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How to compute (directly from the definition) the derivative of matrix-valued function $M^{-1}(t)$ with respect to $t$ and recover the standard result $-M^{-1}(t)\frac{dM}{dt}M^{-1}(t)$?

A similar question has been asked on this site before involving this computation several times before, but without the restriction the proof come directly from the definition. In this case, I know how to give a proof using the fact that $M^{-1}(t)M(t) = I$, and applying the product rule. However, I would like to give an argument directly from the definition if possible.

A few hours of tinkering have led me nowhere fast - the issue is that the known formula is in terms of the derivative of $M$, and all methods I know of relating $M$ to $M^{-1}$, such as through the adjugate formula seem to be too ugly to recover the formula in question. Either I'm missing something, or this problem is difficult without the slick product-rule approach.

Since this is homework (of course, why else would such an arbitrary restriction be imposed on an otherwise fine argument?), a full solution is probably not necessary.

Rodrigo de Azevedo
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A. Thomas Yerger
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2 Answers2

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$$\left( \mathrm M (t + \mathrm d t) \right)^{-1} = \left( \mathrm M (t) + \dot{\mathrm M} (t) \,\mathrm d t \right)^{-1} = \cdots = \mathrm M^{-1} (t) \color{blue}{- \mathrm M^{-1} (t) \, \dot{\mathrm M} (t) \, \mathrm M^{-1} (t)} \,\mathrm d t$$

Rodrigo de Azevedo
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Let $M \, : \, t \in I \, \rightarrow \, M(t) \in \mathrm{GL}_{n}(\mathbb{R})$ be a differentiable matrix-valued function defined on an open interval $I$ of $\mathbb{R}$ and $\mathrm{Inv} \, : \, M \in \mathrm{GL}_{n}(\mathbb{R}) \, \rightarrow \, M^{-1}$ be the matrix inverse. You are interested in the derivative of:

$$ f = \mathrm{Inv} \circ M $$

It follows from the chain rule that for $t \in I$:

$$ f'(t) = \mathrm{D}_{M(t)} \mathrm{Inv} \cdot M'(t) $$

where $D_{M} \mathrm{Inv} \cdot H$ denotes the value at $H$ of the differential $D_{M} \mathrm{Inv}$ (differential of $\mathrm{Inv}$ at $M$).

You can show that (see this post, for example):

$$ \mathrm{D}_{M} \mathrm{Inv} \cdot H = - M^{-1} H M^{-1}. $$

As a result:

$$ \forall t \in I, \; f'(t) = - M(t)^{-1} M'(t) M(t)^{-1}. $$

pitchounet
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  • The chain rule does not work in this way for non-commuting objects. BTW why this old question popped on the front page? – lcv Dec 17 '19 at 15:28
  • @lcv Where is commutativity needed? For me, there is nothing wrong here. – pitchounet Dec 17 '19 at 16:21
  • $D_M$ Is a matrix, as well as Inv$M'$. Ask yourself why they should go in this order (and not the opposite one for example) – lcv Dec 17 '19 at 16:34
  • @lcv : $\mathrm{D}{M(t)}\mathrm{Inv}$ is a linear operator. It is applied to $M'(t)$. I cannot make sense of " $M'(t) \mathrm{D}{M(t)}\mathrm{Inv}$ ". – pitchounet Dec 17 '19 at 16:44
  • I think I only now understand your notation. My apologies. – lcv Dec 18 '19 at 09:12