Derivative of the equilibrium of a zero-sum game Let $n_1,n_2 \in \mathbb{N}$. Let $\triangle n$ be the $n$-dimensional standard simplex. Let $G \in \mathbb{R}^{n_1 \times n_2}$ be the payoff matrix of a zero-sum normal-form game. Then
\begin{align}
N_1(G) &= \operatorname*{argmax}_{s_1 \in \triangle n_1} \min_{s_2 \in \triangle n_2} G \cdot s_1 \cdot s_2 \\
N_2(G) &= \operatorname*{argmin}_{s_2 \in \triangle n_2} \max_{s_1 \in \triangle n_1} G \cdot s_1 \cdot s_2
\end{align}
are the maxmin and minmax strategies for each player. $N_1$ and $N_2$ are differentiable almost everywhere. How can I obtain the partial derivatives
\begin{align}
\frac{\partial N_1(G)}{\partial G} \qquad
\frac{\partial N_2(G)}{\partial G}
\end{align}
in terms of $G$, $N_1(G)$, and $N_2(G)$? I suspect implicit differentiation and the KKT conditions might be helpful. I know that the derivative of the equilibrium value is (see page 3, column 1)
\begin{align}
\frac{\partial N(G)}{\partial G}
&= \frac{\partial G \cdot N_1(G) \cdot N_2(G)}{\partial G} \\
&= \frac{\partial G : N_1(G) \otimes N_2(G)}{\partial G} \\
&= N_1(G) \otimes N_2(G)
\end{align}
where $:$ is the double dot product and $\otimes$ is the outer product. Assume the solution is not degenerate.
 A: If the game with matrix $\ G\ $ is non-degenerate, then there is a unique square submatrix $\ A\ $ of $\ G\ $ such that the matrix $\ \pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}\ $ is non-singular, and the corresponding subvectors, $\ \overline{N_1(G)}, 
\overline{N_2(G)}\ $, of the minimax strategies $\ N_1(G), N_2(G)\ $, and the value, $\ N(G)\ $ satisfy
\begin{align}
\pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}\pmatrix{\overline{N_2(G)}\\-N(G)}&= \pmatrix{\mathbb{0}\\1}\\
\pmatrix{A^\top&\mathbb{1}\\\mathbb{1}^\top&0}\pmatrix{\overline{N_1(G)}\\-N(G)}&= \pmatrix{\mathbb{0}\\1}
\end{align}
Differentiating the first of these equations with respect to $\ a_{ij}\ $ gives
\begin{align}
\pmatrix{E_{ij}&\mathbb{0}\\\mathbb{0}^\top&0} \pmatrix{\overline{ N_2(G)}\\-N(G)}+ \pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}\pmatrix{\frac{\partial \overline{ N_2(G)}}{\partial a_{ij}}\\-\frac{\partial N(G)}{\partial a_{ij}}}&= \pmatrix{\mathbb{0}\\0}\ ,
\end{align}
where $\ E_{ij}\ $ is the matrix whose $\ ij^\text{th}\ $ entry is $1$, and all of whose other entries are $0$. Therefore,
\begin{align}
\pmatrix{\frac{\partial \overline{ N_2(G)}}{\partial a_{ij}}\\-\frac{\partial N(G)}{\partial a_{ij}}}&= -\pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}^{-1} \pmatrix{E_{ij}&\mathbb{0}\\\mathbb{0}^\top&0} \pmatrix{\overline{ N_2(G)}\\-N(G)}\\
&=  -\pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}^{-1} \pmatrix{E_{ij}&\mathbb{0}\\\mathbb{0}^\top&0} \pmatrix{A&\mathbb{1}\\\mathbb{1}^\top&0}^{-1} \pmatrix{\mathbb{0}\\1}\ ,
\end{align}
and similarly
$$
\pmatrix{\frac{\partial \overline{ N_1(G)}}{\partial a_{ij}}\\-\frac{\partial N(G)}{\partial a_{ij}}}=   -\pmatrix{A^\top&\mathbb{1}\\\mathbb{1}^\top&0}^{-1} \pmatrix{E_{ji}&\mathbb{0}\\\mathbb{0}^\top&0} \pmatrix{A^\top&\mathbb{1}\\\mathbb{1}^\top&0}^{-1} \pmatrix{\mathbb{0}\\1}\  
$$
For entries $\ g_{ij}\ $ of $\ G\ $ lying outside the matrix $\ A\ $, $\ \frac{\partial N_1(G)}{\partial g_{ij}}=$$\frac{\partial N_2(G)}{\partial g_{ij}}=$$\mathbb{0}\ $.
