanks for your answer. This academic uses $\Omega^{-1}=\Omega^{-1/2}\Omega^{-1/2}$ and $\Omega^{-1/2}\Omega\Omega^{-1/2}=I$. I was really hoping to find out what kind of decomposition this is so I could learn more and understand it better. $\endgroup$ – Mathemanic Oct 24 '14 at 4:23
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    $\begingroup$ Since $\Omega$ is SPD, $\Omega^{-1/2}\Omega\Omega^{-1/2}=I$ is only a congruent transformation. Some authors might use the Cholesky decomposition $\Omega^{-1}=\Psi\Psi^T$ instead of the square root simply because they actually want to compute it (note that the Cholesky decomposition is much easier to obtain than the square root) in their algorithms. $\endgroup$ – Algebraic Pavel Oct 24 '14 at 11:02
  • $\begingroup$ I think this topic (math.stackexchange.com/questions/986805/…) answers why $\Omega^{-1/2}\Omega\Omega^{-1/2}=I$. However, I'm still not sure what kind of matrix decomposition into square roots this is, or how it works: $\Omega^{-1}=\Omega^{-1/2}\Omega^{-1/2}$ $\endgroup$ – Mathemanic Oct 25 '14 at 1:09
  • $\begingroup$ Omnomnomnom, if not Cholesky, Eigenvalue, or Jordan... what kind of decomposition is $\Omega^{-1}=\Omega^{-1/2}\Omega^{-1/2}$, where I want to make sure that my $\Omega^{-1/2}$'s are symmetric (which is not necessarily the case with Cholesky decomposition, correct? $\endgroup$ – Mathemanic Oct 26 '14 at 4:54

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