Derivative of trace of a product containing an inverse matrix

Question

What's the derivative of $$f(X)=\text{Tr}(YX^{-1})$$ with respect to $X$, where $X$ and $Y$ are square matrices of the same dimension?

My first attempt is to apply the chain rule as: Let $h(X)=X^{-1}$. So, $f(X)=\text{Tr}(Yh(X))$. Hence,

$$\frac{df(X)}{dX}=\frac{\partial f(x)}{\partial h(X)}\cdot\frac{\partial h(X)}{\partial X}= Y \cdot - X^{-2}=-YX^{2}$$

I'm not sure of my answer.

related: https://math.stackexchange.com/a/3330818/550103 – user550103 Aug 22 '19 at 20:39 — user550103, Aug 22 '19 at 20:39

score 1 · Answer 1 · answered Aug 22 '19 at 20:36

1

An alternative approach is to use a Taylor expansion approach $$ f(X + \varepsilon M) = tr( Y (X + \varepsilon M)^{-1}) = tr( Y X^{-1} (I + \varepsilon M X^{-1})^{-1}) $$ and $$ (I + \varepsilon M X^{-1})^{-1} = I - \varepsilon M X^{-1} + O(\varepsilon^2) $$ which yields $$ f(X + \varepsilon M) = f(X) - \varepsilon tr( Y X^{-1} M X^{-1} ) + O(\varepsilon^2) $$

answered Aug 22 '19 at 20:36

user7440

832

To complete this answer, one may observe that $\Delta f = \text{tr}[(-X^{-1} Y X^{-1})\Delta X]$. – Semiclassical Aug 22 '19 at 20:46
So the right answer is $-X^{-1}YX^{-1}$? – Guess601 Aug 22 '19 at 20:47
@Guess601: yes, the final answer is $-X^{-1} Y X^{-1}$ – user550103 Aug 23 '19 at 13:21
@user550103 You forgot to transpose. – Rodrigo de Azevedo Aug 25 '19 at 20:31
@RodrigodeAzevedo Thank you for the feedback. I was too quick to write that (not my usual layout). But now I think the answer was in different layout. So, I think both transposed and non-transposed are equivalent but in the different layouts. Agreed? – user550103 Aug 26 '19 at 04:01
It should be $$-X^{-T}Y^TX^{-T}$$ – MathLearner Mar 21 '23 at 16:34

Derivative of trace of a product containing an inverse matrix

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