Trace of a endomorphism of a finite dimentional vector space

Let X be a n-dimensional k-vector space over a complete field k and $ \mathcal{A}\in \mathcal{L}(X,X) $ The spectrum of $ \mathcal{A} $ is formed by n eigenvalues $ \lambda_1,\lambda_2,....,\lambda_n \in k $ .
We call $ tr(\mathcal{A})=\sum_{k=1}^n \lambda_k $.
From "Matrix representation of linear operator" there is a matrix $ A=\mathcal{M}(\mathcal{A};\mathcal{B},\mathcal{B}) $ associated with the linear operator and $ (\lambda_i)_{1\leq i\leq n} $ are solutions of equation $ P(\lambda)=det(A-\lambda I_n)=0 $ and after a short calculation we have

$$tr(\mathcal{A})=\sum _{i=1}^n a_{i,i}$$

Now let $ \mathcal{B}\in \mathcal{L}(X,X) $ be another operator and let consider equation


Then there is also a relation between associated matrix


Let $ C=AB $
We have

$$tr(AB)=\sum_ {i=1}^n c_{i,i}=\sum_{i=1}^n\sum_{j=1}^na_{i,j}b_{j,i}=\sum_{j=1}^n\sum_{i=1}^nb_{j,i}a_{i,j}=tr(BA)$$

But $ tr(AB-BA)=0\neq tr(I_n)=n $ so there are no solution for the previous equation on finite dimensional space.

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