Dual basis
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Exercise:
The elements v_^*...v_n^* in V^* form a basis for V^*. This basis is called the dual basis to mathcalB. We denote this basis by mathcalB^*v_^*...v_n^*. Moreover forall lin V^* l_i^n lv_i v_i^*.
Solution:
Proof. Let lin V^*. We claim that abcliste abc l_i^n lv_i v_i^* or in other words we claim: abc forall vin V lv_i^n lv_i v_i^*v. abcliste Indeed both sides of a are linear maps Vlongrightarrow K so it's enough to check that b holds for vv_ vv_...vv_n because v_...v_n form a basis for V. And indeed for vv_k we have: _i^n lv_i v_i^*v_k_i^n lv_i delta_iklv_k The claim just proven shows that v_^*...v_n^* span V^*. Since textdimV^*textdimVn it follows that v_^*...v_n^* form a basis for V^*.
The elements v_^*...v_n^* in V^* form a basis for V^*. This basis is called the dual basis to mathcalB. We denote this basis by mathcalB^*v_^*...v_n^*. Moreover forall lin V^* l_i^n lv_i v_i^*.
Solution:
Proof. Let lin V^*. We claim that abcliste abc l_i^n lv_i v_i^* or in other words we claim: abc forall vin V lv_i^n lv_i v_i^*v. abcliste Indeed both sides of a are linear maps Vlongrightarrow K so it's enough to check that b holds for vv_ vv_...vv_n because v_...v_n form a basis for V. And indeed for vv_k we have: _i^n lv_i v_i^*v_k_i^n lv_i delta_iklv_k The claim just proven shows that v_^*...v_n^* span V^*. Since textdimV^*textdimVn it follows that v_^*...v_n^* form a basis for V^*.