Suppose $B=[v,e]$ is an $n \times 2$ matrix with $v=[v_1,...,v_n]^T$ and $e=[1,...,1]^T$, and $J_{2\times 2}=[(0,1),(1,0)]$, and so $Rank(BJB^T)=2$. How can we prove that $BJB^T$ and $JB^TB$ have the same non-zero eigenvalues? Also, how can we prove that $$det(I-BJB^T)=det(I-JB^TB)$$
Thank you!