Skip navigation.
Home
mathdynamics
Home » Algebra » Vector space

Trace of a endomorphism of a finite dimentional vector space

Submitted by Isoscel on Thu, 11/22/2007 - 19:35.

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

$$\mathcal{A}\mathcal{B}-\mathcal{B}\mathcal{A}=id_X$$

Then there is also a relation between associated matrix

$$AB-BA=I_n$$

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.

Average: 3.3 (3 votes)
‹ Matrix associated to rotation around Oz real axisup
»