Question

Let\(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula\(\left( {\bf{1}} \right)\)for the determinant in section\({\bf{5}}{\bf{.2}}\)to explain why\(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of\(G\)is the product of the characteristic polynomials of\(A\)and\(B\).

Short answer

For any scalar \(\lambda \), the matrix \(G - \lambda I\) has the same partitioned form as \(G\), with \(A - \lambda I\) , and \(B - \lambda I\) as its diagonal blocks.

Step-by-step solution

Step 1: Write the echelon form

Consider \(G = \left( {\begin{aligned}{*{20}{c}}A&X\\0&B\end{aligned}} \right)\).

Assume \(U\) and \(V\) be echelon forms of \(A\) and \(B\) obtained by \(r\) and \(s\) row interchanges.

\(\begin{aligned}{l}\det A &= {\left( { - 1} \right)^r}\det U\\\det B &= {\left( { - 1} \right)^s}\det V\end{aligned}\)

Step 2: Explain why \(\det G = \left( {\det A} \right)\left( {\det B} \right)\)

Using row operations, when\(A\)is reduced to\(U\), then\(G\)is reduced to\(G' = \left( {\begin{aligned}{*{20}{c}}U&Y\\0&B\end{aligned}} \right)\).

Using row operations, when\(B\)is reduced to\(V\), then\(G\)is reduced to\(G'' = \left( {\begin{aligned}{*{20}{c}}U&Y\\0&V\end{aligned}} \right)\).

Since there are \(r + s\) row operations then we get,

\(\begin{aligned}{c}\det G &= \det \left( {\begin{aligned}{*{20}{c}}A&X\\0&B\end{aligned}} \right)\\ &= {\left( { - 1} \right)^{r + s}}\det \left( {\begin{aligned}{*{20}{c}}U&Y\\0&V\end{aligned}} \right)\\ &= {\left( { - 1} \right)^{r + s}}\left( {\det U} \right)\left( {\det V} \right)\\ &= \left( {\det A} \right)\left( {\det B} \right)\end{aligned}\)

Thus, \(\det \left( {G - \lambda I} \right) = \det \left( {A - \lambda I} \right)\det \left( {B - \lambda I} \right)\).

For any scalar \(\lambda \), the matrix \(G - \lambda I\) has the same partitioned form as \(G\), with \(A - \lambda I\) and \(B - \lambda I\) as its diagonal blocks, that is, \(\det \left( {G - \lambda I} \right) = \det \left( {A - \lambda I} \right)\det \left( {B - \lambda I} \right)\).