Expansion of the Kähler metric of Grassmannian / A simple question on linear algebra Here's a dumb question from a physicist. The problem comes from expanding the Kähler potential of the Grassmannian manifold in the vicinity of a certain point. 
Consider two matrices, $\{A^i_j\}_{N\times N}$ and $\{\tilde{A}^\alpha_\beta\}_{M\times M}$, defined as:
\begin{alignat}{9}
\begin{alignedat}{9}
A^i_j = \delta^i_j + \phi^i_\alpha\bar{\phi}^\alpha_j\\
\tilde{A}^\alpha_\beta = \delta^\alpha_\beta + \bar{\phi}^\alpha_i\phi^i_\beta
\end{alignedat}\quad,\qquad
\begin{alignedat}{9}
i,j&=1\ldots N\\
\alpha,\beta&=1\ldots M
\end{alignedat}
\end{alignat}
Here bar denotes Hermitean conjugation.
It's straightforward to show that $\operatorname{Tr{}} \ln(A) = \tilde{\operatorname{Tr{}}} \ln(\tilde{A}) \equiv K$, where the tilde over the trace reminds that the trace is taken over the greek indices. Indeed, since the logarithm is defined through a series, we have:
\begin{equation}
    \operatorname{Tr{}} \ln(A) = \operatorname{Tr{}} f_k (\phi^i_\alpha\bar{\phi}^\alpha_j)^k
    = f_k \operatorname{Tr{}} (\phi^i_\alpha\bar{\phi}^\alpha_j)^k
    = f_k \tilde{\operatorname{Tr{}}} (\bar{\phi}^\alpha_i\phi^i_\beta)^k
    = \tilde{\operatorname{Tr{}}} f_k (\bar{\phi}^\alpha_i\phi^i_\beta)^k
    = \tilde{\operatorname{Tr{}}} \ln(\tilde{A})
\end{equation}
Here $f_k$ are the coefficients in the Taylor expansion on the logarithm.
As I am constructing the small-$\phi$ expansion of the following quantity (the generalisation of the Fubini-Study metric):
$$
G_{i\beta}^{j\alpha}=
\dfrac{\partial}{\partial\phi^i_\alpha}\dfrac{\partial}{\partial\bar{\phi}_j^\beta} K
$$
Depending on which definition of $K$ I'm using, I'm getting different results. The definition above renders:
\begin{alignat}{9}
{}^{(I)}G_{i\beta}^{j\alpha}&=\operatorname{Tr{}} (A^{-1}) \delta_i^j \delta^\alpha_\beta
- \operatorname{Tr{}}(A^{-2}) \phi^j_\beta \bar{\phi}_i^\alpha \\
{}^{(II)}G_{i\beta}^{j\alpha}&=\tilde{\operatorname{Tr{}}} (\tilde{A}^{-1}) \delta_i^j \delta^\alpha_\beta
- \tilde{\operatorname{Tr{}}}(\tilde{A}^{-2}) \phi^j_\beta \bar{\phi}_i^\alpha
\end{alignat}
OK, looks like the quantities $\operatorname{Tr{}} (A^{-1})$  and $\tilde{\operatorname{Tr{}}} (\tilde{A}^{-1})$ should coincide. However:
\begin{alignat}{9}
(A^{-1})^i_j &\approx \delta^i_j - \phi^i_\alpha \bar{\phi}^\alpha_j \\
(\tilde{A}^{-1})^\alpha_\beta &\approx \delta^\alpha_\beta - \bar{\phi}^\alpha_i  \phi^i_\beta
\end{alignat}
Which gives for the traces:
\begin{alignat}{9}
\begin{alignedat}{9}
\operatorname{Tr{}} (A^{-1}) &= N - \phi^i_\alpha \bar{\phi}^\alpha_i \\
\tilde{\operatorname{Tr{}}} (\tilde{A}^{-1}) &= M - \phi^i_\alpha \bar{\phi}^\alpha_i
\end{alignedat}
\quad,\qquad
\begin{alignedat}{9}
\operatorname{Tr{}} (A^{-2}) &= N - 2\phi^i_\alpha \bar{\phi}^\alpha_i \\
\tilde{\operatorname{Tr{}}} (\tilde{A}^{-2}) &= M - 2\phi^i_\alpha \bar{\phi}^\alpha_i
\end{alignedat}
\end{alignat}
Clearly, the expansions ${}^{(I)}G_{i\beta}^{j\alpha}$ and ${}^{(II)}G_{i\beta}^{j\alpha}$ differ, and only match upon replacing $N$ with $M$.
What am I doing wrong?
 A: I already discussed this with you, but given there are upvotes on the question I'll write up an answer. Consider a matrix as a function of a variable $A(u)$. Now how do derivatives act on the powers $A^n$ (which appear as terms in power series expansions) ?
$$\partial_u A^n = A'A^{n-1}+A\,A'A^{n-2}+\dots+A^{n-1}A'$$
Now the derivative matrix $A'$ and $A$ don't necessarily commute, but if we take the derivative inside a trace there is no problem with the naive chain rule.
$$\partial_u\,Tr(A^n)=n\,Tr(A^{n-1}A').$$
But if we take a second derivative we start to see problems because for instance
$$Tr(A'A'A^{n-2})\neq Tr(A'AA'A^{n-3}).$$

In your case to be a little more specific where the manipulations fail I'll start calculating your $G$:
$$Tr(\partial _{\bar{\phi}^\beta_j}(\ln A)^{i'}_{j'})=Tr((A^{-1})^{i'}_{k'}\partial _{\bar{\phi}^\beta_j}A^{k'}_{j'})=Tr((A^{-1})^{i'}_{k'}\phi^{k'}_\beta\delta^j_{j'})=(A^{-1})^j_{k'}\phi^{k'}_\beta$$
Now I was able to use the chain rule because it was in a trace, but now I am left with a sum over a specific row of the inverse matrix. If I take the next derivative,
$$^{(I)}G^{j\alpha}_{i\beta}=(A^{-1})^j_i\delta^{\alpha}_{\beta}+\phi^{k'}_\beta\partial_{{\phi}^i_\alpha}(A^{-1})^j_{k'}.$$
The first term is a bit different from what you had, and the second I can't naively expand it as matrix elements of $A^{-2}$ in an obvious way. But if you expand it in terms of a power series you should be able to show it's equivalent to $^{(II)}G$, given that's how you demonstrated they were equivalent in the first place.
