Sherman-Morrison formula to demonstrate $\frac{\partial x_k}{\partial a_{ij}} = -x_j c_{ki}$

Question

I have the following exercise:

Suppose $A \in \mathbb R^{n\times n} $ is nonsingular, b $\in \mathbb R^n, Ax = b $ and $C=A^{-1} $. Using the Sherman-Morrison formula to show that $$\frac{\partial x_k}{\partial a_{ij}} = -x_j c_{ki}$$

My attempt

$\lim_{\epsilon\to 0} \dfrac{\Big(A+\Big(\epsilon e_i\Big)e_j^T\Big)^{-1}-A^{-1}}{\epsilon}=-\dfrac{\big(A^{-1}e_i\big)(e_j^TA^{-1}\big)}{1+\epsilon\big(e_j^TA^{-1}e_i)}$ from this we can obtain the following expression:

$\dfrac{\partial C}{\partial a_{ij}}=-A^{-1}e_ie_j^TA^{-1}$, since by the expression of the inverse we have $C=\dfrac{1}{\det(A)} \operatorname {Adj}(A)$.

Now:

$\dfrac{\partial x_k}{\partial a_{ij}}=-\sum_{l=1}^{n} (c_{jl}b_l c_{kl})$ as $x=Cb$,and this comes from

$$x_k=\sum_{l=1}^{n}C_{kl}b_l$$ and $$\dfrac{\partial x_k}{\partial a_{ij}}=\sum_{l=1}^{n}\Big(\dfrac{\partial C_{kl}}{\partial a_{ij}}\Big)b_l.$$

The above is all I have so far, I do not know where to apply the Sherman-Morrison identity, it had a small application when taking the limit, but to conclude the statement. I hope you can help me, in advance, thanks

Answer 1

\begin{aligned} \frac{\partial x}{\partial a_{ij}} &=\lim_{h\to0}\frac{1}{h}\left[(A+he_ie_j^T)^{-1}-A^{-1}\right]b\\ &=\lim_{h\to0} \frac{1}{h}\left[\frac{-A^{-1}(he_ie_j^T)A^{-1}}{1+he_j^TA^{-1}e_i}\right]b\quad\text{(by Sherman-Morrison formula)}\\ &=-A^{-1}e_ie_j^TA^{-1}b\\ &=-A^{-1}e_ie_j^Tx\\ &=-x_jA^{-1}e_i\\ &=-x_jc_{\ast i}.\\ \end{aligned} Therefore $\frac{\partial x_k}{\partial a_{ij}}=-x_jc_{ki}$.

Answer 2

I don't know where to use this formula either, but I don't think it's really necessary. All I need is that $$ \frac{\partial a_{lm}}{\partial a_{ij}} = \delta_{li}\delta_{mj}, \qquad\frac{\partial b_l}{\partial a_{ij}} = 0 \qquad \text{and} \qquad \sum_l c_{kl}a_{lm} = \delta_{km}.$$ We have $$\sum_m a_{lm} x_m = b_l $$ Differentiating both sides $\frac{\partial}{\partial a_{ij}}$ we get. $$\sum_m \left(\frac{\partial a_{lm}}{\partial a_{ij}} x_m + a_{lm} \frac{\partial x_m}{\partial a_{ij}} \right) = 0 $$ $$\sum_m \left(\delta_{li}\delta_{mj} x_m + a_{lm} \frac{\partial x_m}{\partial a_{ij}} \right) = 0 $$ $$\delta_{li} x_j + \sum_m a_{lm} \frac{\partial x_m}{\partial a_{ij}} = 0 $$ Multiplaying by $c_{kl}$ and summing over $l$ we get $$\sum_l c_{kl}\delta_{li} x_j + \sum_{l,m} c_{kl}a_{lm} \frac{\partial x_m}{\partial a_{ij}} = 0 $$ $$c_{ki} x_j + \sum_{m} \delta_{km} \frac{\partial x_m}{\partial a_{ij}} = 0 $$ $$ \frac{\partial x_k}{\partial a_{ij}} = -c_{ki} x_j $$