Question: 14. Let A, B, C, D, and I be \(n \times n\) matrices. Use the definition or properties of a determinant to justify the following formulas. Part (c) is useful in applications of eigenvalues (Chapter 5).

1. \(det\left( {\begin{array}{*{20}{c}}A&{\bf{0}}\\{\bf{0}}&{\bf{I}}\end{array}} \right) = det\,A\)
2. \(det\left( {\begin{array}{*{20}{c}}I&{\bf{0}}\\C&D\end{array}} \right) = det\,D\)
3. \(det\left( {\begin{array}{*{20}{c}}A&{\bf{0}}\\C&D\end{array}} \right) = \left( {det\,A} \right)\left( {det\,D} \right) = det\left( {\begin{array}{*{20}{c}}{\bf{A}}&B\\{\bf{0}}&D\end{array}} \right)\)

(a)

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}&0&0& \cdots &0\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}&0&0& \cdots &0\\0&0& \cdots &0&1&0& \cdots &0\\0&0& \cdots &0&0&1& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &0&0&0& \cdots &1\end{array}} \right)\\ = {\left( { - 1} \right)^{2n + 2n}}1\det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}&0&0& \cdots &0\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}&0&0& \cdots &0\\0&0& \cdots &0&1&0& \cdots &0\\0&0& \cdots &0&0&1& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &0&0&0& \cdots &1\end{array}} \right)\\ = \det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}&0&0& \cdots &0\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}&0&0& \cdots &0\\0&0& \cdots &0&1&0& \cdots &0\\0&0& \cdots &0&0&1& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &0&0&0& \cdots &1\end{array}} \right)\end{array}\)

Continue this process to obtain:

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right) = {\left( { - 1} \right)^{n + 1 + n + 1}}1\det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}\end{array}} \right)\\ = {\left( { - 1} \right)^{2\left( {n + 1} \right)}}\det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}\end{array}} \right)\\ = \det \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}\end{array}} \right)\\\det \left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right) = \det A\end{array}\)

(b)

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}1&0& \cdots &0&0&0& \cdots &0\\0&1& \cdots &0&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &1&0&0& \cdots &0\\{{c_{11}}}&{{c_{12}}}& \cdots &{{c_{1n}}}&{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{1n}}}\\{{c_{21}}}&{{c_{22}}}& \cdots &{{c_{2n}}}&{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{c_{n1}}}&{{c_{n2}}}& \cdots &{{c_{nn}}}&{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\\ = {\left( { - 1} \right)^{1 + 1}}1\det \left( {\begin{array}{*{20}{c}}1&0& \cdots &0&0&0& \cdots &0\\0&1& \cdots &0&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &1&0&0& \cdots &0\\{{c_{12}}}&{{c_{13}}}& \cdots &{{c_{1n}}}&{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{13}}}\\{{c_{22}}}&{{c_{23}}}& \cdots &{{c_{2n}}}&{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{c_{n2}}}&{{c_{n3}}}& \cdots &{{c_{nn}}}&{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\\ = \det \left( {\begin{array}{*{20}{c}}1&0& \cdots &0&0&0& \cdots &0\\0&1& \cdots &0&0&0& \cdots &0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &1&0&0& \cdots &0\\{{c_{12}}}&{{c_{13}}}& \cdots &{{c_{1n}}}&{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{13}}}\\{{c_{22}}}&{{c_{23}}}& \cdots &{{c_{2n}}}&{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\{{c_{n2}}}&{{c_{n3}}}& \cdots &{{c_{nn}}}&{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\end{array}\)

Continue this process to obtain:

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right) = {\left( { - 1} \right)^{n + n}}1\det \left( {\begin{array}{*{20}{c}}{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{1n}}}\\{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\\ = {\left( { - 1} \right)^{2n}}\det \left( {\begin{array}{*{20}{c}}{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{1n}}}\\{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\\ = \det \left( {\begin{array}{*{20}{c}}{{d_{11}}}&{{d_{12}}}& \cdots &{{d_{1n}}}\\{{d_{21}}}&{{d_{22}}}& \cdots &{{d_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{d_{n1}}}&{{d_{n2}}}& \cdots &{{d_{nn}}}\end{array}} \right)\\\det \left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right) = \det D\end{array}\)

Note that

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right) = \left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right)\left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right)\\\det \left( {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right) = \det \left( {\left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right)\left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right)} \right)\\ = \left( {\det \left( {\begin{array}{*{20}{c}}A&0\\0&I\end{array}} \right)} \right)\left( {\det \left( {\begin{array}{*{20}{c}}I&0\\C&D\end{array}} \right)} \right)\\\det \left( {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right) = \left( {\det A} \right)\left( {\det D} \right)\end{array}\)

We know that \(\det M = \det {M^T}\). Hence,

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}A&B\\0&D\end{array}} \right) = \det {\left( {\begin{array}{*{20}{c}}A&B\\0&D\end{array}} \right)^T}\\ = \det \left( {\begin{array}{*{20}{c}}{{A^T}}&0\\{{B^T}}&{{D^T}}\end{array}} \right)\\ = \left( {\det {A^T}} \right)\left( {\det {D^T}} \right)\\\det \left( {\begin{array}{*{20}{c}}A&B\\0&D\end{array}} \right) = \left( {\det A} \right)\left( {\det D} \right)\end{array}\)