Subgradient of Minimum Eigenvalue Consider three $N \times N$ Hermitian matrices $A_0$, $A_1$, $A_2$. Consider the function
\begin{align}
f(t_1,t_2)=\lambda_{\text{min}}(A_0+t_1A_1+t_2A_2)
\end{align}
where $\lambda_{\text{min}}$ denotes the minimum eigenvalue. $f(t_1,t_2)$ is clearly a concave function. How do we find the sub-gradient of it? 
 A: Note that  $f(t)=g(A(t))$, where $A(t_1,t_2)=A_0+t_1A_1+t_2A_2$ is affine and $g(B)=\lambda_{min}(B)$ which is concave on $B$. Now, to obtain the supergradient of this function you just need to use the chain rule of convex analysis.
First, it is easy to see that the Jacobian of $A$ is
$$ D A(t) = A^{\ast} = [A_1^T;A_2^T]. $$
The maximum eigenvalue $g$ can be described by
$$ \lambda_{min}(B) = \min\{u^TBu:\,\, \|u\|_2=1 \}. $$
So its superdifferential is given by the convex hull of rank one matrices given by eigenvectors associated to the minimum eigenvalue, i.e.
$$ \partial \lambda_{min}(B) = \bar{co}\{ uu^T:\,\,u^Tu=1,\,\,Bu=\lambda_{min}(B) u \}.$$
Finally, by the chain rule,
$$ \partial f(t) = \langle[A_1^T;A_2^T],\partial \lambda_{min}(At)\rangle
      := \{ (\langle A_1^T,X\rangle,\langle A_2^T,X\rangle):\,\, 
              X\in \bar{co}\{ uu^T:\,\,u^Tu=1,\,\,A(t)u=\lambda_{min}(A(t)) u \}\}.$$
(the inner product above is the trace inner product)
Which is a set of linear operators from $\mathbb{R}^2$ to $\mathbb{R}$, as it should be.
You can derive chain rules for general affine maps: see Hiriart-Urruty & Lemarechal: Fundamentals of Convex analysis, Theorems D.4.2.1 and D.5.1 (for the maximum eigenvalue subdifferential).
