$E$ 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 $\frac{∂λ\_i}{∂E}$
?



Derivative of eigenvalues w.r.t. a tensor
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 $\{\lambda\_1,\lambda\_2,\lambda\_3\}=\{\lambda\_i,\lambda\_j,\lambda\_k\}$.


