Let $A$ be a rank $2$-matrix, i.e. there are vectors $v_1,v_2,w_1,w_2$ such that
$$Ax = \langle v_1,x\rangle v_2 + \langle w_1,x \rangle w_2.$$
I wonder if there is an explicit expression for the Fredholm determinant
$$\operatorname{det}(1+A).$$
On wikipedia we have a Taylor series for the determinant, but the expression is hard to interpret for me with this wedge product here.
If we actually treat $A$ as an $n \times n$ matrix and so treat $v_1,v_2,w_1,w_2$ as elements of $\Bbb C^n$, then we have $$ Ax = (v_2v_1^* + w_2w_1^*)x $$ (where $M^*$ denotes the conjugate-transpose of $M$), so that $$ A = v_2v_1^* + w_2w_1^* = \pmatrix{v_2 & w_2}\pmatrix{v_1 & w_1}^*. $$ From the Sylvester determinant identity, we have $$ \det[I + A] = \det \left[I + \pmatrix{v_1 & w_1}^*\pmatrix{v_2 & w_2} \right] \\ = \det \pmatrix{1 +\langle v_1,v_2\rangle & \langle v_1,w_2\rangle\\ \langle w_1,v_2\rangle & 1 + \langle w_1,w_2\rangle} \\ = (1 +\langle v_1,v_2\rangle)(1 + \langle w_1,w_2\rangle) - \langle v_1,w_2\rangle\langle w_1,v_2\rangle. $$ The same should hold for operators over a Hilbert space.