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$A$ is an $n\times n$ matrix, partitioned as $$A=\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{bmatrix},$$ where $A_{11}$ has dimensions $k\times k$ and $A_{11}$ and $A_{22}$ are nonsingular. I've managed to prove that the inverse is: $$A^{-1}=\begin{bmatrix} (A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1} & -A_{11}^{-1}A_{12}(A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1}\\ -A_{22}^{-1}A_{21}(A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1} & (A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1} \end{bmatrix}$$ However, I am having difficulty showing that $A^{-1}A=I$. For example, if I multiply the first row of $A^{-1}$ times the first column of $A$, I get:

$$(A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1}A_{11}-A_{11}^{-1}A_{12}(A_{22}-A_{21}A_{11}^{-1}A_{12})^{-1}A_{21}$$

This should equal a $k\times k$ identity matrix, but I can't seem to show that this is the case.

Any suggestions?

David
  • 2,262

1 Answers1

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The form of $A^{-1}$ in your question follows from the usual derivation of Schur complement, but you may also directly verify that the formula is indeed a partitioned matrix inversion formula. For convenience, we relabel $A_{11},A_{12},A_{21},A_{22}$ as $X,Y,Z,W$ respectively. Then \begin{align} &\phantom{=}(X-YW^{-1}Z)^{-1}X-X^{-1}Y(W-ZX^{-1}Y)^{-1}Z\\ &= (X-YW^{-1}Z)^{-1}(X-YW^{-1}Z+YW^{-1}Z)-X^{-1}Y(W-ZX^{-1}Y)^{-1}Z\\ &= I+\left[(X-YW^{-1}Z)^{-1}YW^{-1}-X^{-1}Y(W-ZX^{-1}Y)^{-1}\right]Z. \end{align} So, the question is whether the expression inside the pair of square brackets is zero. However, \begin{align} &(X-YW^{-1}Z)^{-1}YW^{-1}=X^{-1}Y(W-ZX^{-1}Y)^{-1}\\ \Leftrightarrow\ &YW^{-1}(W-ZX^{-1}Y)=(X-YW^{-1}Z)X^{-1}Y\\ \Leftrightarrow\ &Y-YW^{-1}ZX^{-1}Y=Y-YW^{-1}ZX^{-1}Y. \end{align}

user1551
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