Linear Independence
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Exercise:
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.
Prove that the functions phi_a :mathbbR_ rightarrow mathbbRquad x mapsto fraca+x for all a in mathbbR_geq are linear indepent. H: Use that a non-zero polynomial only has finitely many zeros.
Solution:
Take a finite number of u_i in mathbbR and pairwise distinct a_i in mathbbR_geq where limits_i^m u_iphi_a_i . For all x in mathbbR_ it must hold _i^m u_i fraca_i+x If we multiply this term with prodlimits_i ^ma_i+x for the first two terms we get: leftu_ fraca_+x+u_ fraca_+xrighta_+xa_+x u_a_+x+u_a_+x In general this results in: _i^m u_i prodlimits_j neq ia_j+x The left side of the is a polynomial so we can now use the h from above and deduce that it is valid for all x in mathbbR. The is valid if x-a_k quad textfor quad leq k leq m. If i neq k it follows prodlimits_j neq ia_j-a_k so we can reformulate the to u_k prodlimits_j neq ka_j-a_k Since we first set that a_i should be pairwise distinct it follows that prodlimits_j neq ka_j-a_kneq so u_k . Furthermore since k is randomly chosen it is clear that u_u_...u_m So phi_ phi_... are linear indepent.