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As part of proving this: Prove $\frac{\partial \rm{ln}|X|}{\partial X} = 2X^{-1} - \rm{diag}(X^{-1})$.

Prove

$$\frac{\partial \ln \lvert X \rvert}{\partial X_{ij}}=\text{tr} \left[X^{-1} \frac{\partial X}{\partial X_{ij}} \right]$$ where $\forall \ p \ \in \ \mathbb{N}, \ X \ \in \ \mathbb{R}^{p \times p}, \ $ X is a positive definite matrix.

Condition: Use adjoint. For no adjoint: Prove $\frac{\partial ln|X|}{\partial X_{ij}}=tr[X^{-1} \frac{\partial X}{\partial X_{ij}}]$ not using adjoint.

Note: just in case any of this notation seems wrong or or something, see matrix cookbook: p.15's (141), p.9's (57) and p.8's (43)

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1 Answers1

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What I tried based on p. 304-306 of Harville:

Since $\frac{\partial |F|}{\partial x_{ij}} = tr[adj(F) \frac{\partial F}{\partial x_{ij}}]$ (apparently this is Jacobi's formula),

where adj(F) is the adjoint of F, the tranpose of the cofactor matrix of F and adj(F) = $|F|F^{-1}$,

we have:

$\frac{\partial ln|X|}{\partial X_{ij}}$

$=\frac{1}{|X|}\frac{\partial |X|}{\partial X_{ij}}$

$=\frac{1}{|X|} tr[adj(X) \frac{\partial X}{\partial x_{ij}}]$

$=\frac{1}{|X|} tr[|X|X^{-1} \frac{\partial X}{\partial x_{ij}}]$

$=\frac{1}{|X|} tr[|X|X^{-1} \frac{\partial X}{\partial x_{ij}}]$

$=\frac{|X|}{|X|} tr[X^{-1} \frac{\partial X}{\partial x_{ij}}]$

$=tr[X^{-1} \frac{\partial X}{\partial x_{ij}}]$

QED

Harville, David. "Matrix Differentiation." Matrix Algebra From a Statistician's Perspective. New York: Springer-Verlag, 1997. 285-332. Print.

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  • 13,459