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I want to calculate the inverse of a real and symmetric matrix when I remove the first row and the first column from it using the Woodbury formula (as is described here). However, in my case, while $A−UV$ is invertible, $−I+VA^{−1}U$ is not. Is there another way to compute the inverse of the updated matrix?

More explanation:

I quote a part of this post here:

$$ A = \begin{bmatrix} a & v_1 & v_2 & v_3 \\ u_1 & & & \\ u_2 & & B & \\ u_3 & & & \end{bmatrix} $$ we can take $U = \begin{bmatrix} 1 & 0 \\ 0 & u_1 \\ 0 & u_2 \\ 0 & u_3 \end{bmatrix}$ and $V = \begin{bmatrix} 0 & v_1 & v_2 & v_3 \\ 1&0&0&0 \end{bmatrix}$, so that: $$ UV = \begin{bmatrix} 0 & v_1 & v_2 & v_3 \\ u_1 & 0 & 0 & 0 \\ u_2 & 0 & 0 & 0 \\ u_3 & 0 & 0 & 0 \end{bmatrix} $$ Then $A - UV = \begin{bmatrix} a & 0 \\ 0 & B \end{bmatrix}$ and provided $a \neq 0$ the update can be done by inverting $(I - VA^{-1}U)$, a $2\times 2$ matrix.

In my example, $a$ can be zero, and matrix $A$ is symmetric, thus $v_i = u_i$. Also the upper left element of $A^{-1}$ is zero. I define $U = \begin{bmatrix} 1 & 0 \\ 0 & u_1 \\ 0 & u_2 \\ 0 & u_3 \end{bmatrix}$ and $V = \begin{bmatrix} b & u_1 & u_2 & u_3 \\ 1&0&0&0 \end{bmatrix}$, $b\neq 0$. So:

$$ A - UV = \begin{bmatrix} a-b & 0 & 0& 0\\ 0& & & \\ 0& & B & \\ 0& & & \end{bmatrix} $$ and thus:

$$ (A - UV)^{-1} = \begin{bmatrix} (a-b)^{-1} & 0 & 0& 0\\ 0& & & \\ 0& & B^{-1} & \\ 0& & & \end{bmatrix} $$

and $$\left(A-UV \right)^{-1} = A^{-1} - A^{-1}U \left(-I+VA^{-1}U \right)^{-1} VA^{-1}$$ but regardless of the value of $b$, the $2\times 2$ matrix $I - VA^{-1}U$ is always not invertible. Is there a trick or another way to solve this problem?

Mah
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    You may find it worthwhile to examine the Wikipedia page on the Schur complement, specifically the computation of $M^{-1}$ in the Background section. – Semiclassical Mar 04 '21 at 13:58
  • Does your matrix $A$ have any known properties, such as being real symmetric, etc.? – hardmath Mar 04 '21 at 18:14
  • @hardmath Yes. I will add it to the question. – Mah Mar 04 '21 at 18:49
  • @Semiclassical I examined the Wikipedia page on the Schur complement following your suggestion but I could figure out a solution. – Mah Mar 04 '21 at 18:56
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    @Mah Note that by the WA-identity, we have $$ \det(A - UV) = \det(A) \det(I - A^{-1}UV) = \det(A) \det(I - VA^{-1}U). $$ It therefore cannot be true that $A - UV$ is invertible but $-I + VA^{-1}U$ is not. – Ben Grossmann Mar 04 '21 at 19:05
  • @BenGrossmann Yes, you are right. I have made a mistake in another part of the solution that caused this problem. Thank you! – Mah Mar 05 '21 at 20:03

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