As part of some homework assignments in a course in continuum mechanics, I'm trying to prove the Sherman-Morrison's formula: $$(B_{ij}+\alpha u_iv_j)^{-1}=B_{ij}^{-1}-\frac{\alpha}{1 + \alpha v_k B_{kl}^{-1}u_l}B_{im}^{-1}u_m v_n B_{nj}^{-1} $$
using the following identity:
$$(\delta_{ij}+\alpha u_iv_j)^{-1}=\delta_{ij}-\frac{\alpha}{1 + \alpha u_k v_k}u_i v_j$$
The purpose of most of the exercises in the course this far has been to make us students comfortable with using indices with Einstein's summation convention. Although I feel a lot more comfortable with it now after working with it for two weeks than when we first started, it still confuses me at times.
My solution:
Set $\alpha u_i = B_{ik} w_k$. This implies that $w_l=\alpha B_{li}^{-1} u_i$. The left hand side of the formula I'm trying to prove can then be written like this:
$$(B_{ij}+\alpha u_iv_j)^{-1}=(B_{ij}+B_{ik} w_k v_j)^{-1}=(B_{im}(\delta_{mj} + w_m v_j))^{-1}=B_{im}^{-1}\left(\delta_{mj} -\frac{1}{1+ w_k v_k} w_m v_j\right)=B_{ij}^{-1}-\frac{\alpha}{1+\alpha B_{kn}^{-1} u_n v_k}B_{mn}^{-1} u_n v_j B_{im}^{-1}$$
This is correct except for the indices in the $B_{mn}^{-1} u_n v_j B_{im}^{-1}$-factor. Compared to the corresponding factor in the given Sherman-Morrison formula, $B_{im}^{-1}u_m v_n B_{nj}^{-1}$, $B_{mn}$ in my expression contains only dummy indices, and $v_j$ has a free index, but this should not be the case.
I would be very thankful if someone could help me find my error, and I wish you all a good day.
Note that $(AB)^{-1}=B^{-1}A^{-1}$. The orders of the factors reverse. Other than that I think you're good.