$\text{det}(I-AB)=\text{det}(I-BA)$ [duplicate]

I want to show that for $n\times n$ matrices $A,B$ that $\text{det}(I-AB)=\text{det}(I-BA)$.

My thought is that since $AB$ and $BA$ have the same eigenvalues, I know they have they same minimal polynomial. If I can further show they have the the same characteristic polynomial, I'm done.

Is that true though? the characteristic polynomials for $A$ and $B$ respectively are $P_A(x)=(x-\lambda_1)^{c_1}...(x-\lambda_j)^{c_j}$ and $P_B(x)=(x-\lambda_1)^{d_1}...(x-\lambda_j)^{d_j}$ where $\lambda_q\neq\lambda_p$, $j\leq n$ and the superscripts are the dimensions of the the generalized eigenspaces. Is it true that $c_k=d_k?$

Solution 1:

(You don't have to use eigenvalues, but is simpler with them)

Hint: $\det(I-AB) = \det\left(\begin{array}{c c} I & 0 \\ A & I-AB \end{array} \right)$