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I'm reading over a proof for the Woodbury matrix identity. However, there is a part of the proof that I'm not sure of (highlighted in red below).

The Woodbury matrix identity states that, given the complex matrices $A \in \mathbb C^{n \times n},B \in \mathbb C^{n \times m},C \in \mathbb C^{m \times m},$ and $D \in \mathbb C^{m \times n}$, and given that $A,C,$ and $A + BCD$ are invertible, then $$ (A + BCD)^{-1} = A^{-1} - A^{-1}B\left(C^{-1} + DA^{-1}B\right)^{-1}DA^{-1} $$ To prove this identity, let $M = \begin{bmatrix} A & B \\ D & -C^{-1}\end{bmatrix}$ and $M^{-1} = \begin{bmatrix} X & W \\ Y & Z\end{bmatrix}$, where $X,W,Y,$ and $Z$ are complex block matrices with appropriate dimensions, such that \begin{align} MM^{-1} = \begin{bmatrix} I & 0 \\ 0 & I\end{bmatrix} &= \begin{bmatrix} A & B \\ D & -C^{-1}\end{bmatrix} \begin{bmatrix} X & W \\ Y & Z\end{bmatrix} \\ &= \begin{bmatrix} AX + BY & AW + BZ \\ DX - C^{-1}Y & DW - C^{-1}Z\end{bmatrix} \end{align} Therefore, \begin{align} AX + BY &= I \tag{1}\label{eq:1a_1st} \\ DX - C^{-1}Y &= 0 \tag{2}\label{eq:1a_2nd} \end{align} From \eqref{eq:1a_1st}, $X = A^{-1}(I - BY)$. Substituting this into \eqref{eq:1a_2nd} yields \begin{align} DX - C^{-1}Y &= 0 \\ DA^{-1}(I - BY) &= C^{-1}Y \\ DA^{-1} - DA^{-1}BY &= C^{-1}Y \\ DA^{-1} &= DA^{-1}BY + C^{-1}Y \\ \color{red}{DA^{-1}} &\begingroup\color{red}=\endgroup \color{red}{(DA^{-1}B + C^{-1})Y} \\ \color{red}{(DA^{-1}B + C^{-1})^{-1}DA^{-1}} &\begingroup\color{red}=\endgroup \color{red}{Y} \end{align} This is the part of the proof that I don't understand. How do we know that the inverse of $(DA^{-1}B + C^{-1})$ exists to proceed as shown above? We are given that $A + BCD$ is invertible, but I'm not sure how this is relevant.

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The proof you have mentioned is poorly written and is overly complicated. Put $X=A^{-1}B$ and $Y=CD$. Then $A+BCD=A(I+XY)$ and $DA^{-1}B + C^{-1}=C^{-1}(YX+I)$. The identity can be rewritten as $$ (I + XY)^{-1} = I - X\left(I + YX\right)^{-1}Y.\tag{1} $$ You are essentially asking why $I+YX$ is invertible when $I+XY$ is invertible. Suppose the contrary that $I+YX$ is singular. Then $YXu=-u$ for some nonzero vector $u$. Hence $v=Xu\ne0$ and $XYv=X(YXu)=X(-u)=-v$, but this is a contradiction because $I+XY$ is nonsingular.

The result also follows from one of the two determinant theorems of Sylvester, which states that when $X$ is $m\times n$ and $Y$ is $n\times m$, we have $\det(I_m+XY)=\det(I_n+YX)$. More generally, the characteristic polynomial of $XY$ is $x^{m-n}$ times the characteristic polynomial of $YX$. That is, $\det(xI_m-XY)=x^{m-n}\det(xI_n-YX)$. You may see this answer for more details.

Once we know that $I+YX$ is invertible, the identity $(1)$ can be easily proved by showing that $(I + XY)\left[I - X\left(I + YX\right)^{-1}Y\right]=I$. But how can one foresee that the inverse of $I+XY$ is in the form of the RHS of $(1)$? See this MO question for a more in-depth discussion.

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  • $\begingroup$ Why does the vector $u$ being in the null space of $I + YX$ imply that $v=Xu \ne 0$? $\endgroup$
    – mhdadk
    Sep 12, 2022 at 19:22
  • $\begingroup$ @mhdadk If $Xu=0$, then $YXu=-u=0$. $\endgroup$
    – user1551
    Sep 12, 2022 at 19:23
  • $\begingroup$ Doesn’t this mean that $X$ must be invertible? If so, how do we know it is? $\endgroup$
    – mhdadk
    Sep 12, 2022 at 19:24
  • $\begingroup$ @mhdadk I don't understand your question. $X$ is not necessarily a square matrix. $\endgroup$
    – user1551
    Sep 12, 2022 at 19:25
  • $\begingroup$ Nevermind, I realized that $Xu = 0 \implies YXu \neq 0$ is the contrapositive of $YXu = 0 \implies Xu \neq 0$. Thanks for the explanation. I just found out that that if $M = \begin{bmatrix} A & B \\ D & -C^{-1}\end{bmatrix}$ and both $A$ and $C$ are invertible, then $A + BCD$, which is the Schur complement of $-C^{-1}$, is invertible iff $-C^{-1} - DA^{-1}B$, which is the Schur complement of $A$, is invertible. This helped my understanding. Proof here. $\endgroup$
    – mhdadk
    Sep 12, 2022 at 19:54

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