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? 



Subgradient of Minimum Eigenvalue
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).


