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I want to determine the Fréchet derivative of eigenvalues over the space of square symmetric matrices $A \in \mathbb{R}^{d \times d}$. I think the result is that if the $p^{\text{th}}$ eigenvalue $\lambda_p(A)$ is associated with eigenvector $v_p(A)$ for matrix A, the answer is

$$(D \lambda_p)_A(H)=v_p^\top (A) H v_p(A)$$

(The notation is a little abusive but I hope it's acceptable.) A heuristic argument for why this is would be that since $\lambda_p(A) = v_p^\top(A) A v_p(A)$,

$$\lambda_p(A + H) \approx v_p^\top(A) (A + H) v_p(A) = v_p^\top(A) A v_p(A) + v_p^\top(A) H v_p(A) = \lambda_p(A) + (D \lambda_p)_A(H)$$

for small $H$. Additionally, if I squint, this answer looks like a formula near the beginning of this paper (published version), which I don't fully understand. These arguments are not convincing, though. Can someone either justify better or give the correct derivative?

cgmil
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    This answer might be helpful. – user1551 Nov 11 '21 at 06:51
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    Suppose $A = I$, so $\lambda = 1$, and every unit vector in $\Bbb R^d$ can serve as $v(A)$ (unit vectors are required by your formula $\lambda = v^TAv$). But unless $H$ is a multiple of $I$, $v^THv$ will vary with the eigenvector $v$, while $(D\lambda)H$ should not. – Paul Sinclair Nov 11 '21 at 16:49
  • @PaulSinclair Yup, I will concede that my answer is wrong in the case of repeated eigenvalues, which was mentioned in the answer user1551 cited as being a difficult case. – cgmil Nov 11 '21 at 18:19
  • @user1551 I basically learned about Fréchet derivatives a couple days ago so I'm still learning the notation. As pointed out by Paul Sinclair my answer cannot be right in the case of repeated eigenvalues, which the answer you linked to also notes is a difficult case. That aside, the answer I give here still looks a lot like the linked answer. Does this suggest my formula is in agreement, but with different notation and with the linked answer having a more rigorous argument? – cgmil Nov 11 '21 at 18:28

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