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To prove the differentiability of a function defined on $M\in M_n$, how should I adapt the definition of differentiability?

$h^{-1}[f(x+h)-f(x)-hL(x)]\to 0$ 

How do I take the reciprocal of a matrix? Surely it can't be taking the inverse.

Also, are there some commonly used ways to show differentiability of a function (defined on the space of matrices) at some matrix?

Thank you.

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    What is "$M_n$" here? In my head it's a matrix space, thus it doesn't contain functions. Perhaps in your head you're associating it with a function space : maybe you should clear it up so that we can better answer.2011-11-23

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For a function defined on a finite dimensional normed vector space (one can also extend this to the Fréchet derivatives on Banach spaces), we can think of the derivative as a linear approximation to the function. Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be Banach spaces, a function $f:X\to Y$ is said to be differentiable at a point $x_0\in X$ if there is a bounded linear transformation $L_{x_0}:X\to Y$ such that the limit

$ \lim_{X\ni h \to 0} \frac{\| f(x_0 + h) - f(x_0) - L_{x_0}(h) \|_Y}{\|h\|_X} = 0 $

Now. In your case, you take $X = M_n$, the space of matrices. You pick a norm $\|\cdot\|$ (any matrix norm will do: since $M_n$ is a finite dimensional vector space, all norms are equivalent). Your function takes values, I guess, in $\mathbb{C}$. So we will just use the usual norm $|\cdot|$ on it. Linear maps between finite dimensional normed spaces are automatically bounded, so we don't have to worry about that adjective.

So be more explicit. Let us choose the Frobenius norm for your matrices. Then the condition for differentiability at $M\in M_n$ is that there exists a linear map $L_{M}: M_n \to \mathbb{C}$ such that

$ \lim_{M_n\ni H \to 0} \frac{ \left| f(M + H) - f(M) - L_M(H) \right|}{\sqrt{\sum_{1\leq i,j\leq n} |h_{ij}|^2}} = 0$

where $(h_{ij})$ are the matrix entries of $H$.