# Matrix Calculus: Derivative of Vectorized Symmetric Positive Definite Matrix w.r.t. its Vectorized Matrix Logarithm

Setup:

Let $k\in{}\mathbb{N},$ and let $\mathrm{M}_{k,k}(\mathbb{}{R})$ denote the set of $k\times{}k$ matrices over the field of real numbers.

Let $X\in{}\mathrm{M}_{k,k}(\mathbb{}{R})$ be a symmetric, positive definite matrix.

Then $X$ has $k$ positive eigenvalues $\lambda_{1},\dots{},\lambda_{k}$ with corresponding eigenvectors $v_{1},\dots{},v_{k}$.

The eigendecomposition/spectral decomposition of $X$ is:

$$X = V\Lambda{}V^{-1} = V\Lambda{}V',$$

where $\Lambda{}=\mathrm{diag}(\lambda_{1},\dots{},\lambda_{k})\in{}\mathrm{M}_{k,k}(\mathbb{}{R})$ is the diagonal matrix with the $k$ eigenvalues on the main diagonal and $V=(v_{1},\dots{},v_{k})\in{}\mathrm{M}_{k,k}(\mathbb{}{R})$ is the matrix whose $k$ columns are the orthonormal eigenvectors.

We define the natural matrix logarithm of $X$, denoted $\log{}(X)$, to be $$\log{}(X)=V\log{}(\Lambda{})V',$$ where $\log{}(\Lambda{})=\mathrm{diag}(\log{}(\lambda_{1}),\dots{},\log{}(\lambda_{k}))\in{}\mathrm{M}_{k,k}(\mathbb{}{R})$.

Question:

What, if it can be found, is the analytical form of the $[k(k+1)/2] \times{} [k(k+1)/2]$ Jacobian matrix

$$\frac{\partial{}\mathrm{vec}(X)}{\partial{}\mathrm{vec}(\log{}(X))'}$$

where $\mathrm{vec}(\cdot{})$ is the half-vectorization operator that stacks the lower triangular part of its square argument matrix.

(Background: This is a recurring problem in multivariate statistics when one adopts a "log-parameterization" of a covariance or precision matrix, which are both, by definition, symmetric and positive (semi-)definite.)

• I have replaced in you very well presented question all "vech"s by "vec"s. Feb 10, 2017 at 19:57
• @JeanMarie: It is quite common that $\mathrm{vech}(\cdot{})$ refers to the half-vectorization operator, whereas $\mathrm{vec}(\cdot{})$ refers to the full vectorization operator. As both $X$ and $\log{}(X)$ are symmetric, I am interested in $\partial{}\mathrm{vech}(X)/\partial{}\mathrm{vech}(\log{}(X))$.
– MiMa
Feb 11, 2017 at 12:49
• Maybe to see this better, note that $\partial{}\mathrm{vech}(X)/\partial{}\mathrm{vech}(\log{}(X))'$, is a $[k(k+1)/2]\times{}[k(k+1)/2]$ matrix, whereas $\partial{}\mathrm{vec}(X)/\partial{}\mathrm{vec}(\log{}(X))'$ would be a $k^{2}\times{}k^{2}$ matrix (But I am quite new to this forum and not fully familiar with the editing/changing questions system yet, so apologies if there are mistakes)
– MiMa
Feb 11, 2017 at 12:57

Let \eqalign{ G &= \log(X) \implies X = e^G \cr } and find the differential of $$X$$ via the power series of the exponential \eqalign{ dX=de^G &= d\,\Bigg[\sum_{i=0}^\infty \frac{G^i}{i!}\Bigg] \cr &= \sum_{i=1}^\infty \frac{1}{i!}\,\sum_{j=0}^{i-1}\,G^{j}\,dG\,G^{i-j-1}\cr } Now apply vectorization \eqalign{ dx &=\Bigg[\sum_{i=1}^\infty \frac{1}{i!}\,\sum_{j=0}^{i-1}G^{i-j-1}\otimes G^{j}\Bigg]\,dg \cr \frac{\partial x}{\partial g} &= \sum_{i=1}^\infty \frac{1}{i!}\,\sum_{j=0}^{i-1}\,\Big[G^{i-j-1}\otimes G^{j}\,\Big] \cr\cr } To change between vectorization and half-vectorization, multipy by Duplication and Elimination matrices of the appropriate dimensions \eqalign{ L_k\,dx &= L_k\Bigg[\frac{\partial x}{\partial g}\Bigg]D_kL_k\,dg \cr dx^\prime &= L_k\Bigg[\frac{\partial x}{\partial g}\Bigg]D_k\,dg^\prime \cr \frac{\partial x^\prime}{\partial g^\prime} &= L_k\Bigg[\frac{\partial x}{\partial g}\Bigg]D_k \cr \cr\cr } P.S.
The reason I used $$G$$ for the log, was because I wanted to use $$L$$ for the elimination matrix.
In the vectorization of the power series I used the fact that $$G$$ is symmetric to omit some transpose operations.