How do I calculate the derivative of matrix?
Matthew Barrera 
I'd like to expand a real, symmetric and positive definite Matrix $M$ into a Taylor series. I'm trying to do
$$ M (T) = M(T_0) + \frac{\partial M}{\partial T} (T-T_0) + \cdots$$
$T$ is a algebraic vector of parameters (e.g. temperatures at finite element nodes). I'm only interested in the first order term, i.e. the derivative of $M$ at $T_0$. My Professor tried to do this, but used unclear notation.
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$\begingroup$The derivative of $M(T)$ is simply the component-wise derivative. In the $2\times2$ case: $$\frac{d}{dT}\left(\matrix{a(T) & b(T) \cr c(T) & d(T)}\right) = \left(\matrix{a'(T) & b'(T) \cr c'(T) & d'(T)}\right)$$
$\endgroup$ $\begingroup$If $M$ is your matrix, then it represents a linear $f\colon \mathbb{R}^n \to \mathbb{R}^n$, thus when you do $M(T)$ by row times column multiplication you obtain a vectorial expression for your $f(T)$. Thus $\frac{\partial M}{\partial T}$ is just the derivative of the vector $MT$, which you do component-wise. Were you looking for something different?
EDIT: I actually see now that you most likely have a vector space of functions, but this doesn't change much at all: see that if $T = (f_1(t),f_2(t))^T$ and $M$ represents a linear homomorphism $F\colon V \to V$, then you have that $\frac{dF}{dt}(f_1(t),f_2(t))^T$ is actually $F(\frac{df_1(t)}{dt}, \frac{df_2(t)}{dt})$. This is actually straight forward to see: just compute $MT$ by row $\times$ column multiplication and then derive with respect to $t$.
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