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My question is: Given a matrix $A$ and its eigenvector $v$ which corresponds to $A$'s maximum eigenvalue, is there a closed form formula to calculate the derivative

$$\frac{\partial(u^Tv)}{\partial A}$$

where $u$ is an unrelated vector?

Please help! Thanks in advance!

Update:

I check matrix cookbook which shows

$$\partial v = (\lambda I-A)^\dagger\partial(A) v$$

where $\lambda$ is the corresponding eigenvalue, $\dagger$ is the symbol of pseudo-inverse. But I still don't know how to calculate the desired derivative.

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Start with the eigenvalue equation, and take differentials $$ \eqalign { 0 &= (A-\lambda I)v \cr &= dA\,v + (A-\lambda I)dv \cr dA\,v &= (\lambda I-A)dv } $$ For notational convenience, let $M=(\lambda I-A)^{\dagger}$.

Then proceeding to the least squares solution, we obtain the cookbook result $$ \eqalign { dv &= M\,dA\,v \cr } $$ From there, we can pre-multiply by $u^T$ to obtain $$ \eqalign { u^Tdv &= u^TM\,dA\,v \cr d(u^Tv) &= p^TdA\,v \cr &= pv^T:dA \cr } $$ where $p=M^Tu\,\,$ and the colon represents the Frobenius product, $\,X\!:\!Y=tr(X^TY)$.

Since $df = (\frac{\partial f}{\partial A}):dA$, the derivative must be $$ \eqalign { \frac{\partial\,(u^Tv)}{\partial A} &= pv^T \cr &= M^{T}uv^T \cr &= (\lambda I-A^T)^{\dagger} \, uv^T } $$

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  • $\begingroup$ Thanks, @greg ! That's what I want. I forget to use the definition of differential. $\endgroup$ – fetcher May 31 '15 at 3:35
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    $\begingroup$ Please be aware that the cookbook result comes with all sorts of restrictions on $A$, i.e. it must be real, symmetric, with distinct eigenvalues. Also the eigenvector must be normalized, $\|v\|=1$. $\endgroup$ – lynn May 31 '15 at 18:49

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