$f: GL(2, \mathbb{R}) \to GL(2, \mathbb{R)}, \ A\mapsto A^{-1}$. Looking for $df(A)$

Question

Let $f: GL(2, \mathbb{R}) \to GL(2, \mathbb{R)}, \ A\mapsto A^{-1}$

What's $df(A) (\in linear(\mathbb{R^4,\mathbb{R^4)\cong \mathbb{R}^{4\times4}}}$?

If I look at $A \in \mathbb{R}^{2 \times 2}$ as a vector:

$\begin{pmatrix}A_{11}\\A_{12}\\A_{21}\\A_{22}\end{pmatrix}$

Then $f(A)=\frac{1}{A_{11}A_{22}-A_{A12}A_{21}} \begin{pmatrix}A_{22}\\-A_{21}\\-A_{12}\\A_{11}\end{pmatrix}$.

So $\frac{\partial f(A)}{\partial A_{11}}=\begin{pmatrix}0\\0\\0\\\frac{-A_{22}}{(A_{11}A_{22}-A_{12}A_{21})^2}\end{pmatrix}$

$\frac{\partial f(A)}{\partial A_{12}}=\begin{pmatrix}0\\0\\\frac{-A_{21}}{(A_{11}A_{22}-A_{12}A_{21})^2}\\0\end{pmatrix}$

$\frac{\partial f(A)}{\partial A_{21}}=\begin{pmatrix}0\\\frac{-A_{12}}{(A_{11}A_{22}-A_{12}A_{21})^2}\\0\\0\end{pmatrix}$

$\frac{\partial f(A)}{\partial A_{22}}=\begin{pmatrix}\frac{-A_{11}}{(A_{11}A_{22}-A_{12}A_{21})^2}\\0\\0\\0\end{pmatrix}$

So is $df(A)$ the row-vector consisting of this 4 matrices?

Answer 1

$df(A)$ is a linear application from $M(2,\mathbb R)$ into $M(2,\mathbb R)$, where $M(2,\mathbb R)$ is the vector space of $2 \times 2$ real matrices. If you denote an element (a matrix) of $M(2,\mathbb R)$ as a vector with $4$ dimensions, then $df(A)$ is the $4 \times 4$ matrix whose columns are the vectors you computed.

Answer 2

You're not wrong(at least in principle, I didn't check the calculations), but you can also consider that if $g(A)=A$ is the identity map on the matrix space $\mathbb R^{n\times n}$, then the product $f(A)g(A) = I$, so that by product rule, $$0 = d(fg)(A)H = (df(A)H) g(A) + f(A)(dg(A)H) = df(A) HA +A^{-1}H,$$ giving $$ df(A) H= -A^{-1}H A^{-1}.$$