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Considering [[parallel algorithms]], we can compute <math>(\mathbf A^T \mathbf A)^{-1}</math> and
<math>(\mathbf B^T \mathbf B)^{-1}</math> in parallel. Then, we finish to compute <math>(\mathbf A^T \mathbf P_B^{\perp} \mathbf A)^{-1}</math> and <math>(\mathbf B^T \mathbf P_A^{\perp} \mathbf B)^{-1}</math> also in parallel.
== Block matrix inversion ==
Let a block matrix be
:<math>\begin{bmatrix}
A & B \\
C & D
\end{bmatrix}
.</math>
We can get an inverse formula by combining the previous results in.<ref>{{Cite web |url=http://ccrma.stanford.edu/~jos/lattice/Block_matrix_decompositions.html|title=Block matrix decompositions |last=Smith |first=Julius O |archiveurl=http://web.archive.org/web/20080915181439/http://ccrma.stanford.edu/~jos/lattice/Block_matrix_decompositions.html |archivedate=2008-09-15}}</ref>
:<math>\begin{bmatrix}
A & B \\
C & D
\end{bmatrix}^{-1}
=
\begin{bmatrix}
(A - BD^{-1}C)^{-1} & -A^{-1}B(D - CA^{-1}B)^{-1} \\
-D^{-1}C(A - BD^{-1}C)^{-1} & (D - CA^{-1}B)^{-1}
\end{bmatrix}
=
\begin{bmatrix}
S^{-1}_D & -A^{-1}BS^{-1}_A \\
-D^{-1}CS^{-1}_D & S^{-1}_A
\end{bmatrix}
,</math>
where <math>S_A</math> and <math>S_D</math>, respectively, [[Schur complement]]s of <math>A</math>
and <math>D</math>, are defined by <math>S_A = D - C A^{-1}B</math>, and <math>S_D =
A - BD^{-1}C</math>. This relation is derived by using Block Triangular
Decomposition. It is called ''simple block matrix inversion.''<ref>[http://ieeexplore.ieee.org/xpls/abs_all.jsp?isnumber=30419&arnumber=1399280&count=249&index=181 S. Jo, S. W. Kim and T. J. Park, "Equally constrained affine projection algorithm," ''in Conference Record of the Thirty-Eighth Asilomar Conference on Signals, Systems and Computers,'' vol. 1, pp. 955–959, Nov. 7–10, 2004.]</ref>
Now we can obtain the inverse of the symmetric block matrix:
:<math>
\begin{bmatrix}
\mathbf A^T \mathbf A & \mathbf A^T \mathbf B \\
\mathbf B^T \mathbf A & \mathbf B^T \mathbf B
\end{bmatrix}^{-1}
=
\begin{bmatrix}
(\mathbf A^T \mathbf A-\mathbf A^T \mathbf B(\mathbf B^T \mathbf B)^{-1}\mathbf B^T \mathbf A)^{-1}
& -(\mathbf A^T \mathbf A)^{-1}\mathbf A^T \mathbf B(\mathbf B^T \mathbf B-\mathbf B^T \mathbf A(\mathbf A^T \mathbf A)^{-1}\mathbf A^T \mathbf B)^{-1}
\\
-(\mathbf B^T \mathbf B)^{-1}\mathbf B^T \mathbf A(\mathbf A^T \mathbf A-\mathbf A^T \mathbf B(\mathbf B^T \mathbf B)^{-1}\mathbf B^T \mathbf A)^{-1}
& (\mathbf B^T \mathbf B-\mathbf B^T \mathbf A(\mathbf A^T \mathbf A)^{-1}\mathbf A^T \mathbf B)^{-1}
\end{bmatrix}
</math>
:::<math>
=
\begin{bmatrix}
(\mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}
& -(\mathbf A^T \mathbf A)^{-1}\mathbf A^T \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}
\\
-(\mathbf B^T \mathbf B)^{-1}\mathbf B^T \mathbf A(\mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}
& (\mathbf B^T \mathbf P_A^{\perp} \mathbf B)^{-1}
\end{bmatrix}
</math>
Since the block matrix is symmetric, we also have
:<math>
\begin{bmatrix}
\mathbf A^T \mathbf A & \mathbf A^T \mathbf B \\
\mathbf B^T \mathbf A & \mathbf B^T \mathbf B
\end{bmatrix}^{-1}
=
\begin{bmatrix}
(\mathbf A^T \mathbf P_B^{\perp} \mathbf A)^{-1}
&
-(\mathbf A^T \mathbf P_B^{\perp} \mathbf A)^{-1}
\mathbf A^T \mathbf B(\mathbf B^T \mathbf B)^{-1}
\\
-(\mathbf B^T \mathbf P_A^{\perp} \mathbf B)^{-1}
\mathbf B^T \mathbf A (\mathbf A^T \mathbf A)^{-1}
& (\mathbf B^T \mathbf P_A^{\perp} \mathbf B)^{-1}
\end{bmatrix}.
</math>
Then, we can see how the Schur complements are connected to the projection matrices of the symmetric, partitioned matrix.
== See also ==
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