Derivative of an eigenvector with respect to his own 3x3 real symmetric matrix

$\mathbf{C}$ is a real, positive-definitive 3x3 symmetric matrix (I am thinking about the right Cauchy-Green tensor in solid mechanics). We perform eigendecomposition and get:

$$\mathbf{C} = \sum_{i=1}^3 \lambda_i^2 \mathbf{N_i} \otimes \mathbf{N_i}$$

where $\lambda_i^2$ are its eigenvalues and $\mathbf{N_i}$ are its eigenvectors.

My question is, is there a closed form solution for $\frac{\partial \mathbf{N_i}}{\partial \mathbf{C}}$?

PS: I mostly care about in Cartesian coordinate system, where the definition of the derivative of two tensors are: $$(\frac{\partial \mathbf{A}}{\partial \mathbf{B}})_{ijkl} = \frac{\partial A_{ij}}{\partial B_{kl}}$$

\begin{align*} x_+(t) &:= \begin{pmatrix} \cos\frac1t \\ \sin\frac1t \end{pmatrix}, \quad x_-(t) := \begin{pmatrix} -\sin\frac1t \\ \cos\frac1t \end{pmatrix}, \quad \lambda_\pm(t) = \pm e^{-\frac1{t^2}},\\ A(t) &:= (x_+(t),x_-(t)) \begin{pmatrix} \lambda_+(t) & 0 \\ 0 & \lambda_-(t) \end{pmatrix} (x_+(t),x_-(t))^{-1} \\ &= e^{-\frac1{t^2}}\begin{pmatrix} \cos\frac2t & \sin\frac2t \\ \sin\frac2t & -\cos\frac2t \end{pmatrix}. \end{align*} Here $t\mapsto A(t)$ and $t\mapsto \lambda_\pm(t)$ are smooth, whereas the eigenvectors cannot be chosen continuously.
• Thank you Peter! So when $\lambda_1 = \lambda_2$ the eigenvectors are not differentiable. That said, do you think a closed form expression exist for cases where all eigenvalues are unique? Apr 13 '15 at 19:00