- #1
jostpuur
- 2,116
- 19
I have some reason to believe that
[tex]
\det(\textrm{id} + AB) = \det(\textrm{id} + BA)
[/tex]
is true even when AB and BA are not the same size. In other words, A and B are not neccessarily square matrices.
For example, if
[tex]
A = \big(A_1,\; A_2\big),\quad\quad\quad
B = \left(\begin{array}{c} B_1 \\ B_2 \\ \end{array}\right)
[/tex]
then
[tex]
\det(\textrm{id} + AB) = 1 + A_1B_1 + A_2B_2
[/tex]
and
[tex]
\det(\textrm{id} + BA) = \det\left(\begin{array}{cc}
1 + B_1A_1 & B_1 A_2 \\
B_2 A_1 & 1 + B_2 A_2 \\
\end{array}\right)
[/tex]
[tex]
= (1 + B_1A_1)(1 + B_2A_2) - A_1A_2B_1B_2 = 1 + B_1A_1 + B_2A_2
[/tex]
Anyone knowing how to prove the general case?
[tex]
\det(\textrm{id} + AB) = \det(\textrm{id} + BA)
[/tex]
is true even when AB and BA are not the same size. In other words, A and B are not neccessarily square matrices.
For example, if
[tex]
A = \big(A_1,\; A_2\big),\quad\quad\quad
B = \left(\begin{array}{c} B_1 \\ B_2 \\ \end{array}\right)
[/tex]
then
[tex]
\det(\textrm{id} + AB) = 1 + A_1B_1 + A_2B_2
[/tex]
and
[tex]
\det(\textrm{id} + BA) = \det\left(\begin{array}{cc}
1 + B_1A_1 & B_1 A_2 \\
B_2 A_1 & 1 + B_2 A_2 \\
\end{array}\right)
[/tex]
[tex]
= (1 + B_1A_1)(1 + B_2A_2) - A_1A_2B_1B_2 = 1 + B_1A_1 + B_2A_2
[/tex]
Anyone knowing how to prove the general case?