Calculating second derivatives of eigenvectors of a matrix with some degenerate eigenvalues

Question

Given real symmetric matrix $\mathbf{M}$ with eigenvalues $\lambda_i$ and eigenvectors $\mathbf{v}_i$, the derivative of an eigenvector is $$\dot{\mathbf{v}}_i = \sum_{j \ne i} \frac{\mathbf{v}_j \mathbf{v}_j^T}{\lambda_i - \lambda_j} \dot{\mathbf{M}} \mathbf{v}_i$$

This is obviously not defined when $\lambda_i$ is degenerate. However, even if $\mathbf{M}$ contains degenerate eigenvalues, it is still possible to evaluate $\dot{\mathbf{v}}_i$ so long as $\lambda_i$ is not itself degenerate.

My question is whether it is possible in this case to evaluate the second derivative, $\ddot{\mathbf{v}}_i$. Applying the chain rule to the above expression for $\dot{\mathbf{v}}_i$, I obtain $$\ddot{\mathbf{v}}_i = \sum_{j \ne i} \left[ \frac{\dot{\mathbf{v}}_j \mathbf{v}_j^T + \mathbf{v}_j \dot{\mathbf{v}}_j^T}{\lambda_i - \lambda_j} \dot{\mathbf{M}} \mathbf{v}_i - \frac{\left(\dot{\lambda}_i - \dot{\lambda}_j\right)\mathbf{v}_j \mathbf{v}_j^T}{\left(\lambda_i - \lambda_j\right)^2} \dot{\mathbf{M}} \mathbf{v}_i + \frac{\mathbf{v}_j \mathbf{v}_j^T}{\lambda_i - \lambda_j} \left(\ddot{\mathbf{M}} \mathbf{v}_i + \dot{\mathbf{M}} \dot{\mathbf{v}}_i\right)\right]$$

This suggests that $\ddot{\mathbf{v}}_i$ is undefined if $\mathbf{M}$ has any degenerate eigenvalues. However, finite difference testing seem to suggest that $\ddot{\mathbf{v}}_i$ is defined so long as $\lambda_i$ is not degenerate.

Is $\ddot{\mathbf{v}}_i$ actually defined in this case? If so, is there an analytical expression for it?

A related side-question: in cases where $\ddot{\mathbf{v}}_i$ is unambiguously defined, is it necessarily orthogonal to $\mathbf{v}_i$? I understand the reasoning behind $\dot{\mathbf{v}}^T \mathbf{v} = 0$, but I'm unsure whether the same logic holds when it comes to the second derivative.

Answer 1

If $\lambda_i$ is simple, then the eigenvalue and eigenvector are as smooth as your matrix will allow. You start from $(M-\lambda_i)v_i=0$, $v_i^Tv_i=1$. Differentiating this once yields $(M-\lambda_i)\dot v_i+(\dot M-\dot\lambda_i)v_i=0$, $\dot v_i^Tv_i=0$. From this you calculate $\dot\lambda_i=v_i^T\dot Mv_i$, and the expression for $\dot v_i$ which you gave. Now take another derivative to obtain $(M-\lambda_i)\ddot v_i+2(\dot M-\dot\lambda_i)\dot v_i+(\ddot M-\ddot\lambda_i)v_i=0$, $\ddot v_i^Tv_i+\dot v_i^T\dot v_i=0$. You know everything except $\ddot v_i$ and $\ddot\lambda_i$. If you multiply the first equation with $v_i^T$, you obtain $\ddot\lambda_i$, and subsequently you can find $\ddot v_i$. If $M$ is smooth enough, you can continue the same procedure indefinitely to obtain derivatives of any order. If $M$ is not symmetric, you can do something similar, but you need the eigenvector of the adjoint.