Is need to work with $\frac{d}{d\lambda} (1 - v^{T}(\lambda I - A)^{-1}u)$. Is it true that: $$\frac{d}{d\lambda} (1 - v^{T}(\lambda I - A)^{-1}u) = -v^{T}\frac{d}{d\lambda}(\lambda I - A)^{-1}u$$ Then from $(\lambda I - A)^{-1}(\lambda I - A) = I$ differentiating both sides with respect to $\lambda$ one obtains: $$ \frac{d}{d\lambda} (\lambda I - A)^{-1} (\lambda I - A) + (\lambda I -A)^{-1} I = O$$ hence $$ \frac{d}{d\lambda}(\lambda I - A)^{-1} = - ((\lambda I - A)^{-1})^{2}$$ In general $$ \frac{d^{k}}{d\lambda^{k}} (\lambda I - A)^{-1} = (-1)^{k} k!((\lambda I - A)^{-1})^{k+1}$$ similar to the scalar case $\frac{1}{\lambda - a}$ ?
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1$\begingroup$ Yes, your formula for the $k^{th}$ derivative is correct. $\endgroup$– lynnCommented Apr 14, 2016 at 13:36
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$\begingroup$ Actually, you need to negate the exponent on the last term, $$(-1)^k\,k!\,(\lambda I-A)^{-k-1}$$ $\endgroup$– lynnCommented Apr 15, 2016 at 18:39
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$\begingroup$ of course, thank you for letting me know $\endgroup$– C MariusCommented Apr 16, 2016 at 21:31
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