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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.

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Since \begin{align} A=\left[ \begin{array}{cc} A_n & a_n \\ \end{array} \right] \end{align} we have \begin{align} A^TA=\left[ \begin{array}{cc} A_n^TA_n & A_n^Ta_n \\ a_n^TA_n & a_n^Ta_n \end{array} \right] \end{align} Now applying the formula of the inverse of block matrices (see wiki or the paper "Inverses of 2 x 2 Block Matrices" by TzoN-TZER Lu AND SHENG-HUA Shiou) gives \begin{align} (A^TA)^{-1}=\left[ \begin{array}{cc} * & * \\ * & \left(a_n^Ta_n-a_n^TA_n(A_n^TA_n)^{-1}A_n^Ta_n\right)^{-1} \end{array} \right] \end{align} which answers your question.

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  • $\begingroup$ 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? $\endgroup$ – SAM Dec 7 '15 at 10:38
  • $\begingroup$ This problem has been solved by using the above mentioned method with a permutation matrix. Thanks, Shiyu. $\endgroup$ – SAM Dec 9 '15 at 20:29

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