$\DeclareMathOperator{\rk}{rk}$ $\DeclareMathOperator{\tr}{tr}$
I would be happy if someone could check this proof and tell me any mistakes I shall correct.
Let $V$ be a finite-dimensional vector-space with $\dim(V)>0$. Given an Endomorphism $g: V \to V$ we first want to show, that the determinant of the $n \times n$ transformation matrix $A$ regarding two different bases $v$ and $\tilde{v}$ of $V$ depends on the choice of the bases of $V$.
Let us assume that $\det(A)=\det(g)$ regardless of the choice of the bases of $V$ and let us fix $\det(A)=:r \neq \{0,1\}$.
Since $\det(A)\neq 0 \Rightarrow \rk(A)=n$ we know that we can find bases $w$ and $\tilde{w}$ of $V$ as well as invertible matrices $P$ and $Q$ such that the $(n \times n)$ transformation matrix $A'$ regarding $w$ and $\tilde{w}$ is of the form $P^{-1}AQ=I_n=:A'$
Clearly, $\det(A')=\det(I_n)=1 \neq r = \det(A)$ which is a contradiction to the assumption that $\det(A)=\det(g)$ regardless of the choice of the bases of V.
So $\det(A)$ depends on the choice of the bases of $V$, what was to be shown.
Similarly let us assume, that $\tr(A)=\tr(g)$ regardless of the choice of the bases of $V$ and let us fix $\tr(A)=:t > n$. Again we can find bases $u$ and $\tilde{u}$ as well as invertible matrices R and S such that the $(n \times n)$ transformation matrix $\tilde{A}$ is of the form
$$R^{-1}AS= \begin{pmatrix} 1 & 0 & 0 & \cdots & \cdots & \cdots & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & \vdots \\ \vdots & 0 & \ddots & 0 & 0 & 0 & \vdots \\ \vdots & 0 & \cdots & 1 & 0 & 0 & \vdots \\ \vdots & 0 & \cdots & \cdots & 0 & 0 & 0 \\ \vdots & 0 & \cdots & \cdots & \cdots & \ddots & 0\\ 0 & 0 & \cdots & \cdots & \cdots & 0 & 0\\ \end{pmatrix}$$
We observe that $r$ = $\rk(\tilde{A})=\tr(\tilde{A})$. So clearly $\tr(\tilde{A}) =r \leq n \neq \tr(A) > n$ which is a contradiction to the assumption that $\tr(A)=\tr(g)$ regardless of the choice of the bases of $V$.
So $\tr(A)$ depends on the choice of the bases of $V$.