TeachingLinear Algebra – Matrices – Block matrices
If a matrix is not too small, then it can be divided into several submatrices which leads to a block matrix. If the submatrices have the appropriate sizes then we can manipulate with these submatrices similarly as with usual matrices.
Example: If \(A_{11}=\begin{pmatrix}1&0&-1\\0&1&1\end{pmatrix}\), \(A_{12}=\begin{pmatrix}2&-1\\1&-1\end{pmatrix}\), \(A_{21}=\begin{pmatrix}-3&1&0\end{pmatrix}\) and \(A_{22}=\begin{pmatrix}2&1\end{pmatrix}\), then we have
\[A=\begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix}=\begin{pmatrix}1&0&-1&2&-1\\0&1&1&1&-1\\-3&1&0&2&1\end{pmatrix}.\]Further we have, if \(B_{11}=\begin{pmatrix}0&1\\-1&2\\3&1\end{pmatrix}\), \(B_{12}=\begin{pmatrix}-2\\3\\1\end{pmatrix}\), \(B_{21}=\begin{pmatrix}-1&1\\2&3\end{pmatrix}\) and \(B_{22}=\begin{pmatrix}-1\\-1\end{pmatrix}\), then
\[B=\begin{pmatrix}B_{11}&B_{12}\\B_{21}&B_{22}\end{pmatrix}=\begin{pmatrix}0&1&-2\\-1&2&3\\3&1&1\\-1&1&-1\\2&3&-1\end{pmatrix}.\]Then we have \[A_{11}B_{11}+A_{12}B_{21}=\begin{pmatrix}1&0&-1\\0&1&1\end{pmatrix}\begin{pmatrix}0&1\\-1&2\\3&1\end{pmatrix} +\begin{pmatrix}2&-1\\1&-1\end{pmatrix}\begin{pmatrix}-1&1\\2&3\end{pmatrix}=\begin{pmatrix}-3&0\\2&3\end{pmatrix} +\begin{pmatrix}-4&-1\\-3&-2\end{pmatrix}=\begin{pmatrix}-7&-1\\-1&1\end{pmatrix},\] \[A_{11}B_{12}+A_{12}B_{22}=\begin{pmatrix}1&0&-1\\0&1&1\end{pmatrix}\begin{pmatrix}-2\\3\\1\end{pmatrix} +\begin{pmatrix}2&-1\\1&-1\end{pmatrix}\begin{pmatrix}-1\\-1\end{pmatrix}=\begin{pmatrix}-3\\4\end{pmatrix} +\begin{pmatrix}-1\\0\end{pmatrix}=\begin{pmatrix}-4\\4\end{pmatrix},\] \[A_{21}B_{11}+A_{22}B_{21}=\begin{pmatrix}-3&1&0\end{pmatrix}\begin{pmatrix}0&1\\-1&2\\3&1\end{pmatrix} +\begin{pmatrix}2&1\end{pmatrix}\begin{pmatrix}-1&1\\2&3\end{pmatrix}=\begin{pmatrix}-1&-1\end{pmatrix} +\begin{pmatrix}0&5\end{pmatrix}=\begin{pmatrix}-1&4\end{pmatrix}\]
and
\[A_{21}B_{12}+A_{22}B_{22}=\begin{pmatrix}-3&1&0\end{pmatrix}\begin{pmatrix}-2\\3\\1\end{pmatrix} +\begin{pmatrix}2&1\end{pmatrix}\begin{pmatrix}-1\\-1\end{pmatrix}=\begin{pmatrix}9\end{pmatrix}+\begin{pmatrix}-3\end{pmatrix} =\begin{pmatrix}6\end{pmatrix}.\]Note that
\[AB=\Bigg(\begin{matrix}A_{11}B_{11}+A_{12}B_{21}&A_{11}B_{12}+A_{12}B_{22}\\[2.5mm]A_{21}B_{11}+A_{22}B_{21}&A_{21}B_{12}+A_{22}B_{22}\end{matrix}\Bigg) =\begin{pmatrix}-7&-1&-4\\-1&1&4\\-1&4&6\end{pmatrix}.\]Example: If \(A=\begin{pmatrix}1&1\\1&2\end{pmatrix}\) and \(B=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\), then we have that \(A^{-1}=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}\) and \(B^{-1}=\begin{pmatrix}2&1\\1&1\end{pmatrix}\). We can use this to find the inverse of the \(4\times 4\) matrix
\[M=\begin{pmatrix}1&1&0&0\\1&2&0&0\\1&0&1&-1\\0&1&-1&2\end{pmatrix}=\Bigg(\begin{matrix}A&0\\[2.5mm]I&B\end{matrix}\Bigg).\]Suppose that \(M^{-1}=\Bigg(\begin{matrix}C&D\\[2.5mm]E&F\end{matrix}\Bigg)\), then we have:
\[\Bigg(\begin{matrix}I&0\\[2.5mm]0&I\end{matrix}\Bigg)=I=MM^{-1}=\Bigg(\begin{matrix}A&0\\[2.5mm]I&B\end{matrix}\Bigg) \Bigg(\begin{matrix}C&D\\[2.5mm]E&F\end{matrix}\Bigg)=\Bigg(\begin{matrix}AC&AD\\[2.5mm]C+BE&D+BF\end{matrix}\Bigg).\]This implies that \(AC=I\), \(AD=0\), \(C+BE=0\) and \(D+BF=I\). Since \(A\) is invertible, this implies that \(C=A^{-1}\) and \(D=A^{-1}0=0\). And since \(B\) is invertible too, we have that \(BF=I\) implies that \(F=B^{-1}\) and \(BE=-C\) that \(E=-B^{-1}C=-B^{-1}A^{-1}\). Hence:
\[M^{-1}=\Bigg(\begin{matrix}A^{-1}&0\\[2.5mm]-B^{-1}A^{-1}&B^{-1}\end{matrix}\Bigg) =\begin{pmatrix}2&-1&0&0\\-1&1&0&0\\-3&1&2&1\\-1&0&1&1\end{pmatrix}.\]Theorem: If \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix , then we have:
\[AB=\Bigg(\text{col}_1(A)\;\text{col}_2(A)\;\ldots\;\text{col}_n(A)\Bigg)\begin{pmatrix}\text{row}_1(B)\\\text{row}_2(B)\\\vdots\\\text{row}_n(B)\end{pmatrix} =\text{col}_1(A)\text{row}_1(B)+\text{col}_2(A)\text{row}_2(B)+\cdots+\text{col}_n(A)\text{row}_n(B).\]Last modified on March 29, 2021
