Derivative of Inverse Linear Transformation I am going through Sagle and Walde's Introduction to Lie Groups and Lie Algebras and am in the first chapter reviewing (advanced) calculus. I am mostly OK with the advanced calculus stuff, I have been through most of Widder and some of Spivak. There is an exercise in the differentiation section I don't quite get:

Show the map
$$
f: GL(V) \rightarrow GL(V) : T \rightarrow T^{-1}
$$
is differentiable at P and
$$
Df(P)T = -P^{-1}TP^{-1}.
$$
Thus show $f\in C^{1}(GL(V),End(V))$

I understand the question, however, I don't see how we can end up with a $P^{-1}$ multiplied on both sides. If I say:
$$
\begin{align}
f(P)&=P^{-1} \\
Pf(P)&=I \\
f(P) + PDf(P) &= 0 \\
PDf(P) &= -f(P) \\
PDf(P) &= -P^{-1} \\
Df(P) &= -P^{-2}
\end{align}
$$
With this derivative, we would get
$$
Df(P)T = -P^{-2}T
$$
Additonally, the derivative should be a linear transformation, and so it should have an inverse. In my case, that inverse is $P^{2}$. In the case of what the book claims, I don't see how there can be a single inverse independent of $T$.
I am guessing I am missing something because the book is Volume 51 in the Academic Press series Pure and Applied Mathematics, so I am sure a lot of good eyes were on it. Any clarification would be appreciated.
 A: Your argument is incorrect; you are treating this as if it were single-variable calculus and it is not. $Df(P)$ is a linear transformation from matrices to matrices and so the third line $f(P) + P D f(P) = 0$ where you treat $D f(P)$ as if it were just a matrix is incorrect. (Edit: more elaboration on this in the comments.)
Here is a sketch of a correct argument. The derivative of $f$ at $P$, by definition, is the linear map $Df_P(T)$ (I prefer this notation to yours) which satisfies
$$f(P + \varepsilon) = (P + \varepsilon)^{-1} = P^{-1} + Df_P(\varepsilon) + O \left( \| \varepsilon \|^2 \right)$$
where $\| \cdot \|$ denotes any matrix norm and $\varepsilon$ is a matrix of small norm, thought of as a tangent vector to $P$. Now we compute that
$$\begin{eqnarray*} (P + \varepsilon)^{-1} &=& (P(I + P^{-1} \varepsilon))^{-1} \\
 &=& (I + P^{-1} \varepsilon)^{-1} P^{-1} \\
 &=& \left( I - P^{-1} \varepsilon + \left( P^{-1} \varepsilon \right)^2 \mp \dots \right) P^{-1} \\
 &=& \left( I + P^{-1} \varepsilon + O \left( \| \varepsilon \|^2 \right) \right) P^{-1} \\
 &=& P^{-1} - P^{-1} \varepsilon P^{-1} + O \left( \| \varepsilon \|^2 \right) \end{eqnarray*}$$
which gives $Df_P(\varepsilon) = - P^{-1} \varepsilon P^{-1}$ as desired. In the third line we are using a geometric series expansion and the fourth line takes a bit more work to justify but follows from the same estimate for the usual geometric series together with the submultiplicativity of any matrix norm.
The derivative is only equal to $- P^{-2} \varepsilon$ if $P$ and $\varepsilon$ commute, and its inverse is $Df_{P^{-1}}(\varepsilon) = - P \varepsilon P$.
A: Let $DF_X$ denote the differential of $F$ at $X$, and $DF_X(H)$ its evaluation at $H$. We will also write $D^H_X[F(X)]$ to mean the same, the purpose of this notation being that we are differentiating "with respect to" $X$ and to remove the clutter that $(H)$ causes.
To begin, there is a fact that does not seem to be as well-known is it should be which I have taken to stating as follows:

*

*The derivative of an expression is the sum of the derivatives of it's subexpressions:
$$
  D_X[F(X,X)] = D_{\dot X}[F(\dot X, X)] + D_{\dot X}[F(X, \dot X)].
$$
On the RHS, the undotted $X$ should be "held constant"; it is not differentiated. On the LHS, both $X$s are differentiated.

Going step by step, we can see what goes wrong with your argument. First we apply the subexpression rule to $PP^{-1}$:
$$
  0 = D^H_P[PP^{-1}] = D^H_{\dot P}[\dot PP^{-1}] + D^H_{\dot P}[P\dot P^{-1}].
$$
The function $Q \mapsto QP^{-1}$ is linear, so it is its own differential and we get
$$
  HP^{-1} + D^H_{\dot P}[P\dot P^{-1}].
$$
This is where you went wrong. While it is true that $D^H_{\dot P}[\dot PP^{-1}] = D^H_P[P]P^{-1}$ via the chain rule, it is nonsensical to write $D_P[P]P^{-1}$ since $D_P[P]$ is a function $H \mapsto D^H_P[P]$ and $P^{-1}$ is a matrix. It is equally nonsensical to say that $D_P[P]$ is the identity matrix, which is what you did; it is instead the identity function on matrices. To avoid such pitfalls, it is generally a good idea to always evaluate your differentials on something like $H$.
Continuing with our derivation, we use the chain rule on the second term to get
$$
  0 = HP^{-1} + PD^H_P[P^{-1}].
$$
It now easily follows that
$$
  D^H_P[P^{-1}] = -P^{-1}HP^{-1}.
$$
