Prove that the determinant of a transformation matrix does not depend on the basis

Question

Let $T: \Bbb U \to \Bbb V$ be a linear transformation representable by $A \mathbf x$ in some basis $B$, where $A$ is a matrix and $\mathbf x$ is a member of $\Bbb U$. $\ $Show det(A) does not depend on the basis chosen.

Answer 1

Let $A'$ represent the transformation in some other basis $B'$, and let $V$ be the change of basis matrix from $B'$ to $B$. We then have that $$A = VA'V^{-1}$$ Take the determinant on each side of that equation.

Answer 2

Assuming that $B$ is invertible, the global transform is $BAB^{-1}$, combine that with $\det(XY) = \det(X)\det(Y)$, show that for an arbitrary invertible $C$ that $\det(CAC^{-1}) = \det(BAB^{-1})$.