# Derivatives of operations on eigenvectors with repsect to matrix

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?

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.

• What is the "pseudo-inverse"? – Alex M. May 30 '15 at 15:37
• – fetcher May 30 '15 at 16:57

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 }
• 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, $\|v\|=1$. – lynn May 31 '15 at 18:49