What is the Hessian of the spectral norm? The spectral norm of a symmetric matrix is the absolute value of the top eigenvalue. The gradient of this norm is $uu^T$ where $u$ is the eigenvector associated with that top eigenvalue.
Assume that $A$ has an isolated top eigenvalue. Then it is twice differentiable. What is its Hessian?
Any comments about how one generally computes Hessians of matrix norms are also appreciated! 
 A: Let us arrange all eigenvalues (and corresponding eigenvectors) in ascending order: $|\lambda_1| \leq \cdots |\lambda_{n-1}| < |\lambda_n|$. Then the elements of the desired Hessian are:
$$\frac{\partial^2 \|A\|_2}{\partial A_{kl}\partial A_{ij}} = \frac{\partial^2 \lambda_{n}}{\partial A_{kl}\partial A_{ij}} =  [\frac{\partial u^T_{n}}{\partial A_{kl}}]_i [u_{n}]_j +  [u^T_{n}]_i [\frac{\partial u_{n}}{\partial A_{kl}}]_j$$
Here we use the fact that $A$ is diagonalizable, i.e., the set of eigenvectors $u_i, \ i=1,\cdots, n$ form a orthogonal basis. The basis is used to decompose $\frac{\partial u_n}{\partial A_{kl}} = \sum_{m=1}^n c_m u_m$. To find the coefficients $c_m \in \mathbb{R}$ one could follow the perturbation analysis (https://en.wikipedia.org/wiki/Eigenvalue_perturbation).
We just give the results:
$$\frac{\partial u_{n}}{\partial A_{kl}} = \sum_{m=1}^{n-1} \frac{[u_m]_k[u_n]_l}{\lambda_n - \lambda_m} u_m$$
$$\frac{\partial u^T_{n}}{\partial A_{kl}} = \sum_{m=1}^{n-1} \frac{[u_m]_l[u_n]_k}{\lambda_n - \lambda_m} u_m$$
And the Hessian elements are:
$$\frac{\partial^2 \|A\|_2}{\partial A_{kl}\partial A_{ij}} = \sum_{m=1}^{n-1} \frac{[u_n]_k[u_n]_j[u_m]_l[u_m]_i + [u_n]_l[u_n]_i[u_m]_k[u_m]_j}{
\lambda_n - \lambda_m} $$ 
A: @Andrey's solution is very nice, but I found the wikipedia part a bit hard to read, so for completeness, I am adding a simplified proof here to determine $\frac{\partial [u_i]_l}{\partial A_{st}}$ where $u_i$ is the $i$th eigenvector of $A$.
Consider the $i$th eigenvalue/eigenvector pair of $A$, where $Au_i = \lambda_i u_i$. 
We use the perturbation analysis by taking the derivative of $Au_i=\lambda_iu_i$ via product rule
$$
 \partial A u_i + A\partial u_i = \partial \lambda_i u_i + \lambda_i  \partial u_i.
$$
Now $\partial u_i$ is a vector in $\mathbb R^n$, and we assume the eigenvectors $u_1,...,u_n$ are normalized, and thus form an orthonormal basis for $\mathbb R^n$. Therefore there exists $c_{ik}$ for $k = 1,...,n$ such that 
$$
\partial u_i = \sum_{k=1}^n c_{ik} u_k.
$$
We substitute, left-multiply by $u_j^T$, and simplify with $Au_k = \lambda_ku_k$ to get
$$
 u_j^T\partial A u_i +  c_{ij}\lambda_j  = \partial \lambda_i u_j^Tu_i + \lambda_i  c_{ij}.
$$
When $i = j$, this reduces to 
$$
 u_i^T\partial A u_i     = \partial \lambda_i 
$$
which you can then use to find the gradient of the spectral norm. When $i\neq j$, this reduces to
$$
\frac{ u_j^T\partial A u_i }{\lambda_i-\lambda_j }= c_{ij}.
$$
Expanding out gives
$$
\partial u_i = c_{ii}u_i + \sum_{k\neq i}  \frac{ u_i^T\partial A u_k }{\lambda_i-\lambda_k } u_k
= c_{ii}u_i + \sum_{k\neq i} \sum_{s,t=1}^n  \frac{ \partial A_{s,t}[u_i]_s[u_k]_t }{\lambda_i-\lambda_k } u_k
$$
Thus, we get
$$
\frac{\partial [u_i]_l}{\partial A_{s,t}} =\sum_{k\neq i} \frac{ [u_i]_s[u_k]_t[u_k]_l}{\lambda_i-\lambda_k }
$$
The rest follows from @Andrey's answer.
