# Proof determinant and trace of transformation matrix of an endomorphism regarding different bases depend on the choice of the bases

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. $$\Box$$

• 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 :) – darij grinberg Aug 14 at 16:41
• For the trace, your argument requires a field of characteristic $>n$. I'm wondering how well that claim works for finite fields. – darij grinberg Aug 14 at 16:43
• 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 $>n$.) – darij grinberg Aug 14 at 16:43
• It's easy to find a matrix with arbitrary trace, and it has nothing to do with $A$ being $n \times n$. – darij grinberg Aug 14 at 19:54
• 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. – Dirk Aug 15 at 9:47