# Derivative of eigenpair with respect to matrix

Suppose that $$A$$ is real and symmetric matrix (or tensor) of dimension $$3 \times 3$$, with its spectral decomposition

$$A = \sum_{i=1}^3 \lambda_i\ n_i\otimes n_i$$ where $$\lambda_i$$, $$n_i$$ and $$\otimes$$ denote the eigenvalues, eigenvectors and dyadic product, respectively. Further, let $$a$$ be a vector of the form

$$a = \sum_{i=1}^3 f(\lambda_i)\ n_i$$

with an at least twice-differentiable function $$f(\lambda_i)$$.

I need to compute the derivative (third-order tensor) $$B$$ with its components

$$B_{ijk}=\frac{\partial a_i}{\partial A_{jk}}$$

also in the case that the eigenvalues are not distinct (e.g. $$\lambda_1=\lambda_2\neq \lambda_3$$ or even $$\lambda_1=\lambda_2= \lambda_3$$). I know that this is possible for a matrix $$C$$

$$C = \sum_{i=1}^3 f(\lambda_i)\ n_i\otimes n_i$$

In this case, the derivative $$\partial C/\partial A$$ (fourth order tensor) is always computable, also if the eigenvalues are not distinct (see e.g. R.W. Ogden, Non-Linear Elastic Deformations, 1997, p. 162)

• when you perturb $A$, do you wish to preserve the symmetry of the matrix? Jul 9, 2021 at 18:04

If the eigenvalues are distinct you simply fill in the first order perturbation answer for the eigenvalues and eigenvectors, $$\frac{\partial\lambda_i}{\partial A_{jk}}=(n_i)_{j}(n_i)_{k}(2-\delta_{jk}),$$ $$\frac{\partial n_{i}}{\partial A_{jk}}=\sum_{p\neq i}n_p\frac{(n_p)_j(n_i)_k(2-\delta_{jk})}{\lambda_i-\lambda_p},$$ and then $$\frac{\partial a}{\partial A_{jk}}=\sum_i f'(\lambda_i)n_i\frac{\partial\lambda_i}{\partial A_{jk}}+\sum_i f(\lambda_i)\frac{\partial n_{i}}{\partial A_{jk}}.$$
If two (or more) eigenvalues are identical the eigenvectors $$n_i$$ are not uniquely determined by the matrix $$A$$, and neither is the quantity $$a=\sum_i f(\lambda_i)n_i$$. Now you may choose a particular set of eigenvectors and ask for the derivative $$\partial a/\partial A_{jk}$$ for that particular choice, this will in general diverge when two eigenvalues $$\lambda_p$$ and $$\lambda_q$$ coincide, unless $$(n_p)_j (n_q)_k=0$$.
Note that this complication does not appear if you consider instead the quantity $$C=\sum_i f(\lambda_i)n_i\otimes n_i$$, because $$C=f(A)$$ is uniquely defined by $$A$$, even if the eigenvalues are not distinct.