I am confused about how to show that if $f:V \to V$ is a linear map, then the choice of basis is irrelevant when we compute det$(f)$. (where det f refers to computing the determinant of the matrix representation under that basis) And also how could I show that if we had another linear map, that composition is respected with determinant as well?
Thank you
If $B$ and $C$ are basis for $f$, then: $$[f]_B = [{\rm Id}_V]_{C,B}[f]_C [{\rm Id}_V]_{B,C},$$so using that the first and last matrices in the right side are inverses, and that the determinant of a product is the product of the determinants, applying $\det$ above gives $\det [f]_B = \det [f]_C$. So we define $\det f$ as $\det [f]_B$, where $B$ is any basis for $V$. And this value $\det f$ is well-defined, that is, it does not depend on the choice of basis, because of the computation we did.
If $g$ is another linear operator on $V$, pick your favorite basis $B$ and do:$$\det(f\circ g) = \det [f\circ g]_B = \det( [f]_B[g]_B )= \det [f]_B\det[g]_B = \det f \det g.$$