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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 $(n$ x $n)$ 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 $(n$ x $n)$ 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 $(n$ x $n)$ 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 $V$ what was to be shown. $\Box$

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    $\begingroup$ You need $\det A$ to be distinct from both $0$ and $1$. The claim is false over the $2$-element field $\mathbb{F}_2$ ! It is also false if $\dim V = 0$, but let's not be pedantic :) $\endgroup$ – darij grinberg Aug 14 at 16:41
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    $\begingroup$ For the trace, your argument requires a field of characteristic $>n$. I'm wondering how well that claim works for finite fields. $\endgroup$ – darij grinberg Aug 14 at 16:43
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    $\begingroup$ Other than this, it's a good proof. (But just to make sure, you should explain how to come up with a matrix whose trace is $>n$.) $\endgroup$ – darij grinberg Aug 14 at 16:43
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    $\begingroup$ It's easy to find a matrix with arbitrary trace, and it has nothing to do with $A$ being $n \times n$. $\endgroup$ – darij grinberg Aug 14 at 19:54
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    $\begingroup$ Note that usually, one calls trace and determinant independent of the choice of basis. What that means is that as long as we have $v = \tilde{v}$, both $det$ and $tr$ will not change no matter which basis $v$ you pick. $\endgroup$ – Dirk Aug 15 at 9:47

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