1
$\begingroup$

How can I compute the partial derivatives of the dominant eigenvalue and eigenvectors of a real symmetric matrix $\mathbf{A}$?

In particular, given

$ \mathbf{v}^* = \arg\max_{\mathbf{v}} \mathbf{v}^{\top}\mathbf{A}\mathbf{v},\quad \text{subject to } ~ \|\mathbf{v}\|_2 = 1, $

how can I find $\partial v_{i}^*/\partial A_{jk}$?

I know the power method is the usual way to compute the dominant eigenvalue and eigenvector. Is their any similar algorithm for computing the gradient?

Unlike another question, I am interested in an efficient computational solution rather than an analytical one.

$\endgroup$
2
$\begingroup$

See (68) here: if $\lambda$ is a simple eigenvalue, $Av=\lambda v$ and $v$ is normalized to be orthogonal, $$ \frac{\partial v}{\partial A_{jk}} = (\lambda I - A)^+ E^{jk} v, $$ where $E^{jk}$ is the matrix with $1$ in position $jk$ and $0$ everywhere else. That plus symbol is a Moore-Penrose pseudoinverse.

The formula is easy to compute explicitly because it only requires the eigenvector itself and the pseudoinverse. Moreover, If the eigenvector is simple then you know that $\lambda I-A$ has rank $n-1$, so you don't have a (potentially difficult) rank decision to make.

Also, related answer with a technique that can produce those formulas.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for taking your time and writing a full solution with clear instructions. $\endgroup$ – Taha Feb 21 '18 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.