# Thread: Total derivative of f(A)=A^-1 at I

1. ## Total derivative of f(A)=A^-1 at I

Work in M(n, R), the space of nxn matrices over the real numbers.

f(A)=A^-1

Find the total derivative at I, the identity matrix. That is, find an expression

f(I+H)-f(I)= linear in H + small (quadratic and above) in H

Thoughts: This is pretty easy to do if you assume you can expand (I+H)^-1 binomially, but a later question I have to do tells me to use Taylor's Theorem to prove that formula, so I obviously can't use it in this question.

So, how on earth do I find (I+H)^-1 -I without expanding the bracket out?

Thanks for any help. Also, sorry if this is in the wrong place. Couldn't decide if it belonged here or in Analysis.

2. I'd do it like so:
1. For arbitrary H and epsilon sufficiently close to 0,
$(I + \epsilon H)^{-1}(I+\epsilon H) = I.$

2. Taking a derivative of both sides with respect to epsilon,
$\left( \frac{d}{d\epsilon} \left[ (I+\epsilon H)^{-1} \right]\right)(I+\epsilon H) + (I+\epsilon H)^{-1}\frac{d}{d\epsilon}(I+\epsilon H) = 0$.

3. We can evaluate this equation at epsilon = 0 to get
$\frac{d}{d\epsilon}\left[ (I+\epsilon H)^{-1}\right] \Big\vert_{\epsilon = 0} = -H,$
where the LHS is conveniently the directional derivative of f:
$\nabla f(I) H = -H$.

4. Since this last equation holds for arbitrary H, we have the desired
$\nabla f(I) = -I.$

Of course, there are some technical details left to fill in: for instance in step 1 we assumed that the set of invertible matrices is open, and in step 3, that matrix inversion is smooth (follows from Cramer's Rule).