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[Background: Trying to build up my math base knowledge (self-taught) to follow an explanation on page 291 of this document, dealing with spherical parametrization to estimate unknown variance-covariance matrices. The author starts by giving a concrete covariance matrix].

We have the following matrix corresponding to a variance-covariance matrix:

$$\begin{bmatrix}1&1&1\\1&5&5\\1&4&14\end{bmatrix}$$ with a Cholesky decomposition,

$$L^{T}L=\begin{bmatrix}1&0&0\\1&2&0\\1&2&3\end{bmatrix}\begin{bmatrix}1&1&1\\0&2&2\\0&0&3\end{bmatrix}.$$

I need help with an explanation for the step that follows: The spherical coordinates of the elements of the Cholesky matrix are given as $l_i$, with $i$ corresponding to the first $i$ elements of the column $L_i$ of the Cholesky:

$[L_i]_1 = [l_i]_1 \,\,cos([l_i]_2)$

$[L_i]_2 = [l_i]_1\,\, sin([l_i]_2) \,\,cos([l_i]_3)$

$\dots$

$[L_i]_{i-1} = [l_i]_1 \,\,sin([l_i]_2)\,...\, cos([l_i]_i)$

$[L_i]_i = [l_i]_1 \,\,sin([l_i]_2)\,...\,sin([l_i]_i) $

What is the meaning of the "spherical coordinates"? And if at all possible, why is then that $\Sigma_{ii} = [l_i]_1^2$ and $\rho_{1i} = cos([l_i]_2)$, $i= 2,...,\,n$, where $\rho_{ij}$ denotes the correlation coefficient between $X_i$ and $X_j$?

This last, nested question may become self-evident after understanding spherical coordinates in this context.

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OK... the symbols are more disorienting than what it really is. The spherical coordinates of the Cholesky upper triangular matrix, $L$ can be understood in 3-dimensional space and then extrapolated to higher dimensions.

Let's look at the $L$ matrix as bound column vectors $L_i$ so that in the example above, there would be three column vectors ($L_1$, $L_2$ and $L_3$). The spherical coordinates in this case would require three parameters ($l_i$) for each of the column vectors: $[l_i]_1$, $[l_i]_2$ and $[l_i]_3$, denoting respectively: 1. The length of the column vector or norm ($||v||$ or $|v|)$ symbolized by $[l_i]_1$; 2. The angle between the column vector and the $x$ axis $[l_i]_2$ (named $\Omega$ on the illustration below); and 3. The angle between the projection of the column vector on the $(x, y)$ plane and the $y$ axis (name $\Phi$ in the illustration):

enter image description here

In this way the spherical projections boil down to basic trigonometry. The missing part was the "mystery" surrounding the $l_i$ parameters.

Solving it now, the first column of $L$, $L_1$, is simply $[1\,0\,0]^T$ and the parameters are $[l_1]_1 = 1$ because this is clearly the length of the vector; and $[l_1]_2 =\Omega= 0$ , because the vector lies in the $x$ axis. Therefore, the spherical coordinates are: $[L_1]_1= 1 \,\cdot cos([l_1]_2)\, = 1\,\cdot cos(0) =1$.

The second column vector has a norm of $\sqrt {1^2+2^2}$, which is $[l_2]_1$. In this case, $\Omega = [l_2]_2=arccos(\frac{1}{\sqrt{5}})=1.107$; and $\Phi = [l_2]_3=0$, because the vector is not in the $(y,z)$ plane. Just to check, the first entry of the second column vector, $[L_2]_1 = [l_2]_1\cos([l_2]_2)= \sqrt{5}\cdot\frac{1}{\sqrt{5}}=1$, and the second entry is, $[L_2]_2 = [l_2]_1\sin([l_2]_2\cos([l_2]_3)= \sqrt{5}\cdot sin(1.107)\cdot cos(0)=2.$

The third column vector, $L_3$, has parameters: $[l_3]_1=\sqrt{14}$; $[l_3]_2=\Omega=arccos\frac{1}{\sqrt{14}}=1.3$; and $[l_3]_3=\Phi=arctg\frac{3}{2}=0.983$. Verifying, the spherical coordinates would be: $[L_3]_1=\sqrt{14}\cdot cos(1.3)=\sqrt{14}\frac{1}{sqrt{14}}=1$; $[L_3]_2=\sqrt{14}\cdot sin(1.3)\cdot cos(0.983)=2$; and $[L_3]_3=\sqrt{14}\cdot sin(1.3)\cdot sin(0.983) =3.$

Therefore, the spherical coordinates of $L$ are:

$\begin{bmatrix}1&\sqrt{5}&\sqrt{14}\\ 0&1.107&1.3\\ 0&0&0.983\end{bmatrix}.$

As for the second part of the question, $\Sigma_{ii} = [l_i]_1^2$ because it is the square of the norm of the column vector in the Cholesky matrix, which is simply a dot product of any column vector of the Cholesky with itself.

$\rho_{1i} = cos([l_i]_2)$, $i= 2,...,\,n$, where $\rho_{ij}$ because the first column vector in the Cholesky is of the form $[[l_1]_1,0,0,...,n]$ and hence, the value of the angle $\Omega$ is zero, meaning that the angles of the cosines of the $\Omega$ angles of the other columns are really a projection of the any other column vector on the first one, or a dot product.

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