Let $T$ be a linear operator on $F^n$. Define $D_T(a_1,...a_n)=det(Ta_1,...,Ta_n).$
(a) Show that $D_T$ is an alternating n-linear function. (b) If $B$ is any ordered basis for $F^n$ and $A$ is the matrix of $T$ in the ordered basis ${B}$, show that $det(A)=c$
For part (a), since the determinant function is is an alternating n-linear function, then $D_T$ is again an alternating linear function. I'm worried that there is something more to show.
For part (b) here is what I have got so far: $[T]_B=A$. Let $I=\{e_1,...,e_n\},$ the standard basis for $F^n$. Then $\{Te_1,...,Te_n\}$ is a matrix with regards to the standard basis. Now $det(A)=det(\{Te_1,...,Te_n\})=det(T)det(e_1,...,e_n)=c\cdot1$. Hence $det(A)=c$.
I'm not feeling confident in my proofs and would like a review.
For (a): you still have to show, or maybe just remark, that multilineary and alternating work because $\;T\;$ is a linear map...
For (b): Suppose $\;D\;$ is the matrix representing $\;T\;$ wrt the standard basis, then there exists invertible $\;P\;$ such that $\;A=P^{-1}DP\;$ , so using the product theorem for determinants we get
$$\det A=\det\left(P^{-1}DP\right)=\det(P)^{-1}\det D\det(P)=\det(D)$$
What you just proved in (b) is that the determinant (like the trace) of a square matrix is an invariant under similarity.