Linear Algebra – Show That A_{22} = A_{21}A_{11}^{-1}A_{12} for Given Matrix A

generalized-inverselinear algebramatricespseudoinverseschur-complement

I'm currently trying to solve the following problem.

$A$ is $n \times p$ matrix with rank $r < \text{min}(n,p)$ and $A$ is partitioned as follows. $$A =
\begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{bmatrix}
$$
where $A_{11}$ is $(r \times r)$ invertible matrix. Show that the following $(p \times n)$ matrix $G$ is g-inverse of $A$ (not Moore-Penrose g-inverse, just g-inverse).
$$G = \begin{bmatrix}
A_{11}^{-1} & O \\
O & O \\
\end{bmatrix}$$

The calculation $AGA$ is straightforward;
$$AGA = \begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{21}A_{11}^{-1}A_{12}\\
\end{bmatrix}$$

Then, it remains to show that $A_{22} = A_{21}A_{11}^{-1}A_{12}$. But I found it difficult to generalize this in all $n \times p$ matrices.

Some hints regarding this approach would be appreciated. Thank you.

Best Answer

As $A=\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{bmatrix}$ and rank($A$)$=r =$ rank($A_{11}$). $[A_{11} ~ A_{12}]$ has full row rank, which implies $[A_{21}~ A_{22}]$ are in the row space of $[A_{11} ~ A_{12}]$. Hence $\exists T~s.t.~[A_{21}~ A_{22}]=T[A_{11} ~ A_{12}]$.

For similar reasoning, $\begin{bmatrix} A_{12}\\ A_{22} \end{bmatrix} = \begin{bmatrix} A_{11}\\ A_{21} \end{bmatrix}S$ for some $S$, because $\begin{bmatrix} A_{11}\\ A_{21} \end{bmatrix}$ has full column rank.

In particular, we deduce $A_{21}=TA_{11}, A_{22}=TA_{12}, A_{12}=A_{11}S, A_{22}=A_{21}S.$ Recall $A_{11}^{-1}$ is invertible, $T=A_{21}A_{11}^{-1}$, which further imples $A_{22}=A_{21}A_{11}^{-1}A_{12}$.

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