Det(1+AB) = det(1+BA)

1. Jun 20, 2013

jostpuur

I have some reason to believe that

$$\det(\textrm{id} + AB) = \det(\textrm{id} + BA)$$

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

$$A = \big(A_1,\; A_2\big),\quad\quad\quad B = \left(\begin{array}{c} B_1 \\ B_2 \\ \end{array}\right)$$

then

$$\det(\textrm{id} + AB) = 1 + A_1B_1 + A_2B_2$$

and

$$\det(\textrm{id} + BA) = \det\left(\begin{array}{cc} 1 + B_1A_1 & B_1 A_2 \\ B_2 A_1 & 1 + B_2 A_2 \\ \end{array}\right)$$
$$= (1 + B_1A_1)(1 + B_2A_2) - A_1A_2B_1B_2 = 1 + B_1A_1 + B_2A_2$$

Anyone knowing how to prove the general case?

2. Jun 20, 2013

micromass

Staff Emeritus
3. Jun 20, 2013

jostpuur

So much names to be known.