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Eigenvalues and direct sum

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Exercise

https://texercises.com/exercise/eigenvalues-and-direct-sum/

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Exercise:
Let V be a vector space over K and Tin textEndV. Let lambda_...lambda_rin K be pairwise distinct eigenvalues of T. Then textEig_Tlambda_+...+textEig_Tlambda_r is a direct .

Solution:
Proof. Ase u_+...+u_ru_'+...+u_r' where u_i u_i'in textEig_Tlambda_iquad forall leq ileq r. We need to show that u_iu_i'quad forall i iff u_i-u_i'quad forall i. We have: u_-u_'+...+u_r-u_r'quad *. Ase by contradiction that exists i s.t. u_i-u_i'neq . Remove from the list u_-u_'...u_i_l-u_i_l' of non-zero vectors and by * we still have u_i_-u_i_'+...+u_i_l-u_i_l'quad **. Now forall k u_i_k-u_i_k' is either an eigenvector of lambda_i or but by our asptions u_i_k-u_i_k'neq . Recall that eigenvectors corresponding to pairwise distinct eigenvalues are linearly indepent vectors and by ** their is contradiction.

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Tags	eigenvalue, eigenvector, eth, fs23, lineare algebra, proof
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Difficulty	(3, default)
Points	0 (default)
Language	(English)
Type	Proof
Creator	rk
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