Derivative of eigenvalues w.r.t. a tensor

$E$ is a real, positive-definitive 3x3 symmetric tensor (I am thinking about the strain tensor in solid mechanics). We perform eigendecomposition and get:

$$E_p=\sum_{i=1}^{3}λ_iN_i⊗N_i$$

into its principal components, where $λ_i$ are its eigenvalues and $N_i$ are its eigenvectors. My question is, is there a closed form solution for $\frac{∂λ_i}{∂E}$ ?

Let $p(\lambda)$ be the characteristic polynomial $p(\lambda)=\det(E-\lambda I)$. Then $p(\lambda)=(\lambda_1(E)-\lambda)(\lambda_2(E)-\lambda)(\lambda_3(E)-\lambda)$. Differentiate in $E$ and then set $\lambda=\lambda_i(E)$: $$\frac{\partial \lambda_i}{\partial E} = \frac{1}{(\lambda_j-\lambda_i)(\lambda_k-\lambda_i)} \left.\frac{\partial \det(E-\lambda I)}{\partial E}\right|_{\lambda=\lambda_i(E)}$$ if $\{\lambda_1,\lambda_2,\lambda_3\}=\{\lambda_i,\lambda_j,\lambda_k\}$.
• Doing the calculation I find a difference in the sign of the term $\frac{1}{(\lambda_i-\lambda_j)(\lambda_i-\lambda_k)}$ but i do not understand the reason.Regarding the term $\frac{\partial det(E-\lambda I)}{\partial E}$ which equation can I use. I know how to calculate $\frac{\partial det(E)}{\partial E}$ but in this case it is different. – F.Danzi Jan 12 '17 at 14:04
I think it is way better using the formula $$\frac{\partial\lambda_i}{\partial E}\cdot F=\frac{N_i^TFN_i}{N_i^TN_i}\,.$$ Just a comment tfollowing Kato's book). The differential $\frac{\partial\lambda_i}{\partial E}$ might not exist, when $\lambda_i$ has multiplicity $\ge2$. Nevertheless, if the symmetric tensor $E$ depends analytically upon one parameter (that is $s\mapsto E(s)$ is analytic over an interval $J$), then $s\mapsto\lambda(s)$ can be determined analytically.