Derivative of eigenvectors of a matrix with respect to its components Suppose that $B$ is a real, positive-definitive symmetric ($3\times3$) matrix (more accurately, $B$ is a tensor) with distinct eigenvalues, and that we can write it as 
$$
B= \sum_{i=1}^3 \lambda_{i}(n_{i}\otimes n_{i}),
$$
where $ n_{i} $ and $ \lambda_{i} $ are the unit eigenvectors and eigenvalues of $B$, and $\otimes$ is the dyadic product.
I need to calculate the derivative of eigenvectors of $B$ with respect to its components; in other words:
 $$
\frac{\partial n_{i}}{\partial B}
$$
I read some sources such as Kato's Perturbation theory but I couldn't figure out how to solve this, so I'm looking for pointers to a closed form solution.
 A: first order perturbation theory in $\epsilon$ on the matrix B with elements $B_{nm}+\epsilon \delta_{nk}\delta_{ml}$ tells you that
$$\frac{d\vec{n}_i}{dB_{kl}}=\sum_{j\neq i}\frac{(\vec{n}_j)_k (\vec{n}_i)_l}{\lambda_i-\lambda_j}\vec{n}_j$$
with $(\vec{n}_i)_l$ the $l$-th component of the vector $\vec{n}_i$.
A: If $B$ depends on a single parameter $t$ then derivating  with respect to $t$  the equality
$$ B n_i =\lambda_i n_i $$
we deduce
$$\dot{B} n_i +B\dot{n}_i=\dot{\lambda}_i n_i +\lambda_i\dot{n}_i. $$
Here we assume that $\Vert n_i\Vert =1$.  Hence $\dot{n}_i\perp n_i$, $\forall i$. Taking the inner product of  the above equality with $n_i$  and observing that
$$ (B\dot{n}_i, n_i)=(\dot{n}_i, Bn_i)=\lambda_i(\dot{n}_i,n_i)=0 \tag{1} $$
we deduce
$$\boxed{\dot{\lambda}_i=(\dot{B}n_i,n_i).} $$
This determines $\dot{\lambda}_i$ in terms of $\dot{B}$. 
Next, we take the inner product of (1)  with $n_j$, $j\neq i$.  Using the fact that $B$ is symmetric we deduce
$$(\dot{B}n_i, n_j)+(\dot{n}_i, Bn_j)=\lambda_i (\dot{n}_i, n_j) $$
so that
$$(\dot{B}n_i, n_j)+\lambda_j(\dot{n}_i, n_j)=\lambda_i (\dot{n}_i, n_j) $$
This shows that
$$(\dot{n}_i, n_j)=\frac{1}{\lambda_i-\lambda_j}(\dot{B}n_i, n_j), $$
that is
$$\dot{n}_i=\sum_{j\neq i} \frac{1}{\lambda_i-\lambda_j}(\dot{B}n_i, n_j) n_j.\tag{2} $$
If we let the parameter  $t$ to be  entry $b_{k\ell}$, $k\leq \ell$,  of $B$  with respect to some fixed orthonormal basis we deduce
$$\boxed{\frac{\partial n_i}{\partial b_{k\ell}}=\sum_{j\neq i} \frac{1}{\lambda_i-\lambda_j}\Biggl(\frac{\partial B}{\partial b_{k\ell}}n_i, n_j\Biggr) n_j.} $$
This is equivalent with the formula of Carlo Beenakker.
