Let's say I have a function $f$ which maps any invertible $n\times n$ matrix to its inverse. How do I calculate the derivative of this function?
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Implicitly.${}$ – Ben Grossmann Oct 07 '14 at 17:52
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2Are you asking about the partial derivatives of the image $f(M)$ entries with respect to the entries of $M$, the matrix being inverted? There are various senses of differentiating a matrix-valued function, the simplest of which is for a matrix considered as a function of a single parameter, where the last section treats the specific case of a matrix inverse. – hardmath Oct 07 '14 at 17:55
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2I'm asking about the linear approximation, a linear transformation from $n\times n$ matrices to $n\times n$ matrices. – user181600 Oct 07 '14 at 17:57
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To calculate the directional derivatives of the map, at a fixed matrix $A$:
Suppose you change entry $i,j$ of your matrix $A$ by adding a small $t$ (small enough that the result is still invertible). Let $E_{i,j}$ be the matrix with a $1$ in the $i,j$th position and $0$ elsewhere.
Note that
$$ (A + tE_{i,j})(A + tE_{i,j})^{-1} = I $$
so on one hand, the derivative with respect to $t$ is zero (as the identity is constant), on the other hand, the derivative can be written
$$ \frac{d(A + tE_{i,j})}{dt}(A + tE_{i,j})^{-1} + (A + tE_{i,j})\frac{d(A + tE_{i,j})^{-1}}{dt} = 0 $$
so $$ \frac{d(A + tE_{i,j})^{-1}}{dt} = -(A + tE_{i,j})^{-1}\frac{d(A + tE_{i,j})}{dt}(A + tE_{i,j})^{-1}\\ = -(A + tE_{i,j})^{-1}E_{i,j}(A + tE_{i,j})^{-1} $$
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$$(X+\delta)^{-1} = [X\cdot ( I + X^{-1} \cdot \delta)]^{-1}= ( I + X^{-1} \cdot \delta)^{-1}\cdot X^{-1} = \\ =(I - X^{-1} \delta + X^{-1} \delta X^{-1} \delta - \ldots ) \cdot X^{-1}= X^{-1} - X^{-1} \delta X^{-1} + X^{-1} \delta X^{-1} \delta X^{-1} - \ldots $$
so the linear part of the variation is $-X^{-1} \delta X^{-1}$.
We also have the Taylor series expansion at $X$.
orangeskid
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