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I want to invert the following matrix: $$ I_m - B (B^T B)^{-1} B^T $$ where $B$ is $m \times n$ with $m > n$ and $B$ has full column rank. The Woodbury matrix inverse identity states: $$ (A + UCV)^{-1} = A^{-1} - A^{-1} U (C^{-1} + V A^{-1} U)^{-1} V A^{-1} $$ Now if we identify: \begin{align} A &= I_m \\ U &= B \\ C &= -(B^T B)^{-1} \\ V &= B^T \end{align} then we have \begin{align} (I_m - B (B^T B)^{-1} B^T)^{-1} &= I_m - B (-B^T B + B^T B)^{-1} B^T \end{align} but this leads to the inverse of a singular matrix. So this seems to indicate that the Woodbury formula won't work on my matrix.

Does anyone know of another way to invert my matrix?

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    $\begingroup$ Are you sure the matrix is invertible? $\endgroup$ Commented Dec 28, 2019 at 17:55
  • $\begingroup$ Your matrix is not invertible so there is no way to invert it $\endgroup$ Commented Dec 28, 2019 at 18:04

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Consider the example: let \begin{align} B = \begin{pmatrix} 1 & 0\\ 0 & 1\\ 0 & 0 \end{pmatrix} \end{align} then \begin{align} B^TB = I_2 \end{align} which means \begin{align} B(B^TB)^{-1}B^T = BB^T= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}. \end{align} Hence it follows \begin{align} I-B(B^TB)^{-1}B^T = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix} \end{align} which is not invertible.

In fact, $P=B(B^TB)^{-1}B^T$ is a projection matrix onto the column space of $B$ and $I-P$ projects onto the orthogonal complement.

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