Use the result of Exercise 8 to prove that, if $A$ and $B$ are $n\times n$ matrices over the field $F$, then $AB$ and $BA$ have precisely the same characteristic values in $F$.
My attempt: We claim $\det (I-AB)=0_F$$\iff$$\det (I-BA)=0_F$. Proof: By exercise 8 section 6.2, $I-AB$ is invertible$\iff$$I-BA$ is invertible. By theorem 4 section 5.4, $\det (I-AB)=0_F$$\iff$$I-AB$ is not invertible$\iff$$I-BA$ is not invertible$\iff$$\det (I-BA)=0_F$. Our desired result.
We need to show $c\in F$ is eigenvalue of $AB$$\iff$$c\in F$ is eigenvalue of $BA$, i.e. $\det (cI-AB)=0_F$$\iff$$\det (cI-BA)=0_F$. If $c=0_F$. By theorem 3 section 5.3, $\det (0_FI-AB)=\det (-AB)=\det (-BA)=\det (0_FI-BA)$. Thus $\det (0_FI-AB)= \det (0_FI-BA)$. If $c\neq 0_F$. $(\Rightarrow)$ Suppose $\det (cI-AB)=0_F$. By elementary property of determinant, $\det (cI-AB)=\det [c(I-c^{-1}AB)]=c^n \det (I-c^{-1}AB)=0_F$. Which implies $c^n=0_F$ or $\det (I-c^{-1}AB)=0_F$. Since $c\neq 0_F$, we have $c^{n}\neq 0_F$. So $\det (I-c^{-1}AB)=0_F$. By above claim, $\det (I-c^{-1}BA)=0_F$ (since $M_{n\times n}(F)$ is linear algebra over $F$). Multiply both side by $c^n$, we have $c^n \det (I-c^{-1}BA)=\det [c(I-c^{-1}BA)]=\det (cI-BA)=0_F$. Thus $\det (cI-BA)=0_F$. $(\Leftarrow)$ Conversely suppose $\det (cI-BA)=0_F$. Proof is similar to $(\Rightarrow)$. Is my proof correct?
