Eigenvalues and linear independence
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Exercise:
Let V be a vector space over K and T:Vrightarrow V a linear map. Suppose that lambda_...lambda_nin K are eigenvalues of T which are pairwise distinct i.e. lambda_ineq lambda_jquad forall ineq j. Let v_...v_n be eigenvectors for lambda_...lambda_n respectively. Then v_...v_n are linearly indepent.
Solution:
Proof with induction on n. n: v_ is an eigenvector for lambda_. By definition v_neq Longrightarrow v_ is linearly indepent. ngeq : Suppose the proposition holds for any list of n eigenvalues and eigenvectors. We'll prove now that the proposition holds also for a list of n+ eigenvalues and eigenvectors. Let lambda_...lambda_n lambda_n+ be pairwiese distinct eigenvalues of T and v_...v_nv_n+ a given list of eigenvectors of lambda_...lambda_n lambda_n+ respectively. We need to prove that v_...v_nv_n+ are linearly indepent. Ase that a_v_+...+a_nv_n+a_n+v_n+quad*. We need to show that a_...a_na_n+. Apply T to *: a_Tv_+...+a_nTv_n+a_n+Tv_n+ But Tv_ilambda_iv_i Longrightarrow forall i a_ lambda_ v_+...+a_n lambda_n v_n+a_n+ lambda_n+ v_n+quad ** Consider now **-lambda_n+*: a_ lambda_-lambda_n+ v_+...+a_n lambda_n-lambda_n+ v_n+a_n+ lambda_n+-lambda_n+ v_n+ By the induction hypothesis v_...v_n are linearly indepent Longrightarrow a_lambda_-lambda_n+...a_nlambda_n-lambda_n+ Going back to * we get a_n+v_n+. But also v_n+neq because v_n+ is an eigenvector Longrightarrow a_n+.
Let V be a vector space over K and T:Vrightarrow V a linear map. Suppose that lambda_...lambda_nin K are eigenvalues of T which are pairwise distinct i.e. lambda_ineq lambda_jquad forall ineq j. Let v_...v_n be eigenvectors for lambda_...lambda_n respectively. Then v_...v_n are linearly indepent.
Solution:
Proof with induction on n. n: v_ is an eigenvector for lambda_. By definition v_neq Longrightarrow v_ is linearly indepent. ngeq : Suppose the proposition holds for any list of n eigenvalues and eigenvectors. We'll prove now that the proposition holds also for a list of n+ eigenvalues and eigenvectors. Let lambda_...lambda_n lambda_n+ be pairwiese distinct eigenvalues of T and v_...v_nv_n+ a given list of eigenvectors of lambda_...lambda_n lambda_n+ respectively. We need to prove that v_...v_nv_n+ are linearly indepent. Ase that a_v_+...+a_nv_n+a_n+v_n+quad*. We need to show that a_...a_na_n+. Apply T to *: a_Tv_+...+a_nTv_n+a_n+Tv_n+ But Tv_ilambda_iv_i Longrightarrow forall i a_ lambda_ v_+...+a_n lambda_n v_n+a_n+ lambda_n+ v_n+quad ** Consider now **-lambda_n+*: a_ lambda_-lambda_n+ v_+...+a_n lambda_n-lambda_n+ v_n+a_n+ lambda_n+-lambda_n+ v_n+ By the induction hypothesis v_...v_n are linearly indepent Longrightarrow a_lambda_-lambda_n+...a_nlambda_n-lambda_n+ Going back to * we get a_n+v_n+. But also v_n+neq because v_n+ is an eigenvector Longrightarrow a_n+.