Let $I:=I_n, n\in\mathbb N$ be the unit matrix of dimension $n$ and $A,B\in\mathbb R^{n\times k}$, such that $C:=I_k-B^TA$ is invertable.Show this identity: $$(I_n-AB^T)^{-1}=I_n+AC^{-1}B^T$$ and find a similiar identity for $M-AB^T$ with an invertable matrix $M\in\mathbb R^{n\times n}$ with $C=I_k-B^TM^{-1}A$.
Start: To prove $$(I_n-AB^T)(I_n+AC^{-1}B^T)=I_n$$ but my problem here is to find the expression for $C^{-1}$. I do not know how to reshape it, so that the $C^{-1}$ fall away.
$$(I_n-AB^T)(I_n+AC^{-1}B^T) = (I_n+AC^{-1}B^T) - (AB^T+AB^TA C^{-1}B^T)\\=I_n -AB^T +A(I_k-B^TA) C^{-1}B^T.$$ Now we employ the definition $C= I_k - B^T A$ and obtain $$ (I_n-AB^T)(I_n+AC^{-1}B^T)= I_n -A B^T +A B^T = I_n.$$