# About the diagonal entries of an inverse matrix

Suppose that $\bf{A}$ is a full-rank $N \times N$ matrix, ${\bf{a}}_n$ is the $n$-th column of $\bf{A}$ and ${\bf{A}}_n$ is the submatrix obtained by deleting ${\bf{a}}_n$ out of $\bf{A}$.

How to prove $${\left[ {{{\left( {{\bf{A}}^H{{\bf{A}}}} \right)}^{ - 1}}} \right]_{nn}} = \frac{1}{{{\bf{a}}_n^H{{\bf{a}}_n} - {\bf{a}}_n^H{{\bf{A}}_n}{{\left( {{\bf{A}}_n^H{{\bf{A}}_n}} \right)}^{ - 1}}{\bf{A}}_n^H{{\bf{a}}_n}}}$$ where ${\left( \cdot \right)^H}$ stands for Hermitian transpose, ${\left[ \cdot \right]_{nn}}$ denotes the $n$-th diagonal element.

Thanks a lot.

• Thanks for your reply. But, we should not restrict $n=N$ in our proof. Actually, the equation holds for any $n = 1, \ldots, N$. So, how to prove this general case? – SAM Dec 7 '15 at 10:38